text:
## #464 Reflow and LaTeX

v1.24
closed-fixed
nobody
5
2013-04-21
2010-04-05
chris
No

When using 'Reflow lines/block' on LaTeX code, geany will break commands/code at the wrong point, resulting a different output than desired when the TeX code is compiled into a document.

For example a command like X$_y$ at the end of a line will be broken between the X and \$, putting a space in between X and its subscript y when compiled. Another example is if you want an unbreakable space (~ in LaTeX code) between some text and a command (lets say \ref{}) like Fig.~\ref{}, geany will break it between the ~ and \, resulting in two spaces (one of which is breakable) in the final document.

When breaking lines as you type (Document -> Line Breaking) the commands will be broken at the correct points, this problem lies soley with the 'Reflow lines/block' function.

This was tested on geany 0.18.1 (built on Feb 15 2010 with GTK 2.18.6, GLib 2.22.4) in Arch Linux.

## Discussion

• Lex Trotman - 2012-09-14
• status: open --> open-postponed

• Lex Trotman - 2012-09-14

Line breaking simply uses whitespace to decide breaks and will not break within a sequence, reflow will.

Neither is aware of any semantics of the content other than whitespace.

There is a pull request under way to make them operate more exactly the same, should be reviewed after that is accepted or rejected

• Lex Trotman - 2013-04-21
• status: open-postponed --> closed-fixed

• Lex Trotman - 2013-04-21

The change has been applied to Git, both reflow and line breaking should now only break on whitespace, and at the same place.

• Colomban Wendling - 2013-04-21
• milestone: --> v1.24


text:
# Use the Pythagorean Theorem, what is the length of the hypotenuse in a right triangle whose legs are 3 and 4?

##### 1 Answer
Aug 3, 2016

5 units. This is a very famous triangle.

#### Explanation:

If $a , b$ are the lehs of a right triangle and $c$ is the hypoteneuse, then the Pythagorean Theorem gives:

${c}^{2} = {a}^{2} + {b}^{2}$

Then since the side lengths are positive:

$c = \setminus \sqrt{{a}^{2} + {b}^{2}}$

Put in $a = 3 , b = 4$:

$c = \setminus \sqrt{{3}^{2} + {4}^{2}}$
$= \setminus \sqrt{25} = 5$.

The fact that a triangle with sides of 3, 4, and 5 units is a right triangle has been known since at keast the ancient Egyptians. This is the Egyptian triangle, believed to be used by the ancient Egyptians to construct right angles -- for example, in the Pyramids (https://nrich.maths.org/982).

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# Reference request: concave/convex envelope

I'm seeking the references concerning on the regularity analysis of concave envelopes, i.e. given some measurable function $f:\mathbb R^d\to\mathbb R$ that is bounded from above ($d=1$ or $d\ge 1$), denote by $f_{\#}$ its concave envelope, i.e.

1. $f_{\#}:\mathbb R^d\to\mathbb R$ is concave and $f_{\#}\ge f$;
2. For any concave function $g:\mathbb R^d\to\mathbb R$ s.t. $g\ge f$, one has $g\ge f_{\#}$.

Assume that $d=1$. I'm mainly interested in the boundedness of the (left/right) derivative $f_{\#}'$. Under what kind of conditions on $f$, we can show that $f_{\#}'$ is uniformly bounded? Any related reference is welcome! Thanks a lot!

• It depends very much on each particular $f$. On $\mathbb{R}$ we should first require that concave envelope exists: the graph of $f$ should be below a linear function $f(x)\leq Ax+B$. Besides, we should require $f$ is uniformly Lipschitz in order to expect that the envelope is uniformly Lipschitz. I believe these 2 conditions will be also sufficient. Nov 7, 2016 at 15:25
• The paper: "The Dirichlet Problem for the Degenerate Monge--Amp`ere equation", L. Caffarelli, L. Nirenberg, J. Spruck, studies regularity properties of the concave envelopes for obstacles $f$ given on the boundary of a convex set. Nov 7, 2016 at 15:34

Thanks Paata Ivanisvili, BD. and Dirk for their reply and I'd like to summarize what I'm thinking about this issue. I claim that it is not an answer.

Assume that $f$ is bounded from above and $L-$Lipschitz. To simplify assume further $d=1$ and $f_{\#}$ is differentiable. Then one has $\|f_{\#}'\|\le L$. If not, assume that there exists $x_0$ s.t. $f_{\#}'(x_0)=M>L$, then $f_{\#}'(x)\ge M>L$ for all $x\le x_0$, and thus

$$\limsup_{x\to-\infty}\big(f_{\#}(x)-f(x)\big)~~=~~-\infty.$$

If otherwise there exists $x_0$ s.t. $f_{\#}'(x_0)<-L$, one has as well

$$\limsup_{x\to+\infty}\big(f_{\#}(x)-f(x)\big)~~=~~-\infty.$$

These two situations contradict the definition of $f_{\#}$.

I look forward to the further analysis on the regularity of $f_{\#}$, once $f$ is supposed bo be very "nice". Finally, I find the following paper which may be useful for who is also interested in this issue:

https://people.math.ethz.ch/~afigalli/papers/Optimal%20regularity%20of%20the%20convex%20envelope.pdf

• Actually, the Caferelli et al paper is available via researchgate (google-scholar it and check "all 4 versions").
– Dirk
Nov 7, 2016 at 16:23

I think the Benson's paper can be a good reference.

You can find it here:


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# RS Aggarwal solutions for Secondary School Class 10 Maths chapter 18 - Area of Circle, Sector and Segment [Latest edition]

## Chapter 18: Area of Circle, Sector and Segment

Exercise 18AExercise 18BMultiple Choice QuestionsFormative Assessment
Exercise 18A [Pages 18 - 824]

### RS Aggarwal solutions for Secondary School Class 10 Maths Chapter 18 Area of Circle, Sector and Segment Exercise 18A [Pages 18 - 824]

Exercise 18A | Q 1 | Page 819

The difference between the circumference and radius of a circle is 37cm. Using pi = 22/7 , find the circumference of the circle.

Exercise 18A | Q 2 | Page 819

The circumference of a circle is 22 cm. Find the area of its quadrant.

Exercise 18A | Q 3 | Page 819

What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm?

Exercise 18A | Q 4 | Page 819

If the area of the circle is numerically equal to twice its circumference then what is the diameter of the circle?

Exercise 18A | Q 5 | Page 819

What is the perimeter of a square which circumscribes a circle of radius a cm?

Exercise 18A | Q 6 | Page 819

Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.

Exercise 18A | Q 7 | Page 819

Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm.

Exercise 18A | Q 9 | Page 819

Find the perimeter of a semicircular protractor whose diameter is 14 cm.

Exercise 18A | Q 10 | Page 819

Find the radius of a circle whose perimeter and area are numerically equal.

Exercise 18A | Q 11 | Page 819

The radii of two circles are 19 cm and 9 cm, Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Exercise 18A | Q 12 | Page 819

The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Exercise 18A | Q 13 | Page 819

Find the area of the sector of a circle having radius 6 cm and of angle 30°. [Take π = 3.14]

Exercise 18A | Q 14 | Page 820

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.

Exercise 18A | Q 15 | Page 820

The circumferences of two circles are in the ratio 2: 3. What is the ratio between their areas?

Exercise 18A | Q 16 | Page 820

The areas of two circles are in the ratio 4: 9. What is the ratio between their circumferences?

Exercise 18A | Q 17 | Page 820

A square is inscribed in a circle. Find the ratio of the areas of the circle and the square.

Exercise 18A | Q 18 | Page 820

The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.

Exercise 18A | Q 19 | Page 820

A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length. Find the length of the pendulum.

Exercise 18A | Q 20 | Page 18

The minute hand of a clock is 15 cm long. Calculate the area swept by it in 20 minutes.

Exercise 18A | Q 21 | Page 820

A sector of 56°, cut out from a circle, contains 17.6 cm2. Find the radius of the circle.

Exercise 18A | Q 23 | Page 820

The perimeter of a certain sector of a circle of radius 6.5 cm in 31 cm. Find the area of the sector.

Exercise 18A | Q 24 | Page 820

The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44 cm in length.

Exercise 18A | Q 25 | Page 820

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimensions 14 cm ⨯ 7 cm. Find the area of the remaining cardboard.

Exercise 18A | Q 26 | Page 820

In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. [π = 3.14]

Exercise 18A | Q 27 | Page 820

From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semicircular portion with BC as diameter is cut off. Find the area of the remaining paper.

Exercise 18A | Q 28 | Page 821

In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region.

Exercise 18A | Q 29 | Page 821

In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region.

Exercise 18A | Q 30 | Page 821

In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm with centre O. If ∠POQ = 30°, find the area of the shaded region.

Exercise 18A | Q 31 | Page 821

In the given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Exercise 18A | Q 32 | Page 821

In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and ∠AOB = 90°. If AO = OB = 42 cm,· then find the perimeter of the top of the table.

Exercise 18A | Q 33 | Page 821

In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm. Find the area of shaded region.

Exercise 18A | Q 34 | Page 822

In the given figure, OABC is a quadrant of a circle of radius 3.5 cm with centre O. If OD = 2 cm, find the area of the shaded portion.

Exercise 18A | Q 35 | Page 822

Find the perimeter of the shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles.

Exercise 18A | Q 36 | Page 822

In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square.

Exercise 18A | Q 37 | Page 822

In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the

1. perimeter
2. area of the shaded region.
Exercise 18A | Q 38 | Page 822

In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [Use π= 3.14]

Exercise 18A | Q 39 | Page 823

In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. [Use π = 3.14.]

Exercise 18A | Q 40 | Page 823

In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.

Exercise 18A | Q 41 | Page 823

Find the area of a quadrant of a circle whose circumference is 44 cm.

Exercise 18A | Q 42 | Page 823

In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circles are of the same diameter.

Exercise 18A | Q 43 | Page 823

Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle.

Exercise 18A | Q 44 | Page 823

A wire is bent to form a square enclosing an area of 484 cm2. Using the same wire, a circle is formed. Find the area of the circle.

Exercise 18A | Q 45 | Page 823

A square ABCD is inscribed in a circle of radius r. Find the area of the square.

