text:
# 105 14.6 Convection

### Summary

• Discuss the method of heat transfer by convection.

Convection is driven by large-scale flow of matter. In the case of Earth, the atmospheric circulation is caused by the flow of hot air from the tropics to the poles, and the flow of cold air from the poles toward the tropics. (Note that Earth’s rotation causes the observed easterly flow of air in the northern hemisphere). Car engines are kept cool by the flow of water in the cooling system, with the water pump maintaining a flow of cool water to the pistons. The circulatory system is used the body: when the body overheats, the blood vessels in the skin expand (dilate), which increases the blood flow to the skin where it can be cooled by sweating. These vessels become smaller when it is cold outside and larger when it is hot (so more fluid flows, and more energy is transferred).

The body also loses a significant fraction of its heat through the breathing process.

While convection is usually more complicated than conduction, we can describe convection and do some straightforward, realistic calculations of its effects. Natural convection is driven by buoyant forces: hot air rises because density decreases as temperature increases. The house in Figure 1 is kept warm in this manner, as is the pot of water on the stove in Figure 2. Ocean currents and large-scale atmospheric circulation transfer energy from one part of the globe to another. Both are examples of natural convection.

### TAKE-HOME EXPERIMENT: CONVECTION ROLLS IN A HEATED PAN

Take two small pots of water and use an eye dropper to place a drop of food coloring near the bottom of each. Leave one on a bench top and heat the other over a stovetop. Watch how the color spreads and how long it takes the color to reach the top. Watch how convective loops form.

### Example 1: Calculating Heat Transfer by Convection: Convection of Air Through the Walls of a House

Most houses are not airtight: air goes in and out around doors and windows, through cracks and crevices, following wiring to switches and outlets, and so on. The air in a typical house is completely replaced in less than an hour. Suppose that a moderately-sized house has inside dimensions$\boldsymbol{12.0\textbf{ m }\times\:18.0\textbf{ m }\times\:3.00\textbf{ m }}$high, and that all air is replaced in 30.0 min. Calculate the heat transfer per unit time in watts needed to warm the incoming cold air by$\boldsymbol{10.0^{\circ}\textbf{C}},$thus replacing the heat transferred by convection alone.

Strategy

Heat is used to raise the temperature of air so that$\boldsymbol{Q=mc\Delta{T}}.$The rate of heat transfer is then$\boldsymbol{Q/t},$where tt is the time for air turnover. We are given that$\boldsymbol{\Delta{T}}$is$\boldsymbol{10.0^{\circ}\textbf{C}},$but we must still find values for the mass of air and its specific heat before we can calculate$\boldsymbol{Q}.$The specific heat of air is a weighted average of the specific heats of nitrogen and oxygen, which gives$\boldsymbol{c=cp\cong1000\textbf{ J/kg}\cdotp^{\circ}\textbf{C}}$from Table 4 (note that the specific heat at constant pressure must be used for this process).

Solution

1. Determine the mass of air from its density and the given volume of the house. The density is given from the density$\boldsymbol{\rho}$and the volume
$\boldsymbol{m=\rho{V}=(1.29\textbf{ kg/m}^3)(12.0\textbf{ m}\times\:18.0\textbf{ m}\times\:3.00\textbf{ m})=836\textbf{ kg}}.$
2. Calculate the heat transferred from the change in air temperature:$\boldsymbol{Q=mc\Delta{T}}$so that
$\boldsymbol{Q=(836\textbf{ kg})(1000\textbf{ J/kg}\cdotp^{\circ}\textbf{C})(10.0^{\circ}\textbf{C})= 8.36\times10^6\textbf{ J}}.$
3. Calculate the heat transfer from the heat$\boldsymbol{Q}$and the turnover time$\boldsymbol{t}.$Since air is turned over in$\boldsymbol{t=0.500\textbf{ h}=1800\textbf{ s}},$the heat transferred per unit time is
$\boldsymbol{\frac{Q}{t}}$$\boldsymbol{=}$$\boldsymbol{\frac{8.36\times10^6\textbf{ J}}{1800\textbf{ s}}}$$\boldsymbol{=\:4.64\textbf{ kW}}.$

Discussion

This rate of heat transfer is equal to the power consumed by about forty-six 100-W light bulbs. Newly constructed homes are designed for a turnover time of 2 hours or more, rather than 30 minutes for the house of this example. Weather stripping, caulking, and improved window seals are commonly employed. More extreme measures are sometimes taken in very cold (or hot) climates to achieve a tight standard of more than 6 hours for one air turnover. Still longer turnover times are unhealthy, because a minimum amount of fresh air is necessary to supply oxygen for breathing and to dilute household pollutants. The term used for the process by which outside air leaks into the house from cracks around windows, doors, and the foundation is called “air infiltration.”

A cold wind is much more chilling than still cold air, because convection combines with conduction in the body to increase the rate at which energy is transferred away from the body. The table below gives approximate wind-chill factors, which are the temperatures of still air that produce the same rate of cooling as air of a given temperature and speed. Wind-chill factors are a dramatic reminder of convection’s ability to transfer heat faster than conduction. For example, a 15.0 m/s wind at$\boldsymbol{0^{\circ}\textbf{C}}$has the chilling equivalent of still air at about$\boldsymbol{-18^{\circ}\textbf{C}}.$

Moving air temperature Wind speed (m/s)
ºC 2 5 10 15 20
5 3 −1 −8 −10 −12
2 0 −7 −12 −16 −18
0 −2 −9 −15 −18 −20
−5 −7 −15 −22 −26 −29
−10 −12 −21 −29 −34 −36
−20 −23 −34 −44 −50 −52
−40 −44 −59 −73 −82 −84
Table 4. Wind-Chill Factors

Although air can transfer heat rapidly by convection, it is a poor conductor and thus a good insulator. The amount of available space for airflow determines whether air acts as an insulator or conductor. The space between the inside and outside walls of a house, for example, is about 9 cm (3.5 in) —large enough for convection to work effectively. The addition of wall insulation prevents airflow, so heat loss (or gain) is decreased. Similarly, the gap between the two panes of a double-paned window is about 1 cm, which prevents convection and takes advantage of air’s low conductivity to prevent greater loss. Fur, fiber, and fiberglass also take advantage of the low conductivity of air by trapping it in spaces too small to support convection, as shown in the figure. Fur and feathers are lightweight and thus ideal for the protection of animals.

Some interesting phenomena happen when convection is accompanied by a phase change. It allows us to cool off by sweating, even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues.

### Example 2: Calculate the Flow of Mass during Convection: Sweat-Heat Transfer away from the Body

The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (This evaporation might occur when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.)

Strategy

Energy is needed for a phase change ($\boldsymbol{Q=mL_{\textbf{v}}}$). Thus, the energy loss per unit time is

$\boldsymbol{\frac{Q}{t}}$$\boldsymbol{=}$$\boldsymbol{\frac{mL_{\textbf{v}}}{t}}$$\boldsymbol{=120\textbf{ W}=120\textbf{ J/s}}.$

We divide both sides of the equation by$\boldsymbol{L_{\textbf{v}}}$to find that the mass evaporated per unit time is

$\boldsymbol{\frac{m}{t}}$$\boldsymbol{=}$$\boldsymbol{\frac{120\textbf{ J/s}}{L_{\textbf{v}}}}.$

Solution

(1) Insert the value of the latent heat from Table 2,$\boldsymbol{L_{\textbf{v}}=2430\textbf{ kJ/kg}=2430\textbf{ J/g}}.$This yields

$\boldsymbol{\frac{m}{t}}$$\boldsymbol{=}$$\boldsymbol{\frac{120\textbf{ J/s}}{2430\textbf{ J/g}}}$$\boldsymbol{=\:0.0494\textbf{ g/s}=2.96\textbf{ g/min}}.$

Discussion

Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages.

Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as clouds form, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere. Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise, where it is colder. More condensation occurs in these colder regions, which in turn drives the cloud even higher. Such a mechanism is called positive feedback, since the process reinforces and accelerates itself. These systems sometimes produce violent storms, with lightning and hail, and constitute the mechanism driving hurricanes.

The movement of icebergs is another example of convection accompanied by a phase change. Suppose an iceberg drifts from Greenland into warmer Atlantic waters. Heat is removed from the warm ocean water when the ice melts and heat is released to the land mass when the iceberg forms on Greenland.

1: Explain why using a fan in the summer feels refreshing!

# Summary

• Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced and generally transfers thermal energy faster than conduction. Table 4 gives wind-chill factors, indicating that moving air has the same chilling effect of much colder stationary air. Convection that occurs along with a phase change can transfer energy from cold regions to warm ones.

### Conceptual Questions

1: One way to make a fireplace more energy efficient is to have an external air supply for the combustion of its fuel. Another is to have room air circulate around the outside of the fire box and back into the room. Detail the methods of heat transfer involved in each.

2: On cold, clear nights horses will sleep under the cover of large trees. How does this help them keep warm?

### Problems & Exercises

1: At what wind speed does$\boldsymbol{-10^{\circ}\textbf{C}}$air cause the same chill factor as still air at$\boldsymbol{-29^{\circ}\textbf{C}}?$

2: At what temperature does still air cause the same chill factor as$\boldsymbol{-5^{\circ}\textbf{C}}$air moving at 15 m/s?

3: The “steam” above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee initially at$\boldsymbol{90.0^{\circ}\textbf{C}}$if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected.

4: (a) How many kilograms of water must evaporate from a 60.0-kg woman to lower her body temperature by$\boldsymbol{0.750^{\circ}\textbf{C}}?$

(b) Is this a reasonable amount of water to evaporate in the form of perspiration, assuming the relative humidity of the surrounding air is low?

5: On a hot dry day, evaporation from a lake has just enough heat transfer to balance the$\boldsymbol{1.00\textbf{ kW/m}^2}$of incoming heat from the Sun. What mass of water evaporates in 1.00 h from each square meter? Explicitly show how you follow the steps in the Chapter 14.3 Problem-Solving Strategies for the Effects of Heat Transfer.

6: One winter day, the climate control system of a large university classroom building malfunctions. As a result,$\boldsymbol{500\textbf{ m}^3}$of excess cold air is brought in each minute. At what rate in kilowatts must heat transfer occur to warm this air by$\boldsymbol{10.0^{\circ}\textbf{C}}$(that is, to bring the air to room temperature)?

7: The Kilauea volcano in Hawaii is the world’s most active, disgorging about$\boldsymbol{5\times10^5\textbf{ m}^3}$of$\boldsymbol{1200^{\circ}\textbf{C}}$lava per day. What is the rate of heat transfer out of Earth by convection if this lava has a density of$\boldsymbol{2700\textbf{ kg/m}^3}$and eventually cools to$\boldsymbol{30^{\circ}\textbf{C}}?$Assume that the specific heat of lava is the same as that of granite.

8: During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by$\boldsymbol{2.00^{\circ}\textbf{C}}.$What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is$\boldsymbol{1050\textbf{ kg/m}^3}?$

9: A person inhales and exhales 2.00 L of$\boldsymbol{37.0^{\circ}\textbf{C}}$air, evaporating$\boldsymbol{4.00\times10^{-2}\textbf{ g}}$of water from the lungs and breathing passages with each breath.

(a) How much heat transfer occurs due to evaporation in each breath?

(b) What is the rate of heat transfer in watts if the person is breathing at a moderate rate of 18.0 breaths per minute?

(c) If the inhaled air had a temperature of$\boldsymbol{20.0^{\circ}\textbf{C}},$what is the rate of heat transfer for warming the air?