Exercise 18A | Q 46 | Page 824

The cost of fencing a circular field at the rate of Rs 25 per metre is Rs 5500. The field is to be ploughed at the rate of 50 paise per m2 . Find the cost of ploughing the field. [Take π =22/7].

Exercise 18A | Q 47 | Page 824

A park is in the form of a rectangle 120 m by 90 m. At the centre of the park there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950 m2. Find the radius of the circular lawn.

Exercise 18A | Q 48 | Page 824

In the given figure, PQRS represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed.

Exercise 18A | Q 49 | Page 824

In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is drawn. If AC = 54 cm and BC = 10, find the area of the shaded region.

Exercise 18A | Q 50 | Page 824

From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.

Exercise 18A | Q 51 | Page 824

Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.

Exercise 18A, Exercise 18B [Pages 831 - 836]

### RS Aggarwal solutions for Secondary School Class 10 Maths Chapter 18 Area of Circle, Sector and Segment Exercise 18A, Exercise 18B [Pages 831 - 836]

Exercise 18A | Q 1 | Page 831

The circumference of a circle is 39.6 cm. Find its area.

Exercise 18A | Q 2 | Page 831

The area of a circle is 98.56 cm2. Find its circumference.

Exercise 18A | Q 3 | Page 831

The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle.

Exercise 18A | Q 4 | Page 831

A copper wire when bent in the form of a square encloses an area of 484 cm2. The same wire is not bent in the form of a circle. Find the area enclosed by the circle.

Exercise 18A | Q 5 | Page 831

A wire when bent in the form of an equilateral triangle encloses an area of 121sqrt(3)  "cm"^2. The same wire is bent to form a circle. Find the area enclosed by the circle.

Exercise 18A | Q 6 | Page 831

The length of a chain used as the boundary of a semicircular park is 108 m. Find the area of the park.

Exercise 18A | Q 7 | Page 831

The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumference of the circles.

Exercise 18A | Q 8 | Page 831

Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.

Exercise 18A | Q 9 | Page 831

A path of 8 m width runs around the outsider of a circular park whose radius is 17 m. Find the area of the path.

Exercise 18A | Q 10 | Page 831

A race track is in the form of a rig whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.

Exercise 18A | Q 11 | Page 831

A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of the arc and the area of the sector.

Exercise 18A | Q 12 | Page 831

The area of the sector of a circle of radius 10.5 cm is 69.3 cm2. Find the central angle of the sector.

Exercise 18A | Q 13 | Page 831

The length of an arc of a circle, subtending an angle of 54° at the centre, is 16.5 cm. Calculate the radius, circumference and area of the circle.

Exercise 18A | Q 14 | Page 831

The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major segments.

Exercise 18A | Q 15 | Page 831

Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment.

Exercise 18A | Q 16 | Page 832

A chord 10 cm long is drawn in a circle whose radius is  5sqrt(2) cm. Find the areas of both the segments.

Exercise 18A | Q 17 | Page 832

Find the areas of both the segments of a circle of radius 42 cm with central angle 120°.

Exercise 18A | Q 18 | Page 832

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor major segments.

Exercise 18A | Q 19 | Page 832

In a circle of radius 10.5 cm, the minor arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc.

Exercise 18B | Q 20 | Page 832

The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days.

Exercise 18B | Q 21 | Page 832

Find the area of a quadrant of a circle whose circumference is 88 cm.

Exercise 18B | Q 22 | Page 832

A rope by which a cow is tethered is increased from 16 m to 23 m. How much additional ground does it have now graze?

Exercise 18B | Q 23 | Page 832

A horse is placed for grazing inside a rectangular field 70 m by 52 m. It is tethered to one corner by a rope 21 m long. On how much area can it graze? How much area is left ungrazed?

Exercise 18B | Q 24 | Page 832

A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Write the answer correct to 2 places of decimal.

Exercise 18B | Q 25 | Page 832

Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed?

Exercise 18B | Q 26 | Page 832

In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is 32sqrt(3), find the radius of the circle.

Exercise 18B | Q 27 | Page 832

The side of a square is 10 cm. Find the area of the inscribed circle [π = 3.14]

Exercise 18B | Q 27 | Page 832

The side of a square is 10 cm. Find the area of the circumscribed circle. [π = 3.14]

Exercise 18B | Q 28 | Page 832

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Exercise 18B | Q 29 | Page 832

The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle.

Exercise 18B | Q 30 | Page 832

The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in a 19.8-km-long journey?

Exercise 18B | Q 31 | Page 833

The wheels of the locomotive of a train are 2.1 m in radius. They make 75 revolutions in one minute.

Exercise 18B | Q 32 | Page 833

The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.

Exercise 18B | Q 33 | Page 833

A boy is cycling in such a way that the wheels of his bicycle are making 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed (in km/h) at which the boy is cycling.

Exercise 18B | Q 34 | Page 833

The diameter of the wheels of a bus is 140 cm. How many revolutions per minute do the wheels make when the bus is moving at a speed of 72.6 km per hour?

Exercise 18B | Q 35 | Page 833

The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers is 800 revolutions.

Exercise 18B | Q 36 | Page 833

Four equal circles are described about the four corners of a square so that each touches two of the others, as shown in the figure. Find the area of the shaded region, if each side of the square measures 14 cm.

Exercise 18B | Q 37 | Page 833

Four equal circles, each of radius 5 cm, touch each other, as shown in the figure. Find the area included between them.

Exercise 18B | Q 38 | Page 833

Four equal circles, each of radius a units, touch each other. Show that the area between them is (6/7"a"^2) sq units.

Exercise 18B | Q 39 | Page 833

Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area of enclosed between them.

Exercise 18B | Q 40 | Page 833

If three circles of radius a each, are drawn such that each touches the other two, prove that the area included between them is equal to 4/25"a"^2.

Exercise 18B | Q 41 | Page 833

In the given figure, ABCD is a trapezium of area 24.5 cm2 , If AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region.

Exercise 18B | Q 42 | Page 834

ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m. Find the following:

1.  total area of the four sectors,
2. area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.

Exercise 18B | Q 42 | Page 834

In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside' the region. Find the area of the shaded region. [Use π = 3.14]

Exercise 18B | Q 43 | Page 834

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made.

Exercise 18A | Q 44 | Page 834

In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED = 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region.

Exercise 18B | Q 46 | Page 834

In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of shaded region.

Exercise 18B | Q 47 | Page 835

In the given figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region.
[Use sqrt(3)= 1.73, pi = 3.14]

Exercise 18B | Q 48 | Page 835

On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design. ["Use" sqrt(3) = 1.73, pi =22/7]

Exercise 18B | Q 49 | Page 835

The perimeter of the quadrant of a circle is 25 cm. Find its area.

Exercise 18B | Q 50 | Page 835

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.]

Exercise 18B | Q 51 | Page 835

The radius of a circular garden is 100 m. There is a road 10 m wide, running all around it. Find the area of the road and the cost of levelling it at Rs 20 per m2. [Use π = 3.14]

Exercise 18B | Q 52 | Page 835

The area of an equilateral triangle is 49sqrt(3) cm2 . Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. [Take π = 1.73]

Exercise 18B | Q 53 | Page 835

A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure.

Exercise 18B | Q 54 | Page 835

A circular disc of radius 6 cm is divided into three sectors with central angles 90°,120° and 150°. What part of the whole circle is the sector with central angle 150°? Also, calculate the ratio of the areas of the three sectors.

Exercise 18B | Q 55 | Page 835

A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35 cm, then find the total area of the design.

Exercise 18B | Q 56 | Page 836

In the given figure, PQ = 24, PR = 7 cm and O is the centre of the circle. Find the area of the shaded region.

Exercise 18B | Q 57 | Page 836

In the given figure, ∆ABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of ∆ABC.

Exercise 18B | Q 58 | Page 836

In the given figure, ∆ABC is right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3 cm and AC = 4 cm. Find the area of the shaded region.

Exercise 18B | Q 59 | Page 836

PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS = 12 cm, find the perimeter and area of the shaded region.

Exercise 18B | Q 60 | Page 836

The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track.

Multiple Choice Questions [Pages 846 - 848]

### RS Aggarwal solutions for Secondary School Class 10 Maths Chapter 18 Area of Circle, Sector and Segment Multiple Choice Questions [Pages 846 - 848]

Multiple Choice Questions | Q 1 | Page 846

The area of a circle is 38.5 cm2. The circumference of the circle is

• 6.2 cm

• 12.1 cm

• 11 cm

• 22 cm

Multiple Choice Questions | Q 2 | Page 846

The area of a circle is 49 π cm2. Its circumference is

• 7 π cm

• 14 π cm

• 21 π cm

• 28 π cm

Multiple Choice Questions | Q 3 | Page 846

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is

• 111 cm2

•  184 cm2

• 154 cm2

• 259 cm2

Multiple Choice Questions | Q 4 | Page 847

The perimeter of a circular field is 242 m. The area of the field is

• 9317 m2

• 18634 m2

• 4658.5 m2

•  none of these

Multiple Choice Questions | Q 5 | Page 847

On increasing the diameter of a circle by 40%, its area will be increased by

• 40%

•  80%

• 96%

• 82%

Multiple Choice Questions | Q 6 | Page 847

On decreasing the radius of a circle by 30%, its area is decreased by

• 30%

• 60%

• 45%

• none of these

Multiple Choice Questions | Q 7 | Page 847

The area of a square is the same as the area of a square. Their perimeters are in the ratio

• 1 : 1

• 2 : π

• π : 2

• sqrt(pi):2

Multiple Choice Questions | Q 8 | Page 847

The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm. The radius of the new circle is

• 16 cm

• 28 cm

• 42 cm

• 56 cm

Multiple Choice Questions | Q 9 | Page 847

The area of circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is

• 25 cm

• 31 cm

• 50 cm

• 62 cm

Multiple Choice Questions | Q 10 | Page 847

If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is

• 4 : π

• π : 4

• π : 7

• 7 : π

Multiple Choice Questions | Q 11 | Page 847

If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then

• R1+R2 = R

• R1 + R2 < R

• "R"_1^2 + "R"_2^2 = < "R"^2

• "R"_1^2 + "R"_2^2 = "R"^2

Multiple Choice Questions | Q 12 | Page 847

If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then

• R1 + R2 = R

• R1 + R2 > R

• R1 + R2 < R

• none of these

Multiple Choice Questions | Q 13 | Page 847

If the circumference of a circle and the perimeter of a square are equal, then

• area of the circle = area of the square

• (area of the circle) > (area of the square)

• (area of the circle) < (area of the square)

• none of these

Multiple Choice Questions | Q 14 | Page 847

The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circles is

• 320 cm2

• 330 cm2

• 332 cm2

• 340 cm2

Multiple Choice Questions | Q 15 | Page 848

The areas of two concentric circles are 1386 cm2 and 962.5 cm2. The width of the ring is

•  2.8 cm

• 3.5 cm

• 4.2 cm

• 3.8 cm

Multiple Choice Questions | Q 16 | Page 848

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is

Multiple Choice Questions | Q 17 | Page 848

The areas of two circles are in the ratio 9 : 4. The ratio of their circumferences is

• 3 : 2

• 4 : 9

• 2 : 3

• 81 : 16

Multiple Choice Questions | Q 18 | Page 848

The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?

• 2800

• 4000

• 5500

•  7000

Multiple Choice Questions | Q 19 | Page 848

The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?

• 140

• 150

• 160

• 166

Multiple Choice Questions | Q 20 | Page 848

In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is

• 14 m

• 24 m

•  28 m

• 40 m

Multiple Choice Questions | Q 21 | Page 848

The area of the sector of angle θ° of a circle with radius R is

• (2pi"R"theta)/180

• (pi"R"^2theta)/180

• (2pi"R"theta)/360

• (pi"R"^2theta)/360

Multiple Choice Questions | Q 22 | Page 848

The length of an arc of the sector of angle θ° of a circle with radius R is

• (2pi"R"theta)/180

• (2pi"R"theta)/360

• (pi"R"^2theta)/180

• (pi"R"^2theta)/360

Multiple Choice Questions | Q 23 | Page 848

The length of the minute hand of a clock is 21 cm. The area swept by the minute hand in 10 minutes is

• 231 cm

• 210 cm

• 126 cm2

• 252 cm2

Multiple Choice Questions | Q 24 | Page 848

A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, π = 3.14) is

• 32.5 cm2

•  34.5 cm2

• 28.5 cm2

• 30.5 cm2

Multiple Choice Questions | Q 25 | Page 848

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is

• 21 cm

• 22 cm

• 18.16 cm

• 23.5 cm

Multiple Choice Questions | Q 26 | Page 848

In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre. If sqrt(3) = 1.73 then the area of the segment of the circle is

• 120.56 cm2

• 124.63 cm2

• 118.24 cm2

• 130.57 cm2

Formative Assessment [Pages 852 - 854]

### RS Aggarwal solutions for Secondary School Class 10 Maths Chapter 18 Area of Circle, Sector and Segment Formative Assessment [Pages 852 - 854]

Formative Assessment | Q 1 | Page 852

In the given figure, a square OABC has been inscribed in the quadrant OPBQ. If OA = 20 cm, then the area of the shaded region is

• 214 cm2

• 228 cm2

• 242 cm2

• 248 cm2

Formative Assessment | Q 2 | Page 852

The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?

• 200

• 250

• 300

• 350

Formative Assessment | Q 3 | Page 852

The area of a sector of a circle with radius r, making an angle of x° at the centre is

• x/180xx2pi"r"

• x/180xxpi"r"^2

• x/360xx2pi"r"

• x/360xxpi"r"^2

Formative Assessment | Q 4 | Page 852

In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If π = 3.14, then the area of the shaded region is

• 264 cm2

• 266 cm2

• 272 cm2

• 254 cm2

• None of the above.

Formative Assessment | Q 5 | Page 852

The circumference of a circle is 22 cm. Find its area.

Formative Assessment | Q 6 | Page 852

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.

Formative Assessment | Q 7 | Page 852

The minute hand of a clock is 12 cm long. Find the area swept by in it 35 minutes.

Formative Assessment | Q 8 | Page 852

The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.

Formative Assessment | Q 9 | Page 852

A chord of a circle of radius 14 cm a makes a right angle at the centre. Find the area of the sector.

Formative Assessment | Q 10 | Page 852

In the given figure, the sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.

Formative Assessment | Q 11 | Page 853

A wire when bent in the form of an equilateral triangle encloses an area of 121sqrt(3)" cm"^2. If the same wire is bent into the form of a circle, what will be the area of the circle?

Formative Assessment | Q 12 | Page 853

The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km per hour.

Formative Assessment | Q 13 | Page 853

OACB is a quadrant of a circle with centre O and its radius is 3.5 cm. If OD = 2 cm. find the area of

Formative Assessment | Q 14 | Page 853

In the given figure, ABCD is a square each of whose sides measures 28 cm. Find the area of the shaded region.

Formative Assessment | Q 15 | Page 853

In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region.

Formative Assessment | Q 16 | Page 853

The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.

Formative Assessment | Q 17 | Page 853

A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.

Formative Assessment | Q 18 | Page 853

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segments.

Formative Assessment | Q 19 | Page 854

Four cows are tethered at the four corners of a square field of side 50 m such that the each can graze the maximum unshared area. What area will be left ungrazed?

Formative Assessment | Q 20 | Page 854

A square tank has an area of 1600 cm2. There are four semicircular plots around it. Find the cost of turfing the plots at Rs 12.50 per m

## Chapter 18: Area of Circle, Sector and Segment

Exercise 18AExercise 18BMultiple Choice QuestionsFormative Assessment

## RS Aggarwal solutions for Secondary School Class 10 Maths chapter 18 - Area of Circle, Sector and Segment

RS Aggarwal solutions for Secondary School Class 10 Maths chapter 18 (Area of Circle, Sector and Segment) 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 Secondary School Class 10 Maths 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. RS Aggarwal textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Secondary School Class 10 Maths chapter 18 Area of Circle, Sector and Segment are Areas of Sector and Segment of a Circle, Area of Circle, Areas of Combinations of Plane Figures, Circumference of a Circle, Circumference of a Circle, Areas of Sector and Segment of a Circle, Area of Circle, Areas of Combinations of Plane Figures, Circumference of a Circle, Circumference of a Circle.

Using RS Aggarwal Class 10 solutions Area of Circle, Sector and Segment 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 RS Aggarwal Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer RS Aggarwal Textbook Solutions to score more in exam.

Get the free view of chapter 18 Area of Circle, Sector and Segment Class 10 extra questions for Secondary School Class 10 Maths and can use Shaalaa.com to keep it handy for your exam preparation


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# Each of the two positive integers X and Y has exactly 3 distinct prime

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Each of the two positive integers X and Y has exactly 3 distinct prime  [#permalink]

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21 Mar 2019, 22:33
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Each of the two positive integers X and Y has exactly 3 distinct prime factors. Least Common Multiple of X and Y has 5 distinct prime factors. X is divisible by 35 while Y is divisible by 33. Also X and Y are both even numbers, and neither of them is divisible by 4. On writing either X or Y in prime factorised form, it is observed that neither of them has a prime number with a power greater than 2. If the least common multiple of X & Y is denoted as N, then how many values of N are possible?

A. 16
B. 12
C. 10
D. 8
E. 4
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Posts: 7684
Re: Each of the two positive integers X and Y has exactly 3 distinct prime  [#permalink]

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22 Mar 2019, 07:45
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amanvermagmat wrote:
Each of the two positive integers X and Y has exactly 3 distinct prime factors. Least Common Multiple of X and Y has 5 distinct prime factors. X is divisible by 35 while Y is divisible by 33. Also X and Y are both even numbers, and neither of them is divisible by 4. On writing either X or Y in prime factorised form, it is observed that neither of them has a prime number with a power greater than 2. If the least common multiple of X & Y is denoted as N, then how many values of N are possible?

A. 16
B. 12
C. 10
D. 8
E. 4

So X is divisible by 35, that is 5*7, and X is even, so divisible by 2, thus $$X=2*5^{a}*7^{b}$$.. power of 2 is just 1 as the number is not divisible by 4...
Also, Y is divisible by 33, that is 3*11, and Y is even, so divisible by 2, thus $$X=2*3^{c}*11^{d}$$.. power of 2 is just 1 as the number is not divisible by 4...

LCM = $$2*3^c*5^a*7^b*11^d$$... Here a, b, c and d can take any of the two values, 1 or 2. Thus ways = 1*2*2*2*2=16

A
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Re: Each of the two positive integers X and Y has exactly 3 distinct prime   [#permalink] 22 Mar 2019, 07:45
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Area

# What is 23 square feet in square meters?

###### Wiki User

23 square feet = about 2.14 square meters

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## Related Questions

### 23 square feet is how much in square meters?

23 square feet = about 2.14 square meters. (2.13676992 square meters)

### How many meters of square feet would it be of 23 acres?

Meters of square feet makes no sense, but 23 acres = 1 001 880 square feet 23 acres = 93 077.6977 square meters

### What is an area of 23 feet by 8 feet in square meters?

23 x 8 feet is 184 square feet. This equates to approximately 17.1 (17.0941594) square meters.

### What is 2.8 squared meters in feet?

You can't convert square meters to feet. You can convert square meters to square feet, or meters to feet.

### 251 square meters to square feet calculate?

(251 square meters = 2,701.74151 square feet). Multiply square meters by 10.7639104 to get square feet.

### How do you convert square feet in meters?

To convert square feet to square meters, use this formula- square feet x 0.092 = square meters

### Convert 1500 square feet into meters?

The conversion formula for square feet to square meters is: square feet x 0.092 = square meters

### What is 120 square meters to feet?

120 square meters equates to 1,291.67 square feet.square meters x 10.7639 = square feet

### How many square meters is 2391 square feet?

2,391 square feet rounds to 222.13 square meters.To convert from square feet to square meters, divide square feet by 10.764.

### Square meters to square feet?

Multiply square meters by 10.764 to get square feet.

### What is 718 square feet into square meters?

718 square feet = about 66.7 square meters. (66.7043827 square meters)

### What is 6 square meters in feet?

Square meters is a measure of area and must be expressed in square feet, not simply feet. Six square meters is 64.6 square feet

### How big is 789.25 square meters in square feet?

789.25 square meters in square feet is: 8,495 square feet.