(d) Discuss the total rate of heat transfer as it relates to typical metabolic rates. Will this breathing be a major form of heat transfer for this person?

10: A glass coffee pot has a circular bottom with a 9.00-cm diameter in contact with a heating element that keeps the coffee warm with a continuous heat transfer rate of 50.0 W

(a) What is the temperature of the bottom of the pot, if it is 3.00 mm thick and the inside temperature is$\boldsymbol{60.0^{\circ}\textbf{C}}?$

(b) If the temperature of the coffee remains constant and all of the heat transfer is removed by evaporation, how many grams per minute evaporate? Take the heat of vaporization to be 2340 kJ/kg.

### Solutions

1: Using a fan increases the flow of air: warm air near your body is replaced by cooler air from elsewhere. Convection increases the rate of heat transfer so that moving air “feels” cooler than still air.

Problems & Exercises

1:

10 m/s

3:

$\boldsymbol{85.7^{\circ}\textbf{C}}$

5:

1.48 kg

7:

$\boldsymbol{2\times10^4\textbf{ MW}}$

9:

(a) 97.2 J

(b) 29.2 W

(c) 9.49 W

(d) The total rate of heat loss would be$\boldsymbol{29.2\textbf{ W }+\:9.49\textbf{ W}=\:38.7\textbf{ W}}.$While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person.


text:
## 08 November 2007

### The birthday paradox

Finding the Odds of Synchronized Couples, from The Numbers Guy.

The problem is as follows: how many opposite-sex couples need to be in a room for the probability that there are couples Mr. and Mrs. A, and Mr. and Mrs. B, such that Mr. A and Mr. B have the same birthday and Mrs. A and Mrs. B have the same birthday, to be more than one half?

This is a variation on the classic "birthday problem", which asks this for individuals. The answer in the case of individuals is 23; for couples it's 431. This number 431 is given in the article without justification, so I wanted to explain it. For same-sex couples, the answer is about 304. (There are less possible pairs of birthdays for same-sex couples because you can no longer distinguish between the two members of the couple.)

Let's say there are n possible types of objects (say, birthdays that a person can have) and objects of these types (people with different birthdays) come along one by one. Their types (birthdays) are chosen uniformly at random. The probability that the first person is not the same type as any of the people before em is, of course, 1. The probability that the second person is not the same type as any of the people before em is 1-1/n. The probability that the third person is not the same type as either of the first two people -- given that the first two people are of different types -- is 1-2/n. And in general the probability that the kth person is not the same type as any of the people preceding em is

But recalling that (1-x) ≈ e-x, we can rewrite this approximately as

which is, again approximately, exp(-k2/2n).

To have this probability be less than one half, then, we need

For birthdays of ordinary Earth people, n = 365, and we get k > 22.49, that is, we need at least 23 people (since people come in integers) to have probability 1/2 that two of them have the same birthday. For birthdays of opposite-sex couples, n = 3652, and this gives k > 430; for birthdays of same-sex couples, n = 365(365+1)/2, and k > 305.

Incidentally, the birthday problem is often called the "birthday paradox" because a lot of people are surprised by the result -- if you have 23 people in a room then there's a fifty-fifty chance that two of them have the same birthday. I don't find it surprising any more, so I won't call it a paradox.

(Somewhat more surprisingly, I don't seem to have blogged about the birthday paradox before!)

It would be interesting to note whether the birthdays of couples are actually evenly distributed; having them not be evenly distributed might be evidence for something. The obvious thing is astrology -- but it wouldn't surprise me if there are earthly reasons why people born around the same time of year are more likely to be attracted to each other. (I have no idea what the mechanism for this would be, though, and I suspect this isn't the case.)

#### 1 comment:

Max said...

You introductory paragraph leaves out the critical 'more than half' part of the sentence. :)

format: plain


text:
Share

# RD Sharma solutions for Class 11 Mathematics Textbook chapter 31 - Mathematical reasoning [Latest edition]

Textbook page

## Chapter 31: Mathematical reasoning

Ex. 31.1Ex. 31.2Ex. 31.3Ex. 31.4Ex. 31.5Ex. 31.6

#### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 31 Mathematical reasoning Exercise 31.1 [Page 3]

Ex. 31.1 | Q 1.01 | Page 3

Find out the following  sentence is a statement and is not. Justify your answer.

Listen to me, Ravi !

Ex. 31.1 | Q 1.02 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Every set is a finite set.

Ex. 31.1 | Q 1.03 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Two non-empty sets have always a non-empty intersection.

Ex. 31.1 | Q 1.04 | Page 3

Find out the sentence are statement and are not. Justify your answer.

The cat pussy is black.

Ex. 31.1 | Q 1.05 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Are all circles round?

Ex. 31.1 | Q 1.06 | Page 3

Find out the sentence are statement and are not. Justify your answer.

All triangles have three sides.

Ex. 31.1 | Q 1.07 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Every rhombus is a square.

Ex. 31.1 | Q 1.08 | Page 3

Find out the sentence are statement and are not. Justify your answer.

x2 + 5 | x | + 6 = 0 has no real roots.

Ex. 31.1 | Q 1.09 | Page 3

Find out the sentence are statement and are not. Justify your answer.

This sentence is a statement.

Ex. 31.1 | Q 1.1 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Is the earth round?

Ex. 31.1 | Q 1.11 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Go !

Ex. 31.1 | Q 1.12 | Page 3

Find out the sentence are statement and are not. Justify your answer.

The real number x is less than 2.

Ex. 31.1 | Q 1.13 | Page 3

Find out the sentence are statement and are not. Justify your answer.

There are 35 days in a month.

Ex. 31.1 | Q 1.14 | Page 3

Find out the sentence are statement and are not. Justify your answer.

Mathematics is difficult.

Ex. 31.1 | Q 1.15 | Page 3

Find out the sentence are statement and are not. Justify your answer.

All real numbers are complex numbers.

Ex. 31.1 | Q 1.16 | Page 3

Find out the sentence are statement and are not. Justify your answer.

The product of (−1) and 8 is 8.

Ex. 31.1 | Q 2 | Page 3

Give three examples of sentences which are not statements. Give reasons for the answers.

#### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 31 Mathematical reasoning Exercise 31.2 [Pages 6 - 7]

Ex. 31.2 | Q 1.1 | Page 6

Write the negation of the statement:
Banglore is the capital of Karnataka.

Ex. 31.2 | Q 1.2 | Page 6

Write the negation of the statement:

It rained on July 4, 2005.

Ex. 31.2 | Q 1.3 | Page 6

Write the negation of the statement:

Ravish is honest.

Ex. 31.2 | Q 1.4 | Page 6

Write the negation of the statement:

The earth is round.

Ex. 31.2 | Q 1.5 | Page 6

Write the negation of the statement:

The sun is cold.

Ex. 31.2 | Q 2.1 | Page 7

All birds sing.

Ex. 31.2 | Q 2.2 | Page 7

Some even integers are prime.

Ex. 31.2 | Q 2.3 | Page 7

There is a complex number which is not a real number.

Ex. 31.2 | Q 2.4 | Page 7

I will not go to school.

Ex. 31.2 | Q 2.5 | Page 7

Both the diagonals of a rectangle have the same length.

Ex. 31.2 | Q 2.6 | Page 7

All policemen are thieves.

Ex. 31.2 | Q 3.1 | Page 7

Are the pair of statement are negation of each other:
The number is not a rational number.
The number x is not an irrational number.

Ex. 31.2 | Q 3.2 | Page 7

Are the pair of statement are negation of each other:

The number x is not a rational number.
The number is an irrational number.

Ex. 31.2 | Q 4.1 | Page 7

Write the negation of the statement:

p : For every positive real number x, the number (x − 1) is also positive.

Ex. 31.2 | Q 4.2 | Page 7

Write the negation of the statement:

q : For every real number x, either x > 1 or x < 1.

Ex. 31.2 | Q 4.3 | Page 7

Write the negation of the statement:

r : There exists a number x such that 0 < x < 1.

Ex. 31.2 | Q 5 | Page 7

Check whether the following pair of statements are negation of each other. Give reasons for your answer.
a + b = b + a is true for every real number a and b.
There exist real numbers a and b for which a + b = b + a.

#### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 31 Mathematical reasoning Exercise 31.3 [Page 14]

Ex. 31.3 | Q 1.1 | Page 14

Find the component statement of the compound statement:
The sky is blue and the grass is green.

Ex. 31.3 | Q 1.2 | Page 14

Find the component statement of the compound statement:

The earth is round or the sun is cold.

Ex. 31.3 | Q 1.3 | Page 14

Find the component statement of the compound statement:

All rational numbers are real and all real numbers are complex.

Ex. 31.3 | Q 1.4 | Page 14

Find the component statement of the compound statement:

25 is a multiple of 5 and 8.

Ex. 31.3 | Q 2.1 | Page 14

For statement, determine whether an inclusive "OR" or exclusive "OR" is used. Give reasons for your answer.

Students can take Hindi or Sanskrit as their third language.

Ex. 31.3 | Q 2.2 | Page 14

For statement, determine whether an inclusive "OR" or exclusive "OR" is used. Give reasons for your answer.

To entry a country, you need a passport or a voter registration card.

Ex. 31.3 | Q 2.3 | Page 14

For statement, determine whether an inclusive "OR" or exclusive "OR" is used. Give reasons for your answer.

A lady gives birth to a baby boy or a baby girl.

Ex. 31.3 | Q 2.4 | Page 14

For statement, determine whether an inclusive "OR" or exclusive "OR" is used. Give reasons for your answer.

To apply for a driving licence, you should have a ration card or a passport.

Ex. 31.3 | Q 3.1 | Page 14

Write the component statement of the compound statement and check whether the compound statement is true or false:

To enter into a public library children need an identity card from the school or a letter from the school authorities.

Ex. 31.3 | Q 3.2 | Page 14

Write the component statement of the compound statement and check whether the compound statement is true or false:

All rational numbers are real and all real numbers are not complex.

Ex. 31.3 | Q 3.3 | Page 14

Write the component statement of the compound statement and check whether the compound statement is true or false:

Square of an integer is positive or negative.i

Ex. 31.3 | Q 3.4 | Page 14

Write the component statement of the compound statement and check whether the compound statement is true or false:

x = 2 and x = 3 are the roots or the equation 3x2 − x − 10 = 0.

Ex. 31.3 | Q 3.5 | Page 14

Write the component statement of the compound statement and check whether the compound statement is true or false:

The sand heats up quickly in the sun and does not cool down fast at night.

Ex. 31.3 | Q 4.1 | Page 14

Determine whether the compound statement are true or false:

Delhi is in India and 2 + 2 = 4.

Ex. 31.3 | Q 4.2 | Page 14

Determine whether the compound statement are true or false:

Delhi is in England and 2 + 2 = 4.

Ex. 31.3 | Q 4.3 | Page 14

Determine whether the compound statement are true or false:

Delhi is in India and 2 + 2 = 5.

Ex. 31.3 | Q 4.4 | Page 14

Determine whether the compound statement are true or false:

Delhi is in England and 2 + 2 =5.

#### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 31 Mathematical reasoning Exercise 31.4 [Page 16]

Ex. 31.4 | Q 1.1 | Page 16

Write the negation of  statement:

For every x ϵ Nx + 3 < 10

Ex. 31.4 | Q 1.2 | Page 16

Write the negation of  statement:

There exists x ϵ Nx + 3 = 10

Ex. 31.4 | Q 2.1 | Page 16

Negate  statement:
All the students completed their homework.