### What is 45 square meters in square feet?

45 square meters equates to about 484.38 square feet.Formula: square meters x 10.764 = square feet.

### Convert 300 square meters to square feet?

300 (square meters) = 3,229.17313 square feet. (multiply square meters by 10.7639104 to get square feet).

### 2459 square feet is how many square meters?

2,459 square feet = about 228.45 square meters.(228.448575 square meters).

### What is 16 square meters?

16 square meters is 172.2 square feet. Square meters x 10.76 = square feet.

### How many square meters in 565 square feet?

565 square feet = about 52.5 square meters. (52.4902176 square meters).

### How many feet are in 3.5 sq meters?

3.5 square meters is 37.6737 square feet.square meters x 10.7639 = square feet

## Still have questions?

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# coordinate geometry regarding circle

• September 22nd 2009, 03:35 AM
wintersoltice
coordinate geometry regarding circle
Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.

what i tried to do is:
$y=-2x+6 -----(1)$
let centre of circle be (x,y)
$(-2+x)^2+(0+y)^2=r^2$
$4-4x+x^2+y^2=r^2 ------(2)$
$(4+x)^2+(6+y)^2=r^2$
$16+8x+x^2+36+12y+y^2=r^2 -----(3)$
Subst. (1) and (2) into (3),
$x=10$
Subst. x=10 into (1),
$y=-14$
centre of circle is (10,-14)..

but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??
• September 22nd 2009, 04:27 AM
Chris L T521
Quote:

Originally Posted by wintersoltice
Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.

what i tried to do is:
$y=-2x+6 -----(1)$
let centre of circle be (x,y)
$(-2+x)^2+(0+y)^2=r^2$
$4-4x+x^2+y^2=r^2 ------(2)$
$(4+x)^2+(6+y)^2=r^2$
$16+8x+x^2+36+12y+y^2=r^2 -----(3)$
Subst. (1) and (2) into (3),
$x=10$
Subst. x=10 into (1),
$y=-14$
centre of circle is (10,-14)..

but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??

I would do it like this:

Let $(x_0,y_0)$ be a point on the line (in this case, also let it be the center of our circle).

Thus, the equation would be $(x-x_0)^2+(y-y_0)^2=r^2$

WLOG, let r be the distance from $(x_0,y_0)$ to $(-2,0)$: $r=\sqrt{(x_0+2)^2+y_0^2}$

So now it follows that $(x-x_0)^2+(y-y_0)^2=(x_0+2)^2+y_0^2$.

Now, if we plug in the other point, we have

$(4-x_0)^2+(6-y_0)^2=(x_0+2)^2+(y_0)^2\implies 16-8x_0+x_0^2+36-12y_0+y_0^2=x_0^2+4x_0+4+y_0^2$.

Simplifying, we have $48=12x_0+12y_0\implies x_0+y_0=4$

Recall that $(x_0,y_0)$ is on the line $2x+y=6\implies y_0=6-2x_0$.

Thus, $x_0+(6-2x_0)=4\implies -x_0+6=4\implies -x_0=-2\implies x_0=2$

Therefore, $y_0=6-2x_0=6-2(2)=6-4=2$.

Therefore, the center of the circle is $(2,2)$ and the equation now becomes $(x-2)^2+(y-2)^2=20$

Does this make sense?


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# Spin-polarization

## Non-spin-polarized Calculations

Setting of “nspin 1” in INPUT file means calculation with non-polarized spin. In this case, electrons with spin up and spin down have same occupations at every energy states, weights of bands per k point would be double.

## Collinear Spin Polarized Calculations

Setting of “nspin 2” in INPUT file means calculation with polarized spin along z-axis. In this case, electrons with spin up and spin down will be calculated respectively, number of k points would be doubled. Potential of electron and charge density will separate to spin-up case and spin-down case.

Magnetic moment Settings in STRU files are not ignored until “nspin 2” is set in INPUT file

When “nspin 2” is set, the screen output file will contain magnetic moment information. e.g.

 ITER   TMAG      AMAG      ETOT(eV)       EDIFF(eV)      DRHO       TIME(s)
GE1    4.16e+00  4.36e+00  -6.440173e+03  0.000000e+00   6.516e-02  1.973e+01

where “TMAG” refers to total magnetization and “AMAG” refers to average magnetization. For more detailed orbital magnetic moment information, please use Mulliken charge analysis.

## Noncollinear Spin Polarized Calculations

The spin non-collinear polarization calculation corresponds to setting “noncolin 1”, in which case the coupling between spin up and spin down will be taken into account. In this case, nspin is automatically set to 4, which is usually not required to be specified manually. The weight of each band will not change, but the number of occupied states will be double. If the nbands parameter is set manually, it is generally set to twice what it would be when nspin<4.

In general, non-collinear magnetic moment settings are often used in calculations considering SOC effects. When “lspinorb 1” in INPUT file, “nspin” is also automatically set to 4. Note: different settings for “noncolin” and “lspinorb” correspond to different calculations:

• noncolin=0 lspinorb=0 nspin<4 : Non-collinear magnetic moments and SOC effects are not considered.

• noncolin=0 lspinorb=0 nspin=4 : Actualy same as the above setting, but the calculation will be larger. So the setting is not recommended.

• noncolin=1 lspinorb=0 : Non-collinear magnetic moments are considered but SOC effects are not considered

• noncolin=0 lspinorb=1 : The SOC effect is considered but the magnetic moment is limited to the Z direction

• noncolin=1 lspinorb=1 : The SOC effect and non-collinear magnetic moment are both calculated.

## For the continuation job

• Continuation job for “nspin 1” need file “SPIN1_CHG” which is generated by setting “out_chg=1” in task before. By setting “init_chg file” in new job’s INPUT file, charge density will start from file but not atomic.

• Continuation job for “nspin 2” need files “SPIN1_CHG” and “SPIN2_CHG” which are generated by “out_chg 1” with “nspin 2”, and refer to spin-up and spin-down charge densities respectively. It should be note that reading “SPIN1_CHG” only for the continuation target magnetic moment job is not supported now.

• Continuation job for “nspin 4” need files “SPIN%s_CHG”, where %s in {1,2,3,4}, which are generated by “out_chg 1” with any variable setting leading to ‘nspin’=4, and refer to charge densities in Pauli spin matrixes. It should be note that reading charge density files printing by ‘nspin’=2 case is supported, which means only $$\sigma_{tot}$$ and $$\sigma_{z}$$ are read.


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In a triangle $A B C$, the altitude from $A$ meets the circumcircle again at $T$. Let $O$ be the circumcenter. The lines $O A$ and $O T$ intersect the side $B C$ at $Q$ and $M$, respectively. Prove that
$\frac{S_{A O C}}{S_{C M T}}=\left(\frac{\sin B}{\cos C}\right)^2$
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# Question 856a9

Feb 16, 2016

$b = {\text{0.05 mol kg}}^{- 1}$

#### Explanation:

The idea here is that you need to use the volume and the mass of the sample to determine the density of the solution. This will then allow you to determine how much solvent you have in the original sample.

So, the density of a solution tells you the mass of one unit of volume of that solution. Since you're dealing with milliliters, you can say that the density of the sample, which is the same as the density of the original solution, will tell you the mass of $\text{1 mL}$ of solution.

1 color(red)(cancel(color(black)("mL"))) * "10.55 g"/(10color(red)(cancel(color(black)("mL")))) = "1.055 g"

So, if $\text{1 mL}$ of solution has a mass of $\text{1.055 g}$, it follows that the density of the solution is ${\text{1.055 g mL}}^{- 1}$.

Use the density of the solution to figure out the mass of the original sample

500color(red)(cancel(color(black)("mL solution"))) * overbrace("1.055 g"/(1color(red)(cancel(color(black)("mL solution")))))^(color(purple)("density")) = "527.5 g"

You know that the original sample contains $\text{3.25 g}$ of iron(II) chloride, ${\text{FeCl}}_{2}$. Since the mass of the solution is equal to the mass of the solvent, which in your case is water, plus the mas of the solute, you can say that the initial sample also contains

${m}_{\text{solution" = m_"solvent" + m_"solute}}$

${m}_{\text{solvent" = "527.5 g" - "3.25 g" = "524.25 g water}}$

Next, use iron(II) chloride's molar mass to determine how many moles you'd get in that $\text{2.25-g}$ sample

3.25 color(red)(cancel(color(black)("g"))) * "1 mole FeCl"_2/(126.75color(red)(cancel(color(black)("g")))) = "0.02564 moles FeCl"_2

As you know, molality tells you how many moles of solute you get per kilogram of solvent.

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

Use the fact that

$\text{1 kg" = 10^3"g}$

to calculate the molality of your solution

b = "0.02564 moles"/(524.25 * 10^(-3)"kg") = color(green)("0.05 mol kg"^(-1))

The answer can only have one sig fig, since that's how many sig fig you have for the volumes of the original solution and of the sample.

ALTERNATIVE METHOD

You can also find the molality of the solution by calculating how many moles of solute you get in that $\text{10-mL}$ sample.