Ex. 31.4 | Q 2.2 | Page 16

Negate of the  statement :

There exists a number which is equal to its square.

#### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 31 Mathematical reasoning Exercise 31.5 [Page 21]

Ex. 31.5 | Q 1.1 | Page 21

Write of the statement in the form "if p, then q".

You can access the website only if you pay a subscription fee.

Ex. 31.5 | Q 1.2 | Page 21

Write of the statement in the form "if p, then q".

There is traffic jam whenever it rains.

Ex. 31.5 | Q 1.3 | Page 21

Write of the statement in the form "if p, then q".

It is necessary to have a passport to log on to the server.

Ex. 31.5 | Q 1.4 | Page 21

Write of the statement in the form "if p, then q".

It is necessary to be rich in order to be happy.

Ex. 31.5 | Q 1.5 | Page 21

Write of the statement in the form "if p, then q".

The game is cancelled only if it is raining.

Ex. 31.5 | Q 1.6 | Page 21

Write of the statement in the form "if p, then q".

It rains only if it is cold.

Ex. 31.5 | Q 1.7 | Page 21

Write of the statement in the form "if p, then q".

Whenever it rains it is cold.

Ex. 31.5 | Q 1.8 | Page 21

Write of the statement in the form "if p, then q".

It never rains when it is cold.

Ex. 31.5 | Q 2.1 | Page 21

State the converse and contrapositive of  statement:

If it is hot outside, then you feel thirsty.

Ex. 31.5 | Q 2.2 | Page 21

State the converse and contrapositive of  statement:

I go to a beach whenever it is a sunny day.

Ex. 31.5 | Q 2.3 | Page 21

State the converse and contrapositive of  statement:

A positive integer is prime only if it has no divisors other than 1 and itself.

Ex. 31.5 | Q 2.4 | Page 21

State the converse and contrapositive of  statement:

If you live in Delhi, then you have winter clothes.

Ex. 31.5 | Q 2.5 | Page 21

State the converse and contrapositive of  statement:

If a quadrilateral is a parallelogram, then its diagonals bisect each other.

Ex. 31.5 | Q 3.1 | Page 21

Rewrite of the  statement in the form "p if and only if q".

p : If you watch television, then your mind is free and if your mind is free, then you watch television.

Ex. 31.5 | Q 3.2 | Page 21

Rewrite of the  statement in the form "p if and only if q".

q : If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular.

Ex. 31.5 | Q 3.3 | Page 21

Rewrite of the  statement in the form "p if and only if q".

r : For you to get an A grade, it is necessary and sufficient that you do all the homework regularly.

Ex. 31.5 | Q 3.4 | Page 21

Rewrite of the  statement in the form "p if and only if q".

s : If a tumbler is half empty, then it is half full and if a tumbler is half full, then it is half empty.

Ex. 31.5 | Q 4.01 | Page 21

Determine the contrapositive of the statement:

If Mohan is a poet, then he is poor.

Ex. 31.5 | Q 4.02 | Page 21

Determine the contrapositive of the statement:

Only if Max studies will he pass the test.

Ex. 31.5 | Q 4.03 | Page 21

Determine the contrapositive of the statement:

If she works, she will earn money.

Ex. 31.5 | Q 4.04 | Page 21

Determine the contrapositive of the statement:

If it snows, then they do not drive the car.

Ex. 31.5 | Q 4.05 | Page 21

Determine the contrapositive of the statement:

It never rains when it is cold.

Ex. 31.5 | Q 4.06 | Page 21

Determine the contrapositive of the statement:

If Ravish skis, then it snowed.

Ex. 31.5 | Q 4.07 | Page 21

Determine the contrapositive of the statement:

If x is less than zero, then x is not positive.

Ex. 31.5 | Q 4.08 | Page 21

Determine the contrapositive of the statement:

If he has courage he will win.

Ex. 31.5 | Q 4.09 | Page 21

Determine the contrapositive of the statement:

It is necessary to be strong in order to be a sailor.

Ex. 31.5 | Q 4.1 | Page 21

Determine the contrapositive of the statement:

Only if he does not tire will he win.

Ex. 31.5 | Q 4.11 | Page 21

Determine the contrapositive of the statement:

If x is an integer and x2 is odd, then x is odd.

#### RD Sharma solutions for Class 11 Mathematics Textbook Chapter 31 Mathematical reasoning Exercise 31.6 [Pages 28 - 29]

Ex. 31.6 | Q 1.1 | Page 28

Check the validity of the statement:

p : 100 is a multiple of 4 and 5.

Ex. 31.6 | Q 1.2 | Page 28

Check the validity of the statement:

q : 125 is a multiple of 5 and 7.

Ex. 31.6 | Q 1.3 | Page 28

Check the validity of the statement:

r : 60 is a multiple of 3 or 5.

Ex. 31.6 | Q 2.1 | Page 28

Check whether the statement are true or not:

p : If x and y are odd integers, then x + y is an even integer.

Ex. 31.6 | Q 2.2 | Page 28

Check whether the statement are true or not:

q : If xy are integers such that xy is even, then at least one of x and y is an even integer.

Ex. 31.6 | Q 3 | Page 29

Show that the statement
p : "If x is a real number such that x3 + x = 0, then x is 0"
is true by
(i) direct method
(ii) method of contrapositive

Ex. 31.6 | Q 4 | Page 29

Show that the following statement is true by the method of contrapositive
p : "If x is an integer and x2 is odd, then x is also odd"

Ex. 31.6 | Q 5 | Page 29

Show that the following statement is true
"The integer n is even if an only if n2 is even"

Ex. 31.6 | Q 6 | Page 29

By giving a counter example, show that the following statement is not true.
p : "If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle".

Ex. 31.6 | Q 7.1 | Page 29

statement are true and false? In each case give a valid reason for saying so

p : Each radius of a circle is a chord of the circle.

Ex. 31.6 | Q 7.2 | Page 29

statement are true and false? In each case give a valid reason for saying so

q : The centre of a circle bisects each chord of the circle.

Ex. 31.6 | Q 7.3 | Page 29

statement are true and false? In each case give a valid reason for saying so

r : Circle is a particular case of an ellipse.

Ex. 31.6 | Q 7.4 | Page 29

statement are true and false? In each case give a valid reason for saying so

s : If x and y are integers such that x > y, then − x < − y.

Ex. 31.6 | Q 7.5 | Page 29

statement are true and false? In each case give a valid reason for saying so

t :  $\sqrt{11}$  is a rational number.

Ex. 31.6 | Q 8 | Page 29

Determine whether the argument used to check the validity of the following statement is correct:
p : "If x2 is irrational, then x is rational"
The statement is true because the number x2 = π2 is irrational, therefore x = π is irrational.

## Chapter 31: Mathematical reasoning

Ex. 31.1Ex. 31.2Ex. 31.3Ex. 31.4Ex. 31.5Ex. 31.6

## RD Sharma solutions for Class 11 Mathematics Textbook chapter 31 - Mathematical reasoning

RD Sharma solutions for Class 11 Mathematics Textbook chapter 31 (Mathematical reasoning) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 11 Mathematics Textbook solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 11 Mathematics Textbook chapter 31 Mathematical reasoning are Mathematically Acceptable Statements, New Statements from Old, Special Words Or Phrases, Contrapositive and Converse, Introduction of Validating Statements, Validation by Contradiction, Difference Between Contradiction, Converse and Contrapositive, Consolidating the Understanding.

Using RD Sharma Class 11 solutions Mathematical reasoning exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer RD Sharma Textbook Solutions to score more in exam.

Get the free view of chapter 31 Mathematical reasoning Class 11 extra questions for Class 11 Mathematics Textbook and can use Shaalaa.com to keep it handy for your exam preparation

S


text:
# Variable Selection in Machine Learning

05 December, 2014

I’ve had a discussion with a colleague on the selection of variables in a model. The discussion boils down to the following question:

Is it better to supply all the variables that you have into the model and risk overfitting or should you start out small and add values to make the model more and more complex?

You should always use a test set to determine the performance of your models and you should apply strategies to prevent it from overfitting, but starting our small and growing the model brings inherit bias into the model. In this document I will provide a mathematical proof of what might be dangerous of this approach.

## Linear Algrebra/Regression Primer

Warning, this document is heavy on math. In this document I’ve assumed that you have some remembrance of college level linear algebra. The following equations should feel readable to you:

\begin{aligned}
M_x & = I_n - X(X'X)^{-1}X' \\
M_x X & = I_nX - X(X'X)^{-1}X'X \\
& = X - XI_n\\
& = 0
\end{aligned}

From statistics you should hopefully feel familiar with the following:

\begin{aligned}
Y & = X\beta + \epsilon  \text{        where    } \epsilon ~ N(0,\sigma) \\
\hat{\beta} & = (X'X)^{-1} X'Y \\
\mathbb{E}(\hat{\beta} ) & = \mathbb{E}\big((X'X)^{-1} X'Y)\big)
\end{aligned}

## Proof

I am going to prove that for linear models you will introduce bias if you use few variables and include more and more as you are building a model and that this will not happen when you start with a lot of variables and reduce. For each case I will show what goes wrong in terms of the expected value of the \beta variables.

### Small to Large Problems

Suppose that the true model is given through:

 Y = X_1\beta_1 + X_2\beta_2 + \epsilon

If we start out with a smaller model, say by only looking at \beta_1 we would estimate for Y = X_1\beta_1 + \epsilon while the whole model should be Y = X_1\beta_1 + X_2\beta_2 + \epsilon. Then our expected value of \beta_1 can be derived analytically.

\begin{aligned}
\mathbb{E}(\beta_1) & = \mathbb{E}\big((X_1'X_1)^{-1} X_1'Y)\big)\\
& = \mathbb{E}\Big((X_1'X_1)^{-1} X_1'\big(X_1\beta_1 + X_2\beta_2 + \epsilon\big)\Big)\\
& = \mathbb{E}\Big((X_1'X_1)^{-1} X_1'X_1\beta_1 + (X_1'X_1)^{-1} X_1'X_2\beta_2 + (X_1'X_1)^{-1} X_1'\epsilon\big)\Big)\\
& = \mathbb{E}\Big(\beta_1 + (X_1'X_1)^{-1} X_1'X_2\beta_2 + (X_1'X_1)^{-1} X_1'\epsilon\big)\Big)  \\
& = \beta_1 + (X_1'X_1)^{-1} X_1'X_2\beta_2 + (X_1'X_1)^{-1} X_1'\mathbb{E}(\epsilon) \\
& = \beta_1 + (X_1'X_1)^{-1} X_1'X_2\beta_2 \\
& \ne \beta_1
\end{aligned}

So our estimate of \beta_1 is biased. This holds for every subset of variables {\beta_1, \beta_2} that make up \beta.

### Large to Small Solution

Suppose that the true model is given through:

 Y = X_1\beta_1 + \epsilon

If we start out with a larger model, say by including some parameters \beta_2 as well while they do not have any influence on the model then we will initially estimate a wrong model Y = X_1\beta_1 + X_2\beta_2 + \epsilon.

#### A lemma in between

Let’s define a matrix M_{X_1} = I_n -X_1(X_1'X_1)^{-1}X_1'. We can use this matrix to get an estimate of \beta_2.