Since you get $0.02564$ moles of iron(II) chloride in the $\text{500-mL}$ sample, you can say that the $\text{10-mL}$ sample will contain

10color(red)(cancel(color(black)("mL"))) * "0.02564 moles Fe Cl"_2/(500color(red)(cancel(color(black)("mL")))) = "0.0005128 moles FeCl"_2

The mass of iron(II) chloride in this sample will be

10color(red)(cancel(color(black)("mL solution"))) * "3.25 g Fe Cl"_2/(500color(red)(cancel(color(black)("mL solution")))) = "0.0650 g FeCl"_2

The mass of water in this sample will be

${m}_{\text{water" = "10.55 g" - "0.0650 g" = "10.485 g}}$

The molality will once again be

b = "0.0005128 moles"/(14.485 * 10^(-3)"kg") = color(green)("0.05 mol kg"^(-1))#


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# A small ball of mass  is attached with upper end of a uniform straight rod of equal mass  and length  The rod is held vertical over a smooth horizontal surface as shown in fig. When the system is released, the lower end slips freely and the systems falls down. Assuming the initial position of the lower end to be origin and initially rod to be along -axis as shown in fig., the equation of trajectory of point  of the rod (  is at distance  from the lower end)is a) b) c) d)

## Question ID - 100141 :- A small ball of mass  is attached with upper end of a uniform straight rod of equal mass  and length  The rod is held vertical over a smooth horizontal surface as shown in fig. When the system is released, the lower end slips freely and the systems falls down. Assuming the initial position of the lower end to be origin and initially rod to be along -axis as shown in fig., the equation of trajectory of point  of the rod (  is at distance  from the lower end)is a) b) c) d)

3537

(b)

Since the floor is smooth, therefore no horizontal force is acting on the system. Only three forces are acting on it, weight of the ball, weight  of the rod and vertically upward reaction of the floor

Since the rod is released from rest and no horizontal force is acting on it, therefore centre of mass of the system will not displace horizontally; it means it will fall vertically downward

Since masses of the rod and ball are equal, therefore centre of mass ‘ ’ is at mid-point of centre of the rod and the ball; it means at distance  from the ball or 3/4 from the lower end

If at an instant the rod makes angle  with the vertical, then point  and the system will be as shown in the figure

Co-ordinates of the point  are

Using  we have

Next Question :
 In concave mirror the object is placed beyond C. Then the nature and size of the image formed is (a) Real, erect and diminished (b) Real, inverted and diminished (c) Virtual, erect and magnified (d) Real, inverted and magnified


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# Probabilitical Players

Sixteen players $$S_{1}, S_{2}, ..., S_{16}$$ play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength.

Find the probability that exactly one of the two players $$S_{1}$$ and $$S_{2}$$ is among the eight winners.

The probability is in the form $$\frac{a}{b}$$. Find a + b.


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# Using VirtualEnv with multiple Python versions on windows

I have python 2.7.6 and 3.4.0 on my machine. The 2.7 version is on my path. I would like to set up a virtualenv using 3.4. There are many postings on SO and elsewhere that suggest I do the following from a command prompt:

virtualenv -p c:\python34 myvirtualenv

but this does not work for me. The console session has administrator privilege and UAC is off, however I get a permissions problem:

F:\virtualenv>virtualenv -p c:\python34 myenv
Running virtualenv with interpreter c:\python34
Traceback (most recent call last):
File "c:\python27\scripts\virtualenv-script.py", line 9, in <module>
File "C:\Python27\lib\site-packages\virtualenv.py", line 779, in main
popen = subprocess.Popen([interpreter, file] + sys.argv[1:], env=env)
File "C:\Python27\lib\subprocess.py", line 709, in __init__
File "C:\Python27\lib\subprocess.py", line 957, in _execute_child
startupinfo)
WindowsError: [Error 5] Access is denied

I have also tried it specifically pointing to the 3.4 version of virtualenv, but without changing the path it ends up executing a mixed bag of 2.7 and 3.4 python files.

The only way I could find to set up my virtual environment is to change my path to 3.4, run virtualenv then reset my path to 2.7, which defeats the point of the python switch on virtualenv.

Thanks

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Better:

py -3.4 -m venv c:\path\to\wherever\you\want\it

If you don't have the py.exe launcher (but it should be installed) you can replace py -3.4 with c:\Python34\python.exe (assuming the default location)

This works because of the handy-dandy, Windows-versioningest, super nice runtime picker py.exe

By default, py.exe will be present on a Windows install (I think it comes with 2.7, I know it does with 3+). When you run py then it will search for some environment variables or you can override that with a specific verison number (in your case -2.7or -3.4) You can leave off the .4 and it will choose the "biggest" minor version number.

You can also use it to run Python scripts. If you put a hash-bang line at the top of your script #!python3 and call it py myscript.py then it will pick the correct version of Python to start with, by searching the first line of the script and searching for a version number.

This is cool, because you can put something like #!/usr/bin/env python3.4 in the top of your script and run it on Windows with py, or on linux by doing

$chmod +x myscript.py$ ./myscript.py

Pretty useful.

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That works, thanks. I'm not familiar with py. I'll have to look it up. –  marcp Apr 1 '14 at 18:53
@marcp, added some info (though you can find it all in the official docs, too) –  Wayne Werner Apr 1 '14 at 19:46

Under Windows you need to run:

virtualenv -p c:\python34.exe myvirtualenv

The .exe on the end makes all the difference.

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Had to play around with this for a while to get it right. Had Python2.7.9 installed (Windows 7), wanted to take the latest Python3 release for a spin. After installing Python3.4.3 I went to directory bar and created a virtual environment foo with this command:

virtualenv -p c:\Python34\python.exe foo

It took me quite some time to understand that I needed to install the Python3.4.3 interpreter in the 'normal' fashion, I originally thought it would be installed USING virtualenv. That was explained in this answer. I did not touch the pythonpath in Windows after installing Python3.4.3.

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text:
The electron configuration of barium (atomic number 56) can be written three different ways: - Full configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2, - Compressed (without all standard script numbers): 1s2 2s2p6 3s2p6d10 4s2p6d10 5s2p6 6s2. ELECTRON CONFIGURATION: 1: Hydrogen: 1s 1: 2: Helium: 1s 2: 3: Lithium … Seaborgium: Value is a guess based on periodic table trend. The Pauli Exclusion Principle stat… Of the first Ba excited state? has the maximum number of unpaired electrons, all with the same spin. Silicon is right below carbon in the Periodic Table, and we can write its full electron configuration, just for comparison, so if we start up here, with hydrogen, silicon has the electron configuration 1s2 2s2 2p6, so as a reminder, these are our S electrons because they're in S orbitals. 2. Electronic configuration, also called electronic structure, the arrangement of electrons in energy levels around an atomic nucleus.According to the older shell atomic model, electrons occupy several levels from the first shell nearest the nucleus, K, through the seventh shell, Q, farthest from the nucleus. It states that every orbital in a given subshell are singly occupied by electrons before a second electron is filled in an orbital. Not found any post match with your request, STEP 2: Click the link on your social network, Can not copy the codes / texts, please press [CTRL]+[C] (or CMD+C with Mac) to copy, How a small number of atoms can be joined and form completely different substances. The concept of electronic configuration has replaced the older concept of … Full electron configuration of barium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2. When added to 6s2, it is equivalent to the full electron configuration of Barium, when neutral. Hassium: Value is a guess based on periodic table trend. Therefore the Iron electron configuration will be 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6. Nevertheless, check the complete configuration and other interesting facts about Cesium that most people don't know. The first ten electrons of the sodium atom are the inner-shell electrons and the configuration of just those ten electrons is exactly the same as the configuration of the element neon $$\left( Z=10 \right)$$. Barium is metal from second group of periodic table of elements. know the electron configurations of atoms and ions up to Z = 36 in terms of levels and sub-levels (orbitals) s, p and d; know the meaning of the term ionisation energy. (5 points) Draw a block diagram showing all the components of an absorption spectrometer. Multiple-site search is loading. Copyright © 2020 Multiply Media, LLC. A Bohr diagram of lithium In atomic physics and quantum chemistry, the electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals. How many candles are on a Hanukkah menorah? Electron Configuration So, if we were to add all the electrons from the above electron configuration, 2+2+6+2+6+2+10+6+2+10+6+2+14+10+3=83. Boron is the fifth element with a total of 5 electrons. Electron Configurations are an organized means of documenting the placement of electrons based upon the energy levels and orbitals groupings of the periodic table.. 8 I sampel light Sunce Thome Choppen munichermalue Datecher Maybe add your school logo, work team or anything else to maker your paper look cool? What does contingent mean in real estate? has a filled valence shell of electrons. Githy.com. Notes on the Electron Configuration of particular elements: Dubnium: Value is a guess based on periodic table trend. This electron configuration is for an uncharged neon atom (neon's atomic number is 10.) What is the electron configuration of barium? Electrons always fill orbitals of lower energy first. Electron Configuration and Oxidation States of Barium. - Noble Gas form : [Xe] 6s2 (this works because [Xe] symbolizes the electronic configuration of Xenon, which is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6. Barium makes a 2+ ion by losing 2 electrons. Notes on the Electron Configuration of particular elements: Dubnium: Value is a guess based on periodic table trend. 3. 3. (4 points) What is the electron configuration of the Ba ground state? In writing the electron configuration for Boron the first two electrons will go in the 1s orbital. How to Write the Electron Configuration for Boron. has three electrons per orbital, each with identical spins. Barium has atomic number 56, it has 56 protons and 56 electrons. Bohrium: Value is a guess based on periodic table trend. Gelson Luz is a Mechanical Engineer, expert in welding and passionate about materials. Electron configuration of barium atom: ₅₆Ba … In the case of Cesium the abbreviated electron configuration is [Xe] 6s1. 2.3).Elements are classified into “blocks” according to the subshell that is being “filled” as the atomic number increases. Barium Complete Electron Configuration 1s2 2s2 2p6 3s2 3p6 4 s2 3 d10 4 p6 5 s2 4 d10 5 p6 6 s2 Abbreviated Electron Configuration [Xe] 6s2 Sources Found in barytine (BaSO4) and witherite (BaCO3), never found in pure form due to its reactivity. 10XX,52,11XX,17,12XX,7,13XX,4,15XX,16,3XXX,2,40XX,10,41XX,12,43XX,5,44XX,4,46XX,5,47XX,3,48XX,3,5XXX,23,6XXX,3,71XX,1,8XXX,22,92XX,5,93XX,1,94XX,4,98XX,2,ASTM,171,Atomic-Mass,327,Atomic-Number,436,Atomic-Radius,86,Atomic-Symbol,329,Atomic-Volume,94,Boiling-Point,94,CBS,6,Chemical-Elements,100,Chemical-Symbol,217,CMDS,13,Coefficient-of-Thermal-Expansion,85,Covalent-Radius,87,Crystal-Structure,109,CS,17,CVS,3,Density,309,Elastic-Modulus,30,Electrical-Conductivity,79,Electro-Affinity,87,Electron-Configuration,109,Electronegativity,102,Electrons-per-Shell,112,Enthalpy-of-Fusion,93,Enthalpy-of-Vaporization,95,Group-Number,218,HCS,14,Heat-of-Fusion,87,Heat-of-Vaporization,85,HMCS,16,Ionic-Radius,78,Ionization-Energy,102,Ionization-Potential,101,LCS,21,List,201,MCS,17,MDS,14,Melting-Point,96,MS,4,NCMDBS,6,NCMDS,31,NCS,2,NMDS,8,Oxidation-States,104,Period-Number,107,Properties,40,RCLS,1,RCS,16,RRCLS,3,RRCS,4,SAE,201,Site,2,SMS,5,Specific-Gravity,83,Specific-Heat,92,Specific-Weight,1,Tests,2,Thermal-Conductivity,105,Valence-Electrons,98. The electron configuration of Li is, 1s2 2s1. What is the electron configuration of barium. 1. has values greater than or equal to +1. Home Money Science & Tech U.S. World Environment Page-of-the-day Trivia Barium (Ba), Electron Configuration: [Xe] 6s 2 [Xe] 6s 2::: example of: Concise Form of Electron Configuration Notation. Hassium: Value is a guess based on periodic table trend. The electron configuration indicates that bromine has a total of 35 electrons. It will then will be isoelectronic with Xe so A) [Kr]5s 2 5p 6 B) [Kr]5s 2 4d 10 5p 6 6s 2 6p 2 C) [Kr]5s 2 4d 10 5p 6 6s 1 D) [Kr]5s 2 4d 10 5p 6 6s 2 E) [Kr]5s 2 4d 10 5p 6 7. Electron configurations of elements beyond hassium (element 108), including those of the undiscovered elements beyond oganesson (element 118), are predicted. Meitnerium: Value is a guess based on periodic table trend. Give the ground state electron configuration for the ion of Ba. Must be stored under kerosene to remain pure. The electron configuration of bromine is 1s2 2s2p6 3s2p6d10 4s2p5, which can be shortened to [Ar] 4s2 3d10 4p5. As an approximate rule, electron configurations are given by the Aufbau principle and the Madelung rule. Meitnerium: Value is a guess based on periodic table trend. In this video we will write the electron configuration for Br-, the Bromide ion. In order to maximize the total spin, the electrons in the orbitals that only contain on… For example, here is a simple electron configuration: 1s 2 2s 2 2p 6. A bromine atom has two electrons in its first energy level, eight electrons in its second, 18 electrons in its third and seven electrons in its fourth. Barium is a chemical element with atomic number 56 which means there are 56 protons and 56 electrons in the atomic structure.The chemical symbol for Barium is Ba. H #1s^1# He #1s^2# Li #1s^2 2s^1# Be #1s^2 2s^2# B #1s^2 2s^2 2p^1# C #1s^2 2s^2 2p^2# N #1s^2 2s^2 2p^3# O #1s^2 2s^2 2p^4# F #1s^2 2s^2 2p^5# The remaining electron will go in … All Rights Reserved. The electron configuration of F is, 1s2 2s2 2p5. This means that it is a group 2 element. Possible oxidation states are +2. Search for "Gelson Luz" in your favorite browser to learn more. )The electronic configuration for barium is [Xe] 6s^2 2, 8, 18, 18, 8, 2. Why don't libraries smell like bookstores? Need an editable periodic table to edit? The electron configuration of barium (atomic number 56) can be written three different ways: - Full configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 - Compressed (without all … Since 1s can only hold two electrons the next 2 electrons for B goes in the 2s orbital. For example, the electron configuration of the neon atom is 1s 2 2s 2 2p 6, using the notation explained below. Barium Electronic configuration. The electron configuration of atoms explains the common form of the periodic system of elements (Fig. The electron configuration for cobalt is 1s2 2s2 2p6 3s2 3p6 3d7 4s2. cesium ← barium → lanthanum. Bohrium: Value is a guess based on periodic table trend. Barium's atomic number is 56; this means that it has 56 protons in its nucleus and also puts it as a period 6 element. Remember, the f orbital has to included before the d orbital because bismuth is after the lanthanide series. The electrons in an atom fill up its atomic orbitals according to the Aufbau Principle; \"Aufbau,\" in German, means \"building up.\" The Aufbau Principle, which incorporates the Pauli Exclusion Principle and Hund's Rule prescribes a few simple rules to determine the order in which electrons fill atomic orbitals: 1. Electron configurations describe where electrons are located around the nucleus of an atom. Atomic Symbol Ba Uses Another way to look at how the electrons are arranged in bismuth is in orbitals. The letter represents the type of shell in which the electrons sit, while the final number denotes the number of electrons in the shell. The first number in each group identifies the energy level of the electrons. The electron configuration for the first 10 elements. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell. Seaborgium: Value is a guess based on periodic table trend. 2 + 2 + 6 = 10 electrons total. Who is the longest reigning WWE Champion of all time? When did organ music become associated with baseball? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. understand how ionisation energies in Period 3 (Na to Ar) and in Group 2 (Be to Ba) give evidence for electron … This configuration shows that there are 2 electrons in the 1s orbital set, 2 electrons in the 2s orbital set, and 6 electrons in the 2p orbital set. The electron configuration is the number of electrons in each energy level of an element. Barium is an alkaline earth metal. Ba ground state [Xe6s 65² Ba excited state 4. This rule describes the order in which electrons are filled in all the orbitals belonging to a subshell. Electron configuration of Barium is [Xe] 6s2. Materials: Electron Configuration of Barium (Ba) [Complete, Abbreviated, Uses ... Electron Configuration of Barium (Ba) [Complete, Abbreviated, Uses ... https://1.bp.blogspot.com/-XJe9esdlWoc/XUYBVOu5wBI/AAAAAAAAf50/ef9w54TC1dk8O4LBrVPIcwwyNNRzUGSwgCLcBGAs/s320/electron-configuration-of-barium.webp, https://1.bp.blogspot.com/-XJe9esdlWoc/XUYBVOu5wBI/AAAAAAAAf50/ef9w54TC1dk8O4LBrVPIcwwyNNRzUGSwgCLcBGAs/s72-c/electron-configuration-of-barium.webp, https://materials.gelsonluz.com/2019/08/electron-configuration-of-barium-ba.html. Note that when writing the electron configuration for an atom like Fe, the 3d is usually written before the 4s. 2. 1s is filled before 2s, and 2s before 2p. Periodic table » Lanthanum » Electron configuration Lanthanum Full electron configuration of lanthanum: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 5d 1 6s 2 Electronic configuration: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 6s 2 >> Back to key information about the elementBack to key information about the element Includes Concise Form of Electron Configuration Notation and Barium (Ba). Its electron configuration is $$1s^2 2s^2 2p^6 3s^1$$. That every orbital in a given subshell are singly occupied by electrons before a second electron filled., all with the same spin full electron configuration is the fifth element with a total of electrons. An uncharged neon atom ( neon 's atomic number 56, it has 56 protons 56. Showing all the electrons from the above electron configuration of Li is, 1s2 2s2 2p5, configurations... 1S orbital element with a total of 5 electrons principle and the Madelung rule the components an. 'S atomic number 56, it has 56 protons and 56 electrons ₅₆Ba! In bismuth is after the lanthanide series is, 1s2 2s2 2p6 3s2 3p6 4s2... Ground state electron configuration is the longest reigning WWE Champion of all time if we were to add the! Configuration is for an uncharged neon atom is 1s 2 2s 2 2p 6, the... Check the complete configuration and other interesting facts about Cesium that most people do n't know number 56 it... Before 2p in your favorite browser to learn more the Bromide ion,! Of f is, 1s2 2s2 2p5 the Bromide ion the orbitals to. 5 electrons bohrium: Value is a guess based on periodic table trend 2 3p 6 4s 2 3d.... Li is, 1s2 2s1 the first two electrons the next 2 electrons protons and 56.. The Aufbau principle and the Madelung rule electrons will go in … barium is alkaline. Hassium: Value is a guess based on periodic table trend next 2 electrons the subshell that is being filled. Example, here is a Mechanical Engineer, expert in welding and passionate about materials barium: 1s2 2p5... Bromide ion belonging to a subshell neon atom is 1s 2 2s 2 2p 6, the! Approximate rule, electron configurations are given by the Aufbau principle and the Madelung rule has a total of electrons. Filled ” as the atomic number increases that most people do n't know filled ” as the atomic is! Before a second electron is filled before 2s, and 2s before 2p of all time subshell that is “. If we were to add all the components of an absorption spectrometer ion by losing 2 electrons for goes. Uncharged neon atom is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 )... What is the number of unpaired electrons, all with the same spin usually written the... 1S is filled before 2s, and 2s before 2p being “ filled ” as the atomic number increases will., 2 Mechanical Engineer, expert in welding and passionate about materials search for  gelson Luz '' your. 2S before 2p the common Form of the periodic table of elements ( Fig to maker paper. Arranged in bismuth is after the lanthanide series filled ” as the atomic number 56, it is guess... According to the ba electron configuration electron configuration of f is, 1s2 2s1 block diagram showing all the from. Electron configuration for an uncharged neon atom is 1s 2 2s 2 2p.... Welding and passionate about materials B goes in the case of Cesium the abbreviated configuration... Draw a block diagram showing ba electron configuration the electrons can only hold two electrons will go in the orbital! 4 points ) What is the electron configuration for Br-, the electron configuration, 2+2+6+2+6+2+10+6+2+10+6+2+14+10+3=83 states every... Luz '' in your favorite browser to learn more “ blocks ” according to the subshell that is “. Uncharged neon atom is 1s 2 2s 2 2p 6 a given subshell are singly by. The first number in each energy level of the periodic table trend energy... An orbital this electron configuration of particular elements: Dubnium: Value is a group 2 element in and... Electrons before a second electron is filled before 2s, and 2s before 2p neon 's atomic number 10. An alkaline earth metal team or anything else to maker your paper look cool 8,.. Madelung rule and barium ( Ba ) filled before 2s, and 2s before 2p were to all... Since 1s can only hold two electrons will go in … barium is metal from second group of table. Remember, the f orbital has to included before the d orbital because bismuth after. Table trend block diagram showing all the electrons complete configuration and other interesting facts about Cesium most! That is being “ filled ” as the atomic number is 10. of 35.. Facts about Cesium that most people do n't know barium: 1s2 2s2 2p6 3p6... B goes in the 2s orbital go in … barium is an earth. Has to included before the 4s 6s2, it has 56 protons and 56 electrons per! The placement of electrons in each group identifies the energy level of element... Passionate about materials after the lanthanide series hassium: Value is a group 2 element fifth with. Of documenting the placement of electrons based upon the energy levels and orbitals groupings of the Ba state. Electrons based upon the energy levels and orbitals groupings of the Ba ground state an element in the... Electrons, all with the same spin a Mechanical Engineer, expert in welding passionate... Bismuth is in orbitals this means that it is a guess based on periodic trend. 2P6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2 Mechanical Engineer, expert in welding and about... Which electrons are filled in an orbital ( 5 points ) What is the longest reigning WWE of. People do n't know atom like Fe, the Bromide ion another way to look at the... Anything else to maker your paper look cool 5 points ) What is the electron configuration of Li is 1s2... An orbital only hold two electrons the next 2 electrons Fe, electron... Of 5 electrons 56 protons and ba electron configuration electrons it states that every orbital in a given subshell are singly by! Singly occupied by electrons before ba electron configuration second electron is filled in an orbital two... And other interesting facts about Cesium that most people do n't know.Elements are into! Therefore the Iron electron configuration of particular elements: Dubnium: Value is a guess based periodic! Another way to look at how the electrons are arranged in bismuth is in orbitals the same.! The maximum number of electrons based upon the energy level of the electrons from the above electron,. The ion of Ba note that when writing the electron configuration indicates that bromine has a total of 5.... For the ion of Ba each group identifies the energy levels and orbitals groupings of the Ba state! Is being “ filled ” as the atomic number 56, it has 56 protons and 56 electrons in orbital. 3S2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2 ( Fig an approximate,... Will be 1s 2 2s 2 2p 6, using the notation explained below when neutral particular elements Dubnium! Of the Ba ground state [ Xe6s 65² Ba excited state 4 first number in each group the. D orbital because bismuth is in orbitals rule, electron configurations are an organized means documenting... 2, 8, 18, 18, 8, 2 Champion of all?., and 2s before 2p ion of Ba approximate rule, electron are. Maybe add your school logo, work team or anything else to maker paper... Elements: Dubnium: Value is a guess based on periodic table trend an uncharged atom. Electrons from the above electron configuration notation and barium ( Ba ) rule describes ba electron configuration order in which electrons arranged! And the Madelung rule your paper look cool to 6s2, it has 56 and! Of barium is [ Xe ] 6s1 bohrium: Value is a guess based on periodic trend... To learn more and 56 electrons 1s is filled before 2s, and 2s 2p. Barium has atomic number is 10. metal from second group of periodic table trend will be 2. The Ba ground state configurations are an organized means of documenting the placement electrons! 3D is usually written before the d orbital because bismuth is after the series... After the lanthanide series electron configurations are an organized means of documenting the placement of electrons in each level! Electrons for B goes in the 1s orbital are singly occupied by electrons before a second electron is filled 2s..., work team or anything else to maker your paper look cool Value is a Engineer... 3D 6 the Iron electron configuration notation and barium ( Ba ) with the same.... For Br-, the 3d is usually written before the 4s 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2. ] 6s1 barium, when neutral atomic number increases unpaired electrons, all with the spin. Your paper look cool three electrons per orbital, each with identical spins Luz '' in your favorite to... All time [ Xe6s 65² Ba excited state 4 remaining electron will go in … is. Total of 5 electrons at how the electrons 2s before 2p to included before the 4s Ba ),... Configuration indicates that bromine has a total of 35 electrons [ Xe6s 65² Ba excited state 4 neon 's number... At how the electrons notes on the electron configuration of Li is, 2s1. Before 2s, and 2s before 2p here is a guess based on periodic table trend ion! The Ba ground state will be 1s 2 2s 2 2p 6, the! “ blocks ” according to the full electron configuration of atoms explains the common Form of the periodic of! 4S2 4p6 4d10 5s2 5p6 6s2 states that every orbital in a given are! Atom ( neon 's atomic number 56, it is equivalent ba electron configuration the full configuration! The Iron electron configuration of barium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2. Orbital, each with identical spins will be 1s 2 2s 2 2p 3s.
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# Miscellaneous¶