Start out with the original formula.

\begin{aligned}
M_{X_1}Y & = M_{X_1}X_1\beta_1 + M_{X_1}X_2\beta_2 + M_{X_1}\epsilon \\
M_{X_1}Y & = M_{X_1}X_2\beta_2 + \epsilon \\
X_2'M_{X_1}Y & = X_2'M_{X_1}X_2\beta_2 + X_2'\epsilon \\
X_2'M_{X_1}Y & = X_2'M_{X_1}X_2\beta_2 \\
\beta_2 & = ( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}Y \\
\end{aligned}

Notice that M_{X_1}X1 = 0 and that M{X_1}\epsilon = \epsilon because of the definition while X_2\epsilon = 0 because \epsilon is normally distributed around zero and orthogonal to any of the explanatory variables.

#### The derivation for large to small

With this definition of \beta_2 we can analyse it to confirm that it should not converge to any other value than zero.

\begin{aligned}
\mathbb{E}(\beta_2) & = \mathbb{E}\big(( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}Y\big) \\
& = \mathbb{E}\big(( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}\big(X_1\beta_1 + \epsilon\big)\big) \\
& = \mathbb{E}\big(( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}X_1\beta_1 + ( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}\epsilon\big) \\
& = \mathbb{E}\big(( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}\epsilon\big) \\
& = ( X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}\mathbb"t{E}(\epsilon) \\
& = 0
\end{aligned}

Notice that ( X2'M{X_1}X_2)^{-1}X2'M{X_1}X_1\beta1 = 0 because M{X_1}X_1 = 0. So we see that \beta_2 is correctly estimated, what about \beta_1?

\begin{aligned}
\mathbb{E}(\beta_1) & = \mathbb{E}\big((X_1'X_1)^{-1} X_1'Y)\big)\\
& = \mathbb{E}\Big((X_1'X_1)^{-1} X_1'\big(X_1\beta_1 + \epsilon\big)\Big)\\
& = \mathbb{E}\Big(\beta_1 + (X_1'X_1)^{-1} X_1'\epsilon\Big)\\
& = \beta_1
\end{aligned}

So in this case we would remove the variables \beta_2 that are not of influence while our estimate of \beta_1 does not have any bias. This is exactly what we want.

# Conclusion

I’ve shown that by starting only a few variables and then adding them to the model has a bias risk in linear models. For other models you might expect the similar problem. A model will try to use any correlation that it can find in the data to find a pattern. To prevent this it might be a good tactic to start with all variables and to use less and less variables until overfitting is no longer an issue and all variables in the model are significant. This way you might prevent assigning too much predictive power to a variable that might need to be reallocated to a variable that isn’t included in the model.


text:
# 3 Tips from Someone With Experience

Oil industry in the Coming Years

Oil Industry is indeed one of the greatest industries that we can generally consider a booming one. With the fact that many machines and even transportation are reliant in this oils makes these oil industry since we all know that it is use in daily basis. Oils are generally a non-renewable resources which has the probability to run out and that is why a need to use it efficiently is indeed need to be considered since we tend to be reliant on these resources. Predictions of the status of oil industries in the coming years will be then thoroughly discussed in this article.

So first of all, one of the greatest thing that you need to take note about these oil industry for the coming years is that the demand for crude oil will be then slow in the next years. The pollution that these cars can emit with the usage of crude oil is indeed very alarming to every people and that is why a decrease in demand for these oil is indeed very possible to come. The oil industry is indeed still existing in the coming years but they are indeed limited to be used due to the fact that they may run out it they are used abusively.

On the other hand, another important thing that you need to keep in mind that will possibly happen in the coming years is to switch from renewable power. With the fact that these using renewable power can be a lot safer for us means that the shifting for the usage is indeed very possible to come at all. The time to switch for these renewable power is indeed maybe long as you imagine since it is not that easy at all. With the fact that we need still to adapt with the change means that these process may not take place just overnight.

The fact that technology will even proceed to more advanced state makes these as one of the most important thing that we can benefit from all of this. Basically, we all know that technology is indeed evolving every day and of course, you need to expect that technology to harness the energy of the sun is indeed possible in this days.

And last but not the least of all is that you need to expect that oil companies will then be more eco-friendly and generally works for the good of the people and planet. And of course, abusive use and drilling of places where it has oils is indeed will be further stop in this time since taking care of our planet is indeed one of the priorities in this modern days.


text:
# L^2 basis of class functions on a compact Lie group that are point-wise small

Consider first the torus group $\mathbb{T}^k$. A natural $L^2$ basis is given by the 1-dimensional complex representations: $(\theta_1, \ldots, \theta_k) \mapsto e^{i \sum_j c_j \theta_j}$ for integral $c_j$'s. This basis has also the nice property that each element in it is point-wise bounded by $1$ in absolute value. In general, an $L^2$ function takes the form $g = \sum_n q_n f_n$ where $f_n$ are the basis functions described above. $g$ can have arbitrary bad $L^\infty$ bound, since the only requirement is that $\sum q_n^2 = 1$, but $\sum_n q_n$ can be $\infty$.

Now my question is, for the unitary or orthogonal group, what is the smallest uniform $L^\infty$ bound one can achieve on an orthogonal basis on the space of class functions? For the unitary groups, any orthogonal basis of class functions forms an orthogonal basis in the space of symmetric functions of the eigenvalues under the inner product such that the Schur polynomials are orthonormal. So the question can be phrased in terms of symmetric functions. For the orthogonal groups, one orthogonal (not normalized) basis of class functions comes from $\prod_{i=1}^n (Tr X^i)^{\alpha_i}$ where for some odd $i$, $\alpha_i$ is odd (I am not entirely sure about this, but see this paper theorem 4).

Edit: what about all $L^2$ functions in general? Can one get a uniform $L^\infty$ bound on the basis of size $e^{\mathcal{O}(n)}$ for $SO(n)$ or $U(n)$?

-
So, unlike the torus, you don't want the basis of characters, since the best pointwise bound you can get for $Tr(\pi(g))$ is $\dim\pi$. For finite groups $G$, you have the basis of normalized characteristic functions of conjugacy classes $C$, which is not good either since it gives you a bound of $\sqrt{|G|/|C|}$. –  Alain Valette Jan 4 '12 at 6:09
@Alain: yes you are right. I guess my stated basis class functions for $SO(n)$ are not really orthogonal. –  John Jiang Jan 4 '12 at 6:26

There is a sharp comparison of sup norm to $L^2$ norm on compact topological groups $K$. (I saw a proof of the version of this for orthogonal groups or spheres in Stein-Weiss "Fourier Analysis on Euclidean Spaces", but the argument succeeds generally. E.g., section 7 of http://www.math.umn.edu/~garrett/m/mfms/notes_c/spheres_I.pdf)

The assertion (as alluded-to by @Alain Valette) is that the sup of the ratio of sup norm to $L^2$ norm is $\sqrt{dim V}$ if the total measure is normalized to 1, where $V$ is a (e.g.) left $K$-stable collection of functions on $K$. This estimate is sharp, in that it is attained.

Indeed, the argument does not depend on the space of functions $V$ being an irreducible repn, and it is easy to do tangible experiments on products of circles to see the phenomenon manifest.

Edit: (Thx @Denis... for noting broken link.) And, yes, the Stein-Weiss trick does not refer to basis elements, but to a subspace, so does not answer the first, original question exactly. Only the question-at-the-end. However, on another hand, the pointwise interaction of orthonormal basis elements is illuminated by the inevitability of that estimate.

-
Thanks a lot for the answer and reference! –  John Jiang Jan 4 '12 at 20:14
As far as I can understand (the link doesn't work for me), you consider the ratio on a linear subspace, which is not the same as the question where one wants the ratio on elements of a fixed basis. E.g. for a finite non-abelian group $G$, you can find an orthonormal basis of the space of functions on $G$ where every function has constant modulus $1$. (Just pick-up a bijection with $\mathbf{Z}/|G|\mathbf{Z}$ and transfer the characters from the latter.) Of course it's not clear what one can hope to do with such a basis... –  Denis Chaperon de Lauzières Jan 5 '12 at 6:53
@Denis: I understand your construction for finite groups now. Can you elaborate on your method for Lie groups? Certainly one can't map $SO(n)$ diffeomorphically onto the torus. –  John Jiang Jan 5 '12 at 19:13

I think for $U(n)$ and similar connected Lie groups you can construct orthonormal bases of the $L^2$ space of class functions which are everywhere of modulus $1$: use the Weyl integration formula to represent the space of conjugacy classes as $[0,\pi]^n$ with an explicit measure, and define functions $f(x)=\varphi(\psi(x))$ where

(1) $\varphi$ runs over an orthonormal basis of $[0,\pi]^n$ with respect to Lebesgue measure, all of which satisfy $|\varphi|=1$ (which can be done using characters);

(2) $\psi$ is a fixed function so that the change of variable formula shows that $\varphi\mapsto f$ is an isometry from the $L^2$-space with respect to Lebesgue measure to the class-function space.

For finite groups, similarly, one can construct an orthonormal basis with maximal $L^{\infty}$-norm equal to $1$ (though usually not with constant modulus).

-
It's not entirely clear to me why this would work. Consider $\varphi_1 (\theta) = e^{i \sum_j a_j \theta_j}$ and $\varphi_2 = e^{i \sum_j b_j \theta_j}$ that are orthogonal on the torus, i.e., $\langle \varphi_1, \varphi_2 \rangle =0$. Let $Jac(\theta)$ be the Jacobian factor for going from the Haar measure on the torus to that on $U(n)$. Then we basically need $int e^{i\sum_j (a_j - b_j) \psi(\theta)} Jac(\psi(\theta)) d(\psi(\theta))$ to vanish for some $\psi$. But unless $\psi$ is linear, the exponential factor doesn't resemble anything that would vanish under integration. –  John Jiang Jan 4 '12 at 20:09
I guess my question is why would composition by $\psi$ yield an isometry between the two function spaces? –  John Jiang Jan 4 '12 at 20:12
Also if your argument works, then the modulus must be constant even on finite groups right? Otherwise the $L^2$ norm would be strictly less than 1. –  John Jiang Jan 4 '12 at 20:13
Thanks for helping by the way! –  John Jiang Jan 4 '12 at 20:15


text:
# Subelliptic $\mathrm{Spin}_{\mathbb{C}}$ Dirac operators, II. Basic estimates

### Abstract

We assume that the manifold with boundary, $X,$ has a Spin${}_{\mathbb{C}}$-structure with spinor bundle $S\mspace{-10mu}/$. Along the boundary, this structure agrees with the structure defined by an infinite order, integrable, almost complex structure and the metric is Kähler. In this case the Spin${}_{\mathbb{C}}$-Dirac operator $\eth$ agrees with $\bar{\partial}_b + \bar{\partial}_b^*$ along the boundary. The induced CR-structure on $bX$ is integrable and either strictly pseudoconvex or strictly pseudoconcave. We assume that $E\to X$ is a complex vector bundle, which has an infinite order, integrable, complex structure along $bX,$ compatible with that defined along $bX.$ In this paper we use boundary layer methods to prove subelliptic estimates for the twisted Spin${}_{\mathbb{C}}$-Dirac operator acting on sections on $S\mspace{-10mu}/ \otimes E$. We use boundary conditions that are modifications of the classical $\bar{\partial}$-Neumann condition. These results are proved by using the extended Heisenberg calculus.