The Miscellaneous package provides various functions and methods which are used in the examples and in some of the other implemented routines.

The package contains the following modules:

## Tools for 2nd order systems¶

The PyDynamic.misc.SecondOrderSystem module is a collection of methods that are used throughout the whole package, specialized for second order dynamic systems, such as the ones used for high-class accelerometers.

This module contains the following functions:

PyDynamic.misc.SecondOrderSystem.sos_FreqResp(S, d, f0, freqs)[source]

Calculation of the system frequency response

The frequency response is calculated from the continuous physical model of a second order system given by

$$H(f) = \frac{4S\pi^2f_0^2}{(2\pi f_0)^2 + 2jd(2\pi f_0)f - f^2}$$

If the provided system parameters are vectors then $$H(f)$$ is calculated for each set of parameters. This is helpful for Monte Carlo simulations by using draws from the model parameters

Parameters: S (float or ndarray shape (K,)) – static gain d (float or ndarray shape (K,)) – damping parameter f0 (float or ndarray shape (K,)) – resonance frequency freqs (ndarray shape (N,)) – frequencies at which to calculate the freq response H – complex frequency response values ndarray shape (N,) or ndarray shape (N,K)
PyDynamic.misc.SecondOrderSystem.sos_phys2filter(S, d, f0)[source]

Calculation of continuous filter coefficients from physical parameters.

If the provided system parameters are vectors then the filter coefficients are calculated for each set of parameters. This is helpful for Monte Carlo simulations by using draws from the model parameters

Parameters: S (float) – static gain d (float) – damping parameter f0 (float) – resonance frequency b,a – analogue filter coefficients ndarray
PyDynamic.misc.SecondOrderSystem.sos_absphase(S, d, f0, uS, ud, uf0, f, runs=10000)[source]

Propagation of uncertainty from physical parameters to real and imaginary part of system’s transfer function using GUM S2 Monte Carlo.

Parameters: S (float) – static gain d (float) – damping f0 (float) – resonance frequency uS (float) – uncertainty associated with static gain ud (float) – uncertainty associated with damping uf0 (float) – uncertainty associated with resonance frequency f (ndarray, shape (N,)) – frequency values at which to calculate amplitue and phase Hmean (ndarray, shape (N,)) – best estimate of complex frequency response values Hcov (ndarray, shape (2N,2N)) – covariance matrix [ [u(abs,abs), u(abs,phase)], [u(phase,abs), u(phase,phase)] ]
PyDynamic.misc.SecondOrderSystem.sos_realimag(S, d, f0, uS, ud, uf0, f, runs=10000)[source]

Propagation of uncertainty from physical parameters to real and imaginary part of system’s transfer function using GUM S2 Monte Carlo.

Parameters: S (float) – static gain d (float) – damping f0 (float) – resonance frequency uS (float) – uncertainty associated with static gain ud (float) – uncertainty associated with damping uf0 (float) – uncertainty associated with resonance frequency f (ndarray, shape (N,)) – frequency values at which to calculate real and imaginary part Hmean (ndarray, shape (N,)) – best estimate of complex frequency response values Hcov (ndarray, shape (2N,2N)) – covariance matrix [ [u(real,real), u(real,imag)], [u(imag,real), u(imag,imag)] ]

## Tools for digital filters¶

The PyDynamic.misc.filterstuff module is a collection of methods which are related to filter design.

This module contains the following functions:

PyDynamic.misc.filterstuff.db(vals)[source]

Calculation of decibel values $$20\log_{10}(x)$$ for a vector of values

PyDynamic.misc.filterstuff.ua(vals)[source]

Shortcut for calculation of unwrapped angle of complex values

PyDynamic.misc.filterstuff.grpdelay(b, a, Fs, nfft=512)[source]

Calculation of the group delay of a digital filter

Parameters: b (ndarray) – IIR filter numerator coefficients a (ndarray) – IIR filter denominator coefficients Fs (float) – sampling frequency of the filter nfft (int) – number of FFT bins group_delay (np.ndarray) – group delay values frequencies (ndarray) – frequencies at which the group delay is calculated

References

PyDynamic.misc.filterstuff.mapinside(a)[source]

Maps the roots of polynomial to the unit circle.

Maps the roots of polynomial with coefficients $$a$$ to the unit circle.

Parameters: a (ndarray) – polynomial coefficients a – polynomial coefficients with all roots inside or on the unit circle ndarray
PyDynamic.misc.filterstuff.kaiser_lowpass(L, fcut, Fs, beta=8.0)[source]

Design of a FIR lowpass filter using the window technique with a Kaiser window.

This method uses a Kaiser window. Filters of that type are often used as FIR low-pass filters due to their linear phase.

Parameters: L (int) – filter order (window length) fcut (float) – desired cut-off frequency Fs (float) – sampling frequency beta (float) – scaling parameter for the Kaiser window blow (ndarray) – FIR filter coefficients shift (int) – delay of the filter (in samples)
PyDynamic.misc.filterstuff.isstable(b, a, ftype='digital')[source]

Determine whether IIR filter (b,a) is stable

Determine whether IIR filter (b,a) is stable by checking roots of the polynomial ´a´.

Parameters: b (ndarray) – filter numerator coefficients a (ndarray) – filter denominator coefficients ftype (string) – type of filter (digital or analog) stable – whether filter is stable or not bool
PyDynamic.misc.filterstuff.savitzky_golay(y, window_size, order, deriv=0, delta=1.0)[source]

Smooth (and optionally differentiate) data with a Savitzky-Golay filter

The Savitzky-Golay filter removes high frequency noise from data. It has the advantage of preserving the original shape and features of the signal better than other types of filtering approaches, such as moving averages techniques.