## Authors

Charles L. Epstein

Department of Mathematics
University of Pennsylvania


text:
Volume 395 - 37th International Cosmic Ray Conference (ICRC2021) - CRI - Cosmic Ray Indirect
The Zettavolt Askaryan Polarimeter (ZAP) mission concept: radio detection of ultra-high energy cosmic rays in low lunar orbit.
A. Romero-Wolf* and Z. Collaboration
Full text: pdf
Pre-published on: July 05, 2021
Published on:
Abstract
Probing the ultra-high energy cosmic ray (UHECR) spectrum beyond the cutoff at ~40 EeV requires an observatory with large acceptance, which is challenging to implement with ground arrays. We present a concept for radio detection of UHECRs impacting the Moon's regolith from low-lunar orbit called the Zettavolt Askaryan Polarimeter (ZAP). ZAP would observe several thousands of events above the cutoff (~40 EeV) with a full-sky field of view to test whether UHECRs originate from Starburst Galaxies, Active Galactic Nuclei, or other sources associated with the matter distribution of the local universe at a distance $>$ 1 Mpc. The unprecedented sensitivity of ZAP to energies beyond 100 EeV would enable a test of source acceleration mechanisms. At higher energies, ZAP would produce the most stringent limits on super heavy dark matter (SHDM) via limits on neutrinos and gamma rays resulting from self-annihilation or decay.
DOI: https://doi.org/10.22323/1.395.0403
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
# Use Gauss' Law to find the magnitude of E in a sphere

1. Sep 25, 2007

### Bluewrath

Problem
Problem 27.56

A sphere of radius R_0 has total charge Q. The volume charge density (C\m^3) within the sphere is $$\rho$$(r) = C/(r^2), where C is a constant to be determined.

Part A
The charge within a small volume dV is dq = $$\rho$$dV. The integral of $$\rho$$dV over the entire volume of the sphere is the total charge Q. Use this fact to determine the constant C in terms of Q and R_0.

Part B
Use Gauss's law to find an expression for the magnitude of the electric field E inside the sphere, r $$\leq$$ R_0.
Express your answer in terms of Q, R_0, r and appropriate constants.

Relevant equations and whatnot :-)
Gauss' Law for a sphere: 4pi(r^2)E = Q_in/$$\epsilon$$_0
Q = $$\rho$$ * V

Solution attempt(s) (and the miserable failure(s))
I've already found Part A, through $$\rho$$ = Q/v; the answer is C = Q/(4*pi*R_0).

For Part B, I've tried the following:

Use Gauss' Law to set up:

E = Q_in/(A*$$\epsilon$$_0)

Define Q_in as:

Q_in / Q = (4/3)*pi*r^3 / (4/3)*pi*((R_0)^3) ==> Q_in = Q*r^3/((R_0)^3)

Define A as:

A = 4*pi*r^2

and plug everything in to get:

E = Qr / ( (R_0)^3) * 4 * pi * $$\epsilon$$_0 )

Which is apparently wrong :-(

So, I figured that they probably were intending to make the students use the solution to Part A, in which we found the C constant of $$\rho$$(r) = C/(r^2).

So, using the same setup and the same A and the same $$\epsilon$$_0 (like that's going to change :-), I defined Q_in as:

Q_in = $$\rho$$(r) * V(r) = (C/r^2) * ((4/3)*pi*r^3) = (Qr/3*(R_0))

Again plugging in all the variables, I got stuck with:

E = Q / ( 3 * 4 * pi * r * (R_0) * $$\epsilon$$_0 )

And that too is also wrong, apparently (unless there's a huge conspiracy against me involving Dr. Randall Knight and quite possibly the US government whom are intent on sabotaging my efforts). And now, I'm lost and have no idea what to do. There's a question in the book that asks for the electric field inside a uniformly charged sphere, and the answer is E = Qr / ( (R_0)^3) * 4 * pi * $$\epsilon$$_0 ), which is what I got at my first try. Am I missing something here? Helpy please! :-D

2. Sep 25, 2007

### Staff: Mentor

The problem is that this is not a uniformly charged sphere. To find the charge enclosed within a Gaussian sphere of radius r, set up the integral of $\rho dV$. (The integral is easy to solve.)

3. Mar 13, 2009

### j88k

Whats the answer for the second part then?


text:
# I Functional Completeness

#### Calculuser

While I was studying Propositional Calculus from Elliott Mendelson's "Introduction to Mathematical Logic," I came across, on page 18, the proof of functional completeness of $\neg$, $\wedge$, $\lor$ logical connectives as shown below:

The part that does not make sense to me starts from "Then $C_{k}$ has . . ." down to the end of the proof.

Question: Why does $C_{k}$ have the $T$ value for this assignment?

Related Set Theory, Logic, Probability, Statistics News on Phys.org

#### jedishrfu

Mentor
Can you elaborate more by which "then $C_k$" you're referring to as there are at least three that I see:
- Then $C_k$ has the value T...
- then $C_k$ is a disjunct of D.
- then $C_k$ is not a disjunct of D

#### Calculuser

By that I meant the following sentences:

"Then $C_{k}$ has the value $T$ for this assignment, whereas every other $C_{i}$ has the value $F$. If $f$ has the value $T$ for row $k$, then $C_{k}$ is a disjunct of $D$. Hence, $D$ would also have the value $T$ for this assignment. If $f$ has the value $F$ for row $k$, then $C_{k}$ is not a disjunct of $D$ and all the disjuncts take the value $F$ for this assignment. Therefore, $D$ would also have the value $F$. Thus, $D$ generates the truth function $f$."

I could not grasp these following steps. For instance, why does $C_{k}$ in the first sentence have to be $T$?

#### stevendaryl

Staff Emeritus
Maybe it would help to work out a particular example. Let $f(x_1, x_2, x_3)$ be "if $x_1$ then $x_2$ else $x_3$". Then the truth table looks like this:

$$\left[ \begin{array} \\ & A_1 & A_2 & A_3 & f(A_1, A_2, A_3) \\ 1 & T & T & T & T \\ 2 & T & T & F & T \\ 3 & T & F & T & F \\ 4 & T & F & F & F \\ 5 & F & T & T & T \\ 6 & F & T & F & F \\ 7 & F & F & T & T \\ 8 &F & F & F & F \end{array} \right]$$

There are 8 possible assignments of values to $A_1, A_2, A_3$

So we have the following values for $C_i$:
$$\left[ \begin{array} \\ & C_j & f \\ 1 & A_1 \wedge A_2 \wedge A_3 & T \\ 2 & A_1 \wedge A_2 \wedge \neg A_3 & T \\ 3 & A_1 \wedge \neg A_2 \wedge A_3 & F \\ 4 & A_1 \wedge \neg A_2 \wedge \neg A_3 & F \\ 5 & \neg A_1 \wedge A_2 \wedge A_3 & T \\ 6 & \neg A_1 \wedge A_2 \wedge \neg A_3 & F \\ 7 & \neg A_1 \wedge \neg A_2 \wedge A_3 & T \\ 8 & \neg A_1 \wedge \neg A_2 \wedge \neg A_3 & F \end{array} \right]$$

Then $D$ is the disjunction of all the $C_i$ where $f$ has value $T$:

$D = C_1 \vee C_2 \vee C_5 \vee C_7 == (A_1 \wedge A_2 \wedge A_3) \vee (A_1 \wedge A_2 \wedge \neg A_3) \vee (\neg A_1 \wedge A_2 \wedge A_3) \vee (\neg A_1 \wedge \neg A_2 \wedge A_3)$

Now, let's augment table 1 to include the values of $C_i$ and $D$. I'm bold-facing those $C_k$ that are in the disjunction $D$.

$$\left[ \begin{array} \\ & A_1 & A_2 & A_3 & f(A_1, A_2, A_3) & \mathbf{C_1} & \mathbf{C_2} & C_3 & C_4 & \mathbf{C_5} & C_6 & \mathbf{C_7} & C_8 & D \\ 1 & T & T & T & T & T & F & F & F & F & F & F & F & T \\ 2 & T & T & F & T & F & T & F & F & F & F & F & F & T \\ 3 & T & F & T & F & F & F & T & F & F & F & F & F & F \\ 4 & T & F & F & F & F & F & F & T & F & F & F & F & F \\ 5 & F & T & T & T & F & F & F & F & T & F & F & F & T \\ 6 & F & T & F & F & F & F & F & F & F & T & F & F & F \\ 7 & F & F & T & T & F & F & F & F & F & F & T & F & T \\ 8 &F & F & F & F & F & F & F & F & F & F & F & T & F \end{array} \right]$$

So what they're saying is that $C_k$ has value $T$ for assignment number $k$, and that $C_k$ has value $F$ for every other assignment. If $C_k$ is in the disjunction $D$, then $D$ has value $T$ for assignment $k$, otherwise $D$ has value $F$. So we can see that $D$ is true exactly the same rows that $f$ is.

#### Calculuser

I noticed my misunderstanding. All is clear now. Thank you very much!

#### Periwinkle

Eliott Mendelson in his book before the referenced page 10-11. examples illustrate operations of the proposition calculus with an electronic circuit diagram. The theorem is that any complicated Truth-function of the A, B, C, ... propositions are given, then this can be done by "and," "or" and denial proposition calculus operations. The Truth-function is a table that assigns a "true" or "false" truth value to the A, B, C, ... propositions. Using this, I tried to illustrate the proof of the theorem.

We can prepare the electronic circuit diagram shown in the figure for the truth function, such that each line in the above table, where the value of the truth function is true, has a through-line. If the value of "A" in a row on the table is true, then we write "A" to the first switch on the through-line, if it is false, then no A; like B, C, and so on. In cases where the value of the truth function is true, the current is flowing from X to Y.

For the row of the table when the value of the truth function is false (in the case of such "true" - "false" distribution of propositions A, B, C, ...) there is no current. Because if we look at any through-line, the current is only on the "true" - "false" distribution of the proposition A, B, C, ..., which distribution is in the appropriate row of the table.

Mendelson's book would have beautiful memories for me. I was 15 when I read the Russian version during the summer break. I learned language and math at the same time. I remember reading the liar paradox first, or, e.g., about Russell's paradox; I first read about Neumann's foundation of set theory (the gloomy Neumann classes that are not sets). Everything seemed mysterious. I could only read a few pages a day because I didn't know the language well enough. I see these things quite differently today. However, if someone asked me which book I would recommend for reading, I wouldn't think for a minute: Ebbinghaus, H.-D., Flum, J., Thomas, Wolfgang, Mathematical Logic, Springer.

Last edited:

"Functional Completeness"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving


text:
# Where's the fallacy

1. Jul 9, 2007

### nicktacik

This one has me stumped.

$$e^{\pi i} = -1$$
$$e^{2 \pi i} = (-1) ^ 2 = 1$$
$$ln(e^{2 \pi i}) = ln(1) = 0$$
$$2 \pi i = 0$$

Or is 2 pi i actually 0, and this does not actually imply that either pi = 0 or i = 0?

Last edited: Jul 9, 2007
2. Jul 9, 2007

### Integral

Staff Emeritus
Since $2 \pi i$ is an imaginary number you must use the definition of LN in the complex plane. You are using the definition for the real number line.

in the complex plane we have:

$$logz = log(r) + i \theta$$

edit (changed my definition of z)
$$r = |z|$$
and
$$\theta = arg(z)$$

Last edited: Jul 9, 2007
3. Jul 9, 2007

### matt grime

More than that - logs of complex numbers are only defined up to multiples of 2pi. One can ought to take the principal branch - this is just a slightly more complicated variation on the square root 'fallacies'.

4. Jul 9, 2007

### Kummer

The mistake is here:
$$\ln e^z = z$$
This is not true for complex numbers.

Note: the other way around:
$$\exp (\ln z) = z, \ z\not =0$$
Is true.

5. Jul 10, 2007

### nicktacik

Ok thanks. As you can see, I haven't taken my complex variables class yet.

6. Jul 10, 2007

### mathwonk

its sort of like saying, (2^2 =4 and (-2)^2 = 4 so 2 = -2.)


text: Where, σ = electrical conductivity. Water Molecule. Heavy water (deuterium oxide, 2 H 2 O, D 2 O) is a form of water that contains only deuterium (2 H or D, also known as heavy hydrogen) rather than the common hydrogen-1 isotope (1 H or H, also called protium) that makes up most of the hydrogen in normal water. Water is the chemical substance with chemical formula H 2 O; one molecule of water has two hydrogen atoms covalently bonded to a single oxygen atom. The measure of the amount of water vapor suspended in the air is what is known as humidity. It is the upward force of a fluid. Additional information This is a resource from the Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry. The molecular formula for water is H 2 O. Electrical conductivity is denoted by the Greek letter ρ. What is the formula mass of water? In conclusion, and because sea water is a solution of salts in water - H2O - you can't really elaborate on a specific chemical formula. Try, for example, the lb/ft3 to kg/m3 converter. In a fluid may be a gas or a liquid, pressure increases with depth. Updated 5 days ago|3/11/2021 3:47:23 PM. Iron oxides or iron carbonates can give a reddish brown colouration to hard water deposits . The specific heat of water is 4190 J/kg∙K. Solution: The relative mass formula of water is M r (H 2 O) = 2 × A r (H) + A r (O) = 2 × 1 + 16 = 18. Next step I have to balance the equation And finally I have to add in the state symbols ( I think I can do that but I'm not sure … Now, on the right side, potassium hydroxide is a molecule; each unit has one potassium atom, one oxygen atom, and one hydrogen atom. So it's formula is KOH. ‘Heavy water’ has a molecular weight of 20 g/mol. Water is an 'asymmetric rotor', which turns out to be the hardest to understand. Calculate the water potential using the formula above. States of Water We know that water isn't always "wet", sometimes it's frozen into ice or snow. Earn a little too. Precautions To Take While Using Water For Baby Formula. eg. In both instances, mass and charge are conserved. The media is a powerful influence because it. Each molecule contains 2 hydrogen atoms and 1 oxygen atom. To about two parts of wine, one part of soda water is added. However, it is still best to boil tap water and let it cool before you use it to make your baby’s formula. This is because of something we may experience in the water called the buoyant force. The chemical formula for water is H2O and its proper chemical name is dihydrogen monoxide. I am describing the reactions and equations of potassium with water. Even caffeinated drinks — such as coffee and soda — can contribute to your daily water intake. How are atomic … The concentration of the solution was found to be 0.04805 M. What is the water of hydration for this hydrate of sodium carbonate? This means that the chemical formula of water is H₂O. They can also make their way into the... Bottled Water. deforestation One oxygen atom with an atomic mass of #15.999" g/mol"# #(2×1.008)+15.999=18.017" g/mol"# Answer link. Question. Ice is the solid form of water that is created when water freezes at a temperature of zero degrees Celsius, so it has the same chemical formula of water, which is H2O. potential energy of water in a system compared to pure water Solution: The heat of fusion for water is applicable here and the equation has to be rearranged to solve it for the mass. bathing river worshipers The formula is: ρ = m/V. Add Your Riddle Here. Mass of hydrate used: 3.90 g. Mass of water of hydration lost: 1.36 g (amount in moles: 0.08 mol) Mass of dehydrated salt remaining: 2.54 g (amount in moles: 0.02 mol) What is x? a decrease in biodiversity Mass of water will be 4.49 gram. Fluoride Concerns. User: The picture below (Black Square, by ... Weegy: According to MLA style, you must include Works Cited at the end of any writing project that contains research. Similarly, water pressure is the term used to describe the strength of water flow through a channel or pipe. Use Heat of Fusion Formula. Water velocity is the speed at which the water flows in a pipe, while the flow rate is the amount of water that is moving in the pipe. Water is an excellent solvent for ionic compounds like salt and for polar compounds like sugar. traveling and trading The Chemical Composition of Sea Water. If H2O is the formula for water, what is the formula for ice? Kasimir Malevich's images had no reference at all to ... How are mla style headers aligned across the top of the page. Water vapor is the gas phase of water, and is invisible. The vinegar solution can clean many different home surfaces and appliances, including countertops, floors, garbage disposals, refrigerators and coffee pots. For example, many fruits and vegetables, such as watermelon and spinach, are almost 100% water by weight. The measure of the amount of water vapor suspended in the air is what is known as humidity. When I realized the lack, I put three examples here (as opposed to their own separate file) and included another in the problems. Previous Riddle. Instead, place the bottle in a container of warm water for a few minutes, or run it under a hot tap. Humans have used the Ganges River, located in India, for all of the following except __________. This means that the water molecule is not linear. I need to show all my work. The specific heat capacity of water vapour at room temperature is also higher than most other materials. Density is mass divided by volume (ρ=m/v), and water was used as the basis for establishing the metric unit of mass, which means a cubic centimeter (1cm 3) of water weighs one gram (1g). 10H 2 O. Water striders, which are small insects, can walk on water as their weight is considerably less to penetrate the water surface. In terms of water, ‘light water’ or H2O has a molecular weight of 18 g/mol. Determining the formula for a hydrate means discovering the number of water molecules that the substance contains. Water contains two parts hydrogen to one part oxygen. When a lake freezes, it freezes at the surface first producing a protective layer which insulates the water below and helps preserves life below. An anhydrate is a hydrate that has lost its water molecules. Buoyancy is an easy concept to understand about pressure in a fluid. What you eat also provides a significant portion. providing a water source for China, Which of the following environmental problems is not associated with human interactions with water in South and Southeast Asia? Search for … D. FAQ. The water pressure formula and SI unit are given as: Formula. Electrolysis is the process of using electricity to produce a chemical reaction. 1 Answer Oscar L. Jun 9, 2016 #18.017" g/mol H"_2"O"# Explanation: Two hydrogen atoms each with an atomic mass of #1.008" g/mol"# Plus. Water freezes at 0°C and boils at 100°C at 1 atmosphere pressure. User: Who applied punishment in Medieval Japanese ... Weegy: Senator Jon Tester of Montana was reelected in 2012. Temperature and Pressure. the MW of Ca(No3)2 is 164 g/mol and the MW of water is 18g/mol so with 1 water attached the MW of the molecule would be 182 g/mol 2 waters 200 g/mol keep adding a water to the formula then take the total mass of the waters in this case i'll take 2 waters 2 waters = 36g/mole divided this by the total mass that molecule would have (200g/mol) and you get the % of water in the molecule. Given this, you know that the conjugate base for water is #OH^-#, the hydroxide ion, and its conjugate acid is the hydronium ion, #H_3O^+#. Don't warm formula in the microwave, which heats things unevenly. Some cases are provided below: Insects walking on water; Floating a needle on the surface of the water. Did you answer this riddle correctly? In living things water allows materials to dissolve and interact in the many chemical reactions which take place in cell that are essential life. Nevertheless, the North Atlantic, South Atlantic, Indian Ocean, and South Pacific areas a… In addition to trace amounts of bicarbonate of soda, many soda waters are also supplemented with other minerals to make them more like all-natural mineral water. Have some tricky riddles of your own? Is it Safe? The density converter will be perfect for you. This is actually the chemical formula for the molecule that makes up water. The chemical formula for water usually is written as H 2 O, but another way to consider the formula is HOH, where a positively charged hydrogen ion (H +) is bonded to a negatively charged hydroxide ion (OH -).This means water has properties of both an acid and a base, where the properties essentially cancel each other out. Water is one of our most plentiful chemicals. Show Answer. The formula for water is H 2 O. mentirosa January 26, 2009 . Water is a unique compound because it has the ability to exist on the earth's surface in all three forms: liquid, solid (ice), and gas (water vapor). Deforestation is not associated with human interactions with water in South and Southeast Asia. Water exists as molecules. As a solid, water forms a hexagonal arrangement between the water molecules and takes up more space than when a liquid. The chemical formula for water vapor is H 2 O. 0 Answers/Comments. Expert answered|houd|Points 1681| Log in for more information. Vinegar, used as a 100 percent solution or mixed with water, is an effective cleaning agent and is environmentally friendly. In addition, beverages such as milk, juice and herbal teas are composed mostly of water. Discard the prepared formula after two hours and an unused bottle of mixed formula after 24 hours, as it is no longer sterile. It doesn’t contain any nutrients, but your baby will get all they need from the formula. 13.0% By definition, a solution's percent concentration by mass, "% m/m", tells you the number of grams of solute present for every "100 g" of the solution. This salinity level is relatively stable everywhere. This means that if you know the number of grams of ethanol dissolved in exactly "100 g" of this solution, you know the solution's percent concentration by mass. Buoyancy Formula What is Buoyancy? Hard water is water that has high mineral content (in contrast with "soft water"). Water is a simple compound with a set of unique properties that are essential for life. The usual chemical formula for water assumes the hydrogen atoms … ‘H’ has a single electron and proton while ‘D’ has an additional neutron. The mass of water is found by weighing before and after heating. Baby formula mixers only mix the water and formula together, leaving you to take care of the proper heating. They provide recreation areas and enhance the surroundings by cleaning air and water. a decrease in the growth of coral reefs. Water used for mixing infant formula must be from a safe water source, as defined by your state or local health department. Water potential is the potential energy of water in a system compared to pure water, when both temperature and pressure are kept the same. Many people, especially those who are not big drinkers, prefer that to straight wine. Like this, there are various examples of surface tension which are found in nature. Here, $$H_f =1500 J$$ Q = 334 C per gram $$\Delta H_f = \frac{q}{m}$$ Then, $$m = \frac {H_f}{q}$$ $$= \frac {1500}{ 334 }$$ = 4.49 gram. D. WINDOWPANE is the live-streaming social network that turns your phone into a live broadcast camera for streaming to friends, family, followers, or everyone. Water vapor is the gas phase of water, and is invisible. what is the chemical formula and name of soda water? Interestingly, the world's oceans have, on average, a salinity level of only 3.5 percent, i.e., 35 grams of dissolved salts per liter of water. Water striders, which are small insects, can walk on water as their weight is considerably less to penetrate the water surface. What Is It Riddles Chemistry Riddles . Water potential at atmospheric pressure is 0. The final water-to-grain ratio has increased to 2.26 qt/lb (18.1 / 8). The SI unit of pressure is Pascal (Pa). Answer: The heat energy transferred to the water is 1676 kJ = 1 676 000 J. If he decides to run again when his current team expires, ... Weegy: Richard Henry Lee introduced the idea that the colonies should declare themselves free and independent states. This means that ice is less dense than water and floats to the top. Each water molecule is made up of two hydrogen atoms and one oxygen atom. Water is a Chemical!? Or maybe do you just need to convert one density unit into another? Now, let's talk about how to analyze the motion of a fluid. flooding Solution: 1) Determine mass of water driven off: 15.67 − 7.58 = 8.09 g of water It’s ultra-clean water, as all contaminants have been removed during the distillation process. Some of which derives from a unit of force per unit area. Hence, velocity of water is dependent on the flow rate. ρ = resistivity. Check also our other articles, for instance, about the water density g/cm3. This answer has been confirmed as correct and helpful. If you are concerned or uncertain about the safety of the tap water: You may use bottled water or bring cold tap water to a rolling boil for 1 minute (no longer), then cool the water to room temperature for no more than 30 minutes before it is used. If H2O is the formula for water, what is the formula for ice? https://study.com/.../what-is-water-vapor-definition-pressure-formula.html The ratio of these elements determines the molecular formula. A. The chemical formula for water is H 2 O and its proper chemical name is dihydrogen monoxide. Distilled water is water that’s boiled into steam, and is then condensed back into its fluid form. Specific Heat Capacity Figure shows water and cooking oil in similar pots and supplied with heat at the same rate. Solved: 60%. Example #1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. Water pressure= ρ g h. SI unit. If there is 5.00 kg of water in the pot, and the temperature is raised by 80.0 K, what is the specific heat of water? Which liquid heats up at a faster rate: water or cooking oil? YES NO . For water vapour at room temperature and pressure, the value of specific heat capacity (Cp) is approximately 1.9 J/g°C. You can calculate the volume based on the height, width, length, filled depth, diameter and radius. Like this, there are various examples of surface tension which are found in nature. So water can be excited rotationally by 2.45 GHz - the rotational behaviour of water as single molecules in the gas phase is very complicated. Water is a V-shaped molecule and is highly polar in nature. Water is a simple compound with a set of unique properties that are essential for life. The water budget formula allows professionals to design a sustainable landscape based on a proper amount of water.