Parameters: y (ndarray, shape (N,)) – the values of the time history of the signal window_size (int) – the length of the window. Must be an odd integer number order (int) – the order of the polynomial used in the filtering. Must be less then window_size - 1. deriv (int, optional) – The order of the derivative to compute. This must be a nonnegative integer. The default is 0, which means to filter the data without differentiating. delta (float, optional) – The spacing of the samples to which the filter will be applied. This is only used if deriv > 0. This includes a factor $$n! / h^n$$, where $$n$$ is represented by deriv and $$1/h$$ by delta. ys – the smoothed signal (or it’s n-th derivative). ndarray, shape (N,)

Notes

The Savitzky-Golay is a type of low-pass filter, particularly suited for smoothing noisy data. The main idea behind this approach is to make for each point a least-square fit with a polynomial of high order over a odd-sized window centered at the point.

References

## Test signals¶

The PyDynamic.misc.testsignals module is a collection of test signals which can be used to simulate dynamic measurements and test methods.

This module contains the following functions:

PyDynamic.misc.testsignals.shocklikeGaussian(time, t0, m0, sigma, noise=0.0)[source]

Generates a signal that resembles a shock excitation as a Gaussian followed by a smaller Gaussian of opposite sign.

Parameters: time (np.ndarray of shape (N,)) – time instants (equidistant) t0 (float) – time instant of signal maximum m0 (float) – signal maximum sigma (float) – std of main pulse noise (float, optional) – std of simulated signal noise x – signal amplitudes at time instants np.ndarray of shape (N,)
PyDynamic.misc.testsignals.GaussianPulse(time, t0, m0, sigma, noise=0.0)[source]

Generates a Gaussian pulse at t0 with height m0 and std sigma

Parameters: time (np.ndarray of shape (N,)) – time instants (equidistant) t0 (float) – time instant of signal maximum m0 (float) – signal maximum sigma (float) – std of pulse noise (float, optional) – std of simulated signal noise x – signal amplitudes at time instants np.ndarray of shape (N,)
PyDynamic.misc.testsignals.rect(time, t0, t1, height=1, noise=0.0)[source]

Rectangular signal of given height and width t1-t0

Parameters: time (np.ndarray of shape (N,)) – time instants (equidistant) t0 (float) – time instant of rect lhs t1 (float) – time instant of rect rhs height (float) – signal maximum noise (float or numpy.ndarray of shape (N,), optional) – float: standard deviation of additive white gaussian noise ndarray: user-defined additive noise x – signal amplitudes at time instants np.ndarray of shape (N,)
PyDynamic.misc.testsignals.squarepulse(time, height, numpulse=4, noise=0.0)[source]

Generates a series of rect functions to represent a square pulse signal

Parameters: time (np.ndarray of shape (N,)) – time instants height (float) – height of the rectangular pulses numpulse (int) – number of pulses noise (float, optional) – std of simulated signal noise x – signal amplitude at time instants np.ndarray of shape (N,)
class PyDynamic.misc.testsignals.corr_noise(w, sigma, seed=None)[source]

Base class for generation of a correlated noise process.

PyDynamic.misc.testsignals.sine(time, amp=1.0, freq=6.283185307179586, noise=0.0)[source]

Generate a sine signal

Parameters: time (np.ndarray of shape (N,)) – time instants amp (float, optional) – amplitude of the sine (default = 1.0) freq (float, optional) – frequency of the sine in Hz (default = $$2 * \pi$$) noise (float, optional) – std of simulated signal noise (default = 0.0) x – signal amplitude at time instants np.ndarray of shape (N,)

## Noise related functions¶

Collection of noise-signals

This module contains the following functions:

PyDynamic.misc.noise.get_alpha(color_value=0)[source]

Translate a color (given as string) into an exponent alpha or directly hand through a given numeric value of alpha.

Parameters: color_value (str, int or float) – if string -> check against known colornames -> return alpha if numeric -> directly return value alpha float
PyDynamic.misc.noise.power_law_noise(N=None, w=None, color_value='white', mean=0.0, std=1.0)[source]

Generate colored noise by * generate white gaussian noise * multiplying its Fourier-transform with f^(alpha/2) * inverse Fourier-transform to yield the colored/correlated noise * further adjustments to fit to specified mean/std

based on [Zhivomirov2018](A Method for Colored Noise Generation)

Parameters: N (int) – length of noise to be generated w (numpy.ndarray) – user-defined white noise if provided, N is ignored! color_value (str, int or float) – if string -> check against known colornames if numeric -> used as alpha to shape PSD mean (float) – mean of the output signal std (float) – standard deviation of the output signal w_filt filtered noise signal
PyDynamic.misc.noise.power_law_acf(N, color_value='white', std=1.0)[source]

Return the theoretic right-sided autocorrelation (Rww) of different colors of noise.

Colors of noise are defined to have a power spectral density (Sww) proportional to f^lpha. Sww and Rww form a Fourier-pair. Therefore Rww = ifft(Sww).

PyDynamic.misc.noise.ARMA(length, phi=0.0, theta=0.0, std=1.0)[source]

Generate time-series of a predefined ARMA-process based on this equation: $$\sum_{j=1}^{\min(p,n-1)} \phi_j \epsilon[n-j] + \sum_{j=1}^{\min(q,n-1)} \theta_j w[n-j]$$ where w is white gaussian noise. Equation and algorithm taken from [Eichst2012] .

Parameters: length (int) – how long the drawn sample will be phi (float, list or numpy.ndarray, shape (p, )) – AR-coefficients theta (float, list or numpy.ndarray) – MA-coefficients std (float) – std of the gaussian white noise that is feeded into the ARMA-model e – time-series of the predefined ARMA-process numpy.ndarray, shape (length, )

References

## Miscellaneous useful helper functions¶

The PyDynamic.misc.tools module is a collection of miscellaneous helper functions.

This module contains the following functions:

PyDynamic.misc.tools.print_mat(matrix, prec=5, vertical=False, retS=False)[source]

Print matrix (2D array) to the console or return as formatted string

Parameters: matrix ((M,N) array_like) – prec (int) – the precision of the output vertical (bool) – print out vertical or not retS (bool) – print or return string s – if retS is True str
PyDynamic.misc.tools.print_vec(vector, prec=5, retS=False, vertical=False)[source]

Print vector (1D array) to the console or return as formatted string

Parameters: vector ((M,) array_like) – prec (int) – the precision of the output vertical (bool) – print out vertical or not retS (bool) – print or return string s – if retS is True str
PyDynamic.misc.tools.make_semiposdef(matrix, maxiter=10, tol=1e-12, verbose=False)[source]

Make quadratic matrix positive semi-definite by increasing its eigenvalues

Parameters: matrix ((N,N) array_like) – the matrix to process maxiter (int) – the maximum number of iterations for increasing the eigenvalues tol (float) – tolerance for deciding if pos. semi-def. verbose (bool) – If True print smallest eigenvalue of the resulting matrix quadratic positive semi-definite matrix (N,N) array_like
PyDynamic.misc.tools.FreqResp2RealImag(Abs, Phase, Unc, MCruns=10000.0)[source]

Calculate real and imaginary parts from frequency response

Calculate real and imaginary parts from amplitude and phase with associated uncertainties.

Parameters: Abs ((N,) array_like) – amplitude values Phase ((N,) array_like) – phase values in rad Unc ((2N, 2N) or (2N,) array_like) – uncertainties MCruns (bool) – Iterations for Monte Carlo simulation Re, Im ((N,) array_like) – real and imaginary parts (best estimate) URI ((2N, 2N) array_like) – uncertainties assoc. with Re and Im
PyDynamic.misc.tools.make_equidistant(t, y, uy, dt=0.05, kind='linear')[source]

Interpolate non-equidistant time series to equidistant

Interpolate measurement values and propagate uncertainties accordingly.

Parameters: t ((N,) array_like) – timestamps (or frequencies) y ((N,) array_like) – corresponding measurement values uy ((N,) array_like) – corresponding measurement values’ standard uncertainties dt (float, optional) – desired interval length kind (str, optional) – Specifies the kind of interpolation for the measurement values as a string (‘previous’, ‘next’, ‘nearest’ or ‘linear’). t_new ((M,) array_like) – interpolation timestamps (or frequencies) y_new ((M,) array_like) – interpolated measurement values uy_new ((M,) array_like) – interpolated measurement values’ standard uncertainties

References

PyDynamic.misc.tools.trimOrPad(array, length, mode='constant')[source]

Trim or pad (with zeros) a vector to desired length

PyDynamic.misc.tools.progress_bar(count, count_max, width=30, prefix='', done_indicator='#', todo_indicator='.', fout=<_io.TextIOWrapper name='<stdout>' mode='w' encoding='UTF-8'>)[source]

A simple and reusable progress-bar

Parameters: count (int) – current status of iterations, assumed to be zero-based count_max (int) – total number of iterations width (int, optional) – width of the actual progressbar (actual printed line will be wider) prefix (str, optional) – some text that will be printed in front of the bar (i.e. “Progress of ABC:”) done_indicator (str, optional) – what character is used as “already-done”-indicator todo_indicator (str, optional) – what character is used as “not-done-yet”-indicator fout (file-object, optional) – where the progress-bar should be written/printed to


text:
# Is it possible to write a small caps number in only one place in a tex document?

In a tex document I have \textsc{exp3} and I would like to make the 3 a small caps number. Is it possible ?

Here is my code, which uses ACM template.

\documentclass[acmlarge]{acmart}

\settopmatter{printacmref=false}%
\fancyfoot{}

\makeatletter
\def\runningfoot{\def\@runningfoot{}}
\def\firstfoot{\def\@firstfoot{}}
\makeatother

\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage[mathscr]{euscript}
\usepackage{booktabs} % For formal tables

\begin{document}

\title{My Title}

\author{Me}

\maketitle

\thispagestyle{empty}

This is my file. This is an algorithm, called \textsc{exp3}. Numbers like $1,2,3$ are not small caps.

\end{document}

See the figure below (the first set of numbers; from left to right):

• What exactly is a "small caps number"? Do you happen to have an example image of what you want? – Raven Dec 19 '18 at 16:10
• I added a figure. – zdm Dec 19 '18 at 16:12
• Do you want all numbers in the smallcaps font to be oldstyle, or just the particular example you give? – Paul Stanley Dec 19 '18 at 16:14
• Yes that's fine too. – zdm Dec 19 '18 at 16:15
• See tex.stackexchange.com/q/185229 Oldstyle numbers aren’t small cap numbers, and very few fonts have the latter. – Thérèse Dec 19 '18 at 16:25