text:
# Math Help - Linear algebra

1. ## Linear algebra

Let V and W be vector spaces, let T : V--->W be linear, and let {w1, w2,...wk} be a linearly indepemdent subset R(t). Prove that if S={v1,v2,...vk} is chosen so that T(vi)=wi for i=1,2,.....k, then S is lenearly independent.

i've been trying to do that problem for ever.. dont know how to! plz help!

2. Look at this.


text:
# Integrals via Infinite Series?

1. Nov 15, 2013

### bolbteppa

On the bottom of page 24 & top of page 25 of this pdf an integral is beautifully computed by breaking it up into an infinite series. Is there any reference where I could get practice in working integrals like these?

2. Nov 16, 2013


text:
Friction simulation in case of body moving with constant velocity and no external force

I am trying to write a simple physics simulation but have troubles with how friction works.

Let's assume a body of mass $m$ is moving horizontally on a plane with initial velocity $v_0$. Can someone please explain me how friction works in this case?

Is there a constant friction force opposite to the direction of motion which suddenly becomes zero when the body velocity becomes zero? In a simulation it would never become exactly zero and thus the body position would oscillate. Would it be reasonable to blend kinematic friction over into static friction at low velocities? How does this work in reality?

This is a tough question to solve in general, but I've seen the following type of approach taken with a few explicit timestepping "contact algorithms" between two bodies in the finite-element type literature if there does not need to be a distinction between the static and dynamic coefficients of friction.

It can be viewed as a way to do what has been mentioned by DilithiumMatrix and David Hammen.

First, we project momentum $p$ forward for each body as if there was no friction, so

$p^{*} = p^{n} + \Delta t f^n$.

If there is no-slip contact, that means at the end of the step the body is moving with the velocity of the combined center of mass (in this case, I assume the ground is not going to move due to the body, so we can just use 0). We then know

$p^{n+1} = p^{*} + \Delta t f^n_c$.

Therefore, the required force on the body for no-slip contact is $f_c = \frac{-p^*}{\Delta t}$ since $p^{n+1} = 0$.

Now, I assume you know the normal force (e.g. due to the weight of the body), in which case you can see if $|\mu f^n_{normal}| > |f_c|$, in which case the object must come to a stop. Otherwise, you calculate the contact force as

$f_{contact} = -\mathrm{sign}(p^n) \mu f_{normal}$

And the final momentum is given by $p^{n+1} = p^* + \Delta t f_{contact}$. Or, altogether

$p^{n+1} = \begin{cases} 0 & \text{ if } |\mu f_{normal}| \ge |\frac{-p^{n}}{\Delta t} - f^n| \\ p^{n} + \Delta t (f^n -\mathrm{sign}(p^n) \mu f_{normal}) & \text{otherwise}\end{cases}$.

Please note that I left it in 1D and this only accounts for the tangential motion. You might be able to set the thresholds for $\mu$ where it shows up as two different values, to get the static and dynamic coefficients of friction but I have not tried that. E.g. since the static friction coefficient is greater than the dynamic one I believe

$p^{n+1} = \begin{cases} 0 & \text{ if } |\mu_{static} f_{normal}| \ge |\frac{- p^{n}}{\Delta t} - f^n| \\ p^{n} + \Delta t (f^n -\mathrm{sign}(p^n) \mu_{dynamic} f_{normal}) & \text{otherwise}\end{cases}$

will work without overshoot but will still stop "too early" if the object was in motion but the no-slip force was between the static and dynamic thresholds.

The simplest and easiest thing to do is chose a threshhold value for the speed. If the speed drops below that value, force the speed to be zero and implement your scheme for when the body is at rest. If you're on a level surface with no horizontal force components, the friction becomes zero.

Like you mentioned, you can't depend on the simulation to get to zero perfectly, but there's nothing wrong, in the limit, with forcing the zero.

If you are just taking into account surface friction (i.e. no air resistance) then yes, the friction force will be constant, opposite to the direction of motion until the object stops moving.

When the object stops moving, static friction keeps it in place. When people refer to the "force of static friction, $F_{fs}$", that is actually a maximum static-friction force. This means that if an external force is applied $|F_e| < |F_{fs}|$, static friction will match it, and prevent acceleration. If the external force exceeds the static friction ($|F_e| > |F_{fs}|$) then the object will start accelerating with a net force ($|F_e| - |F_{fs}|$).

In a simulation the velocity, and thus the kinetic friction should actually become exactly zero at some point, because you should implement your equations such that the friction force can never create a reverse velocity (i.e. you should manually prevent oscillation). One possible implementation would be something like:

Calculate the position at which the object should stop, call that $x_s$. Because you are using some finite time-step, you will tend to eventually reach some position past this value, $x_i > x_s$. In your code, you can check for this to happen, and simply set $x_i = x_s$ and $v_i = 0.0$ instead of integrating the equations of motion, i.e. instead of $v_i = v_{i-1} + a \, \Delta t$ and $x_i = x_{i-1} + v_i \, \Delta t$ (if you were using a simple Euler integration).

Is there a constant friction force opposite to the direction of motion which suddenly becomes zero when the body velocity becomes zero? In a simulation it would never become exactly zero and thus the body position would oscillate.

In reality, there's a distinction between kinetic and static friction. Your simulation of reality needs to reflect this distinction.

Simulations typically have fixed time steps. This is problematic with regard to collisions, and also with regard to things like kinetic / static friction. You need to develop a mechanism that splits the fixed simulation time steps in your simulation into smaller parts so as to capture those apparent discontinuities. In technical terms, your simulation needs a propagator (something that advances state over time) and a solver (something that tells the propagator to slow down). The solver essentially splits the fixed time steps of the simulation into smaller parts.

The solver is quite simple in the case of the simple simulation you are trying to create. Project the velocity forward over the simulation time step just using kinetic friction. You need to solve for when the velocity reaches zero if this would result in the object changing direction. When that would happen is fairly easy assuming a simple model of kinetic friction that makes velocity change linearly with time. (It's not so easy in general. There are papers galore on how to best write a solver.) When the sliding object's velocity gets below some limit, the object stops. An external force is now needed to make the object start moving again.

• Do you know a good reference paper? – Danvil Dec 26 '16 at 23:57
• There is a truly massive review paper on contact algorithms here (I cannot vouch for its correctness, but it seems to be a good place to start if looking for a particular class of algorithms): ipmnet.ru/~burago/papers/cont-e.pdf – neocpp Dec 28 '16 at 0:38
• @Danvil -- I'd recommend a book on physics engines (or game physics) instead of a paper. There are lots of them. – David Hammen Dec 28 '16 at 21:10
• @neocpp: Thank you! That was exactly what I was looking for. – Danvil Dec 29 '16 at 2:35
• @DavidHammen That is a good idea but I would be careful about distinguishing between methods that give accurate results vs just "visually correct" ones. In my experience most of the game-related engines don't solve physical systems to a high degree of fidelity (with good reason, as high fidelity and speed are often at odds). I would hope that a book would go over this, but depending on the application OP has in mind it might be important. – neocpp Dec 29 '16 at 3:44


text:
# Chapter 4 : First Law of Thermodynamics¶

## Example 4.1 Page No : 80¶

In [1]:
import math

# Variables
p = 1.0;			# in MPa
p = p * 10**6;			# in N per m**2
del_v = 1.5;			# in m**3 per min

# Calculations
del_v = del_v * 60;			# in m**3 per h
W = p * del_v;			# in J
W = W * 10**-6;			# in MJ

# Results
print "Work done by the pump upon the water in MJ is : %.0f"%W

Work done by the pump upon the water in MJ is : 90

## Example 4.2 Page No : 81¶

In [1]:
# Variables
# w = 2*g*h
g = 9.81;
m =(0.2+10./1000)*10**3 ;			# in gm
s = 1;			# in cal/gm°C
del_T = 2;			# in °C

# Calculations
H = m * s * del_T;			# in cal
H = H * 10**-3;			# kcal
J = 4.1868 * 1000;
# W= 2*g*h= J*H
h = J*H/(2 * g);			# in m

# Results
print "Height from which the mass should fall in meter is : %.2f"%h

Height from which the mass should fall in meter is : 89.63

## Example 4.3 Page No : 83¶

In [3]:
# Variables
W1 = -25;			# in kJ
W2 = 45;			# in kJ
Q1 = 65;			# in kJ
Q2 = -40;			# in kJ

# Calculations
# del_U = Q - W and but for a cycle del_U = 0, So
# Q = W
# Q1 + Q2 +Q3 = W1 +W2
Q3 = W1 + W2 - Q1 - Q2;			# in kJ

# Results
print "Third Heat transfer in kJ is ",Q3
print "That is Third Heat transfer is of ",abs(Q3)," kJ from the fluid"

Third Heat transfer in kJ is  -5
That is Third Heat transfer is of  5  kJ from the fluid

## Example 4.5 Page No : 98¶

In [3]:
# Variables
m = 1.5;			# in kg
T1 = 90;			# in °C
T1 = T1 + 273;			#in K
T2 = 225;			# in °C
T2 = T2 + 273;			# in K
C_p = 0.24;
C_v = 0.17;

# Calculations
Q = (m * C_p * (T2-T1));			# in kcal
del_U = (m * C_v * (T2-T1));			# in kcal
W = Q - del_U;			# in kcal

# Results
print "The external work done in kcal is : %.1f"%W

The external work done in kcal is : 14.2

## Example 4.7 Page No : 98¶

In [1]:
import math

# Variables
v1 = 0.5;			# in m**3
v2 = 0.125;			# in m**3
p1 = 1.5;			# in bar
p1 = p1 * 10**5;			# in N per m**2
p2 = 9.;			#in bar
p2 = p2 * 10**5;			# in N per m**2
T1 = 100.;			# in °C
T1 = T1 + 273;			# in K
R = 8.31;

# Calculations and Results
# Formula p1*v1= n*R*T1
n= p1*v1/(R*T1);			# in mole
print "Mass of gas in mole is : %.2f"%n

# Part (b)
# Formula p1*v1/T1 = p2*v2/T2
T2 = (p2 * v2 * T1)/(p1 * v1);			# in K
print "Temperature at the end of compression in °C is : %.1f"%(T2-273)

# Part (c)
# Formula p1*v1**n = p2*v2**n
n1= math.log(p2/p1)/math.log(v1/v2)
print "Value of index n of compression is :",round(n1,1)

# Part (d)
F = 3;
C_v =1./2*R*F;
del_U = (n * C_v * (T2-T1));			# in J
print "Increase in internal energy of gas in kJ is : %.1f"%(del_U*10**-3)

# Part(e)
Gamma = 1.67;
Q_12 = n*(Gamma-n1)/(1-n1)*R*(T2-T1)/(Gamma-1);			# in J
Q_12 = Q_12 * 10**-3;			        # in kJ
print "Heat interaction in kJ is : %.2f"%Q_12

if Q_12<0:

# Note: There is some difference between the answer of book and coding . Both the answer is right but accurate answer is of coding.
# Because in the book, the intermediate values taken are appox. and in the coding the values taken are accurate

Mass of gas in mole is : 24.20
Temperature at the end of compression in °C is : 286.5
Value of index n of compression is : 1.3
Increase in internal energy of gas in kJ is : 56.2
Heat interaction in kJ is : -72.24

## Example 4.8 Page No : 99¶

In [10]:
import math

# Variables
p1 = 0.01;			# in N/mm**2
p1 = p1 * 10**3;			# in kN/m**2
p2 = 50.;			# in kN/m**2
v1 = 5.;			# in m**3
v2 = 1.5;			# in m**3
Gamma = 1.4;

# Calculations and Results
# Formula p1*v1**n = p2*v2**n
n= round(math.log(p2/p1)/math.log(v1/v2),2)
print "Part (a)    The value of n is : %.2f"%n
print ("The process followed during air compression is POLYTROPIC");

# Part (b)
print "Part (b)    The law of the process is %.2f p*v**"%n," = constant"

# Part (c)
W= (p1*v1-p2*v2)/(n-1);			# in kNm or (kJ)
print "Part (c)    Work done during the process in kJ is : %.1f"%W
print ("The -ve sign indicates that the work has been done on the system")

# Part (d)
Q = ((Gamma - n)/(Gamma - 1) * W);			# in kJ
print "Part (d)    Heat transfer during the process in kJ is : %.3f"%Q
print ("The -ve sign indicates that the heat is rejected from the system")

Part (a)    The value of n is : 1.34
The process followed during air compression is POLYTROPIC
Part (b)    The law of the process is 1.34 p*v**  = constant
Part (c)    Work done during the process in kJ is : -73.5
The -ve sign indicates that the work has been done on the system
Part (d)    Heat transfer during the process in kJ is : -11.029
The -ve sign indicates that the heat is rejected from the system

## Example 4.9 Page No : 101¶

In [1]:
from scipy.integrate import quad

# Variables
# Relation of specific internal energy of the gas
# U= 1.5*p*v-85 kJ/kg
p1 = 1000;			# in kpa
p2 = 200;			# in pa
v1 = 0.20;			# in m**3
v2 = 1.20;			# in m**3
m = 1.5;			# in kg

# Calculations and Results
U1= 1.5*p1*v1-85;			#  kJ/kg
U2= 1.5*p2*v2-85;			#  kJ/kg
delU= U2-U1;	    		# in kJ
print "Change in internal energy in kJ is",delU
# p1= a+b*v1          (i)
# p2= a+b*v2          (ii)
# From eq(i) and (ii)
b= (p1-p2)/(v1-v2);			# in kN/m**2
a= p1-b*v1;     			# in kN/m**2
print "The value of a in kN/m**2 is ",a
print "The value of b in kN/m**2 is ",b

# Part (c)
# Work done = integration of p w.r.t. v and p = a+b*v1

def f5(v):
return a+b*v

print "Work done in kJ is :",W

# Part (d)
Q= delU+W;			# in kJ
print "The net heat transfer in kJ is : ",Q

Change in internal energy in kJ is 60.0
The value of a in kN/m**2 is  1160.0
The value of b in kN/m**2 is  -800.0
Work done in kJ is : 600.0
The net heat transfer in kJ is :  660.0

## Example 4.10 Page No : 102¶

In [2]:
# Variables
a= 1160.;			# in kN/m**2
b= -800.;			# in kN/m**2

# Calculations
v= -a/(2*b)
Umax= 1.5*(a*v+b*v**2)-85;			# in kJ/kg
# For 1.5 kg mass of gas it is
Umax= Umax*1.5;			# in kJ/kg

# Results
print "The maximum internal energy of the gas in kJ/kg is : ",Umax

The maximum internal energy of the gas in kJ/kg is :  818.625

## Example 4.11 Page No : 103¶

In [8]:
# Exa 4.11
import math

# Variables
T1 = 127.;			# in °C
T1 = T1 + 273;			# in K
R = 287.;
V1 = 300.;			# in m/s
p1 = 2.;			# in MPa
p2 = 0.5;			# in MPa
p1 = p1 * 10**6;			# in Pa
p2 = p2 * 10**6;			# in Pa
C_P = 1.005*10**3;			# in J/ kg-K
Gamma = 1.4;

# Calculations and Results
V2 = math.sqrt(2 * C_P *T1 *(1-(p2/p1)**((Gamma-1)/Gamma)) + V1**2);			# in m/s
print "The exit velocity of air in m/s is : %.3f"%V2

m = 600.;			# in kg/hr
m = m / 3600;			# in kg/sec
v1 = (R * T1)/p1;			# in m**3 per kg
# m = (A1*V1)/v1 = (A2* V2)/v2
A1 = (m * v1)/V1;			# in m**2
A1 = A1 * 10**6;			# in mm**2
print "Inlet area of the nozzle in square milimeter is : %.2f"%A1

T2 = T1*(p2/p1)**((Gamma-1)/Gamma);			# in K
v2 = (R * T2)/(p2);			# in m**3/kg
A2 = (m * v2)/V2;			# in m**2
A2 = A2 * 10**6;			# in mm**2
print "Exit area of nozzel in square milimeter is : %.2f"%A2

The exit velocity of air in m/s is : 594.094
Inlet area of the nozzle in square milimeter is : 31.89
Exit area of nozzel in square milimeter is : 43.35

## Example 4.17 Page No : 106¶

In [13]:
# Variables
W = -1;			# in kWh
W = W * 10**3 * 3600;			# in J
del_U = -5000;			# in kj

# Calculations
del_U = del_U * 10**3;			# in J
Q = del_U + W;			# in J
Q = Q * 10**-6;			# in MJ

# Results
print "Net heat transfer for the system in MJ is : ",Q

Net heat transfer for the system in MJ is :  -8.6

## Example 4.20 Page No : 107¶

In [2]:
# Variables
Q_acb = 84;			#in kJ
W_acb = 32;			# in kJ

# Calculations and Results
#Formula Q_acb = del_U+W_acb where del_U = U_b - U_a;
del_U = Q_acb - W_acb;			# in kJ
# Part (a) Path a b d
W_abd = 10.5;			# in kJ
Q_abd = del_U + W_abd;			# in kJ
print "Heat flows into the system along the path a b d in kJ is : ",Q_abd

# Part (b) curved path b a
W_ba = -(21);			# in kJ
Q_ba = -(del_U) + W_ba;			# in kJ
print "Heat liberated by the system in kJ is : ",(Q_ba)

# Part (c) process a b and d b
W_ad = 10.5;			# in kJ
del_U1 = 42;			# in kJ

W_db = -(42);			# in kJ
del_U2 = 52;			# in kJ
Q_bd = del_U2 + W_db;			# in kJ
print "Heat absorbed in processes bd in kJ is : ",Q_bd

W_db = 0;
W_abd = W_ad + W_db;			# in kJ
print "Heat absorbed in processes in ad and db in kJ is : ",W_abd

Heat flows into the system along the path a b d in kJ is :  62.5
Heat liberated by the system in kJ is :  -73
Heat absorbed in processes ad  in kJ is :  52.5
Heat absorbed in processes bd in kJ is :  10
Heat absorbed in processes in ad and db in kJ is :  10.5

## Example 4.21 Page No : 108¶

In [11]:
# Variables
v1 = 5.;			# in m**3
p1 = 2.;			# in bar
p2 = 6.;			# in bar
p3 = 2.;			# in bar

# Calculations
p1 = p1 * 10**5;			# in N/m**2
p2 = p2 * 10**5;			# in N/m**2
p3 = p3 * 10**5;			# in N/m**2
n = 1.3;
v2 = v1 * ((p1/p2)**(1/1.3));			# in m**3
W1_2 = ((p2 * v2)-(p1 * v1))/(1-n);			# in J
Gamma = 1.4;
v3 = v2 * ((p2/p3)**(1/Gamma));			# in m**3
W2_3 = ((p3 * v3) - (p2 * v2))/(1-Gamma);			# in J
W_net = W1_2 + W2_3;			# in J
W_net = W_net * 10**-3;			# in kJ

# Results
print "net work done in kJ is : %.2f"%W_net

net work done in kJ is : -94.02


text:
# Algebra Examples

Find Pivot Positions and Pivot Columns
Perform the row operation on (row ) in order to convert some elements in the row to .
Replace (row ) with the row operation in order to convert some elements in the row to the desired value .
Replace (row ) with the actual values of the elements for the row operation .
Simplify (row ).
Perform the row operation on (row ) in order to convert some elements in the row to .
Replace (row ) with the row operation in order to convert some elements in the row to the desired value .
Replace (row ) with the actual values of the elements for the row operation .
Simplify (row ).
Perform the row operation on (row ) in order to convert some elements in the row to .
Replace (row ) with the row operation in order to convert some elements in the row to the desired value .
Replace (row ) with the actual values of the elements for the row operation .
Simplify (row ).
Perform the row operation on (row ) in order to convert some elements in the row to .
Replace (row ) with the row operation in order to convert some elements in the row to the desired value .
Replace (row ) with the actual values of the elements for the row operation .
Simplify (row ).
Perform the row operation on (row ) in order to convert some elements in the row to .
Replace (row ) with the row operation in order to convert some elements in the row to the desired value .
Replace (row ) with the actual values of the elements for the row operation .
Simplify (row ).
Perform the row operation on (row ) in order to convert some elements in the row to .
Replace (row ) with the row operation in order to convert some elements in the row to the desired value .
Replace (row ) with the actual values of the elements for the row operation .
Simplify (row ).
Pivot columns are the columns, which contains pivot positions, so those pivot columns are.
A pivot position in a matrix is a position that after row reduction contains a leading . Thus, the leading one in the pivot columns are the pivot positions.
The leading in the pivot columns are the pivot positions
Enter YOUR Problem

Enter the email address associated with your Mathway account below and we'll send you a link to reset your password.

Please enter an email address
Please enter a valid email address
The email address you entered was not found in our system
The email address you entered is associated with a Facebook user
We're sorry, we were unable to process your request at this time

### Mathway Premium

Step-by-step work + explanations
•    Step-by-step work
•    Detailed explanations
•    No advertisements
•    Access anywhere
Access the steps on both the Mathway website and mobile apps
$--.--/month$--.--/year (--%)

### Mathway Premium

Visa and MasterCard security codes are located on the back of card and are typically a separate group of 3 digits to the right of the signature strip.

American Express security codes are 4 digits located on the front of the card and usually towards the right.
This option is required to subscribe.
Go Back

Step-by-step upgrade complete!

Mathway requires javascript and a modern browser.