text:
## Numbers of the form 28(10^n – 1)

Numbers of the form   $28 \,(10^{n} - 1)$

$28 \,(10^1 - 1) \; = \; 252$

$252$   is a palindrome whose square can be expressed as the product of two reversible numbers in two different ways:

$252^2 \; = \; 252\times 252 \; = \; 144\times 441$

$28 \,(10^2 - 1) \; = \; 2772$

$2772^2 = 7683984 \; = \; 2772\times 2772 \; = \; 1584\times 4851$

$28 \,(10^3 - 1) \; = \; 27972$

It is the first palindromic integers containing just the digits 2, 7, and 9, which are divisible by 2, 7, and 9; and, whose square can be expressed as the product of two reversible numbers in two different ways:

$27972^2 \; = \; 782432784 \; = \; 27972\times 27972 \; = \; 15984\times 48951$

$28 \,(10^4 - 1) \; = \; 279972$

$279972^2 \; = \; 78384320784 \; = \; 279972\times 279972 \; = \; 159984\times 489951$

$28 \,(10^5 - 1) \; = \; 2799972$

$2799972^2 \; = \; 7839843200784 \; = \; 2799972\times 2799972 \; = \; 1599984\times 4899951$

$28 \,(10^6 - 1) \; = \; 27999972$

$27999972^2 \; = \; 783998432000784$
$= \; 27999972\times 27999972 \; = \; 15999984\times 48999951$

…………………….

———————————————————-

$2, \; 22, \; 222, \; 2222, \; 22222, \; .....$
are integers of the form   $2 \, (10^n - 1)/9$

$9, \; 99, \; 999, \; 9999, \; 99999, \; .....$
are integers of the form   $(10^n - 1)$

$2\times 9, \; 22\times 99, \; 222\times 999, \; 2222\times 9999, \; ....$
are integers of the form   $2 \, (10^n - 1)^2/9$

$2 \, (10^n - 1)^2/9 \; + \; 10^n$   are :

$(2\times 9) + 10, \; (22\times 99) + 10^2, \; (222\times 999) + 10^3, \; (2222\times 9999) + 10^4, ....$

$28$
$2278$
$222778$
$22227778$
$2222277778$
$222222777778$
$22222227777778$
$2222222277777778$
$222222222777777778$
$22222222227777777778$
…………………….
…………………….

$1/9 \; \times \; (2^{n+1} \; 5^n + 1) \,(10^n + 2)$
OR    $1/9 \; \times \; (2^{2n+1} \; 5^{2n} + 2^n \; 5^{n+1} + 2)$

$9 \,(2 \, (10^n - 1)^2/9 \; + \; 10^n)$
$= \; (2^{n+1} \; 5^n \; + \; 1) \,(10^n \; + \; 2)$
$= \; 2^{2n+1} \; 5^{2n} \; + \; 2^n \; 5^{n+1} \; + \; 2$

giving us:

$252$
$20502$
$2005002$
$200050002$
$20000500002$
$2000005000002$
$200000050000002$
$20000000500000002$
$2000000005000000002$
$200000000050000000002$
…………………….
…………………….

The square of each of these numbers can be expressed as the product of two reversible numbers in two different ways.

——————————————-

$12^2\times 21^2 = 252^2$

$144\times 441 = 252^2$
$114444\times 444411 = 225522^2$
$111444444\times 444444111 = 222555222^2$
$111144444444\times 444444441111 = 222255552222^2$
$111114444444444\times 444444444411111 = 222225555522222^2$
…………………….

$144\times 441 = 252^2$
$144144\times 441441 = 252252^2$
$144144144\times 441441441 = 252252252^2$
$144144144144\times 441441441441 = 252252252252^2$
$144144144144144\times 441441441441441 = 252252252252252^2$
…………………….

$144\times 441 = 252^2$
$10404\times 40401 = 20502^2$
$1004004\times 4004001 = 2005002^2$
$100040004\times 400040001 = 200050002^2$

…………………….

——————————————-

——————————————-

——————————————-

math grad - Interest: Number theory
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### 4 Responses to Numbers of the form 28(10^n – 1)

1. pipo says:

There are more patrons like that:
9*(22 * 99 + 100) = 20502 and 20502*20502 = 10404 * 40401
9*(222 * 999 + 1000) = 2005002 and 2005002*2005002 = 1004004 * 4004001
9*(2222 * 9999 + 10000) = 200050002 and 200050002*200050002 = 100040004 * 400040001

And some more:
144 * 441 = 252 ^2
1584 * 4851 = 2772 ^2
10404 * 40401 = 20502 ^2
12544 * 44521 = 23632 ^2
14544 * 44541 = 25452 ^2
14884 * 48841 = 26962 ^2
15984 * 48951 = 27972 ^2
27648 * 84672 = 48384 ^2
114444 * 444411 = 225522 ^2
144144 * 441441 = 252252 ^2
137984 * 489731 = 259952 ^2
159984 * 489951 = 279972 ^2
409739 * 937904 = 619916 ^2
1022121 * 1212201 = 1113111 ^2
1042441 * 1442401 = 1226221 ^2
1256641 * 1466521 = 1357531 ^2
1004004 * 4004001 = 2005002 ^2
1050804 * 4080501 = 2070702 ^2
1024144 * 4414201 = 2126212 ^2
1141504 * 4051411 = 2150512 ^2
1214404 * 4044121 = 2216122 ^2
2044242 * 2424402 = 2226222 ^2
1044484 * 4844401 = 2249422 ^2
1154844 * 4484511 = 2275722 ^2
1323504 * 4053231 = 2316132 ^2
1236544 * 4456321 = 2347432 ^2
1266944 * 4496621 = 2386832 ^2
1444804 * 4084441 = 2429242 ^2
1354444 * 4444531 = 2453542 ^2
1440144 * 4410441 = 2520252 ^2
1454544 * 4454541 = 2545452 ^2
1585584 * 4855851 = 2774772 ^2
1599984 * 4899951 = 2799972 ^2
3066363 * 3636603 = 3339333 ^2
1376739 * 9376731 = 3592953 ^2
4048144 * 4418404 = 4229224 ^2
2632608 * 8062362 = 4607064 ^2
2763648 * 8463672 = 4836384 ^2

• benvitalis says:

Interesting! Note that
2, 22, 222, 2222, 22222,….. are integers of the form 2*(10^n – 1)/9
9, 99, 999, 9999, 99999,….. are integers of the form (10^n – 1)

2*9, 22*99, 222*999, 2222*9999, …. are integers of the form 2*(10^n – 1)^2/9

2*9 + 10, 22*99 + 10^2, 222*999 + 10^3, 2222*9999 + 10^4, ….

28,2278,222778,22227778,2222277778

1/9 (2^(n+1) * 5^n + 1)(10^n + 2)
OR 1/9 (2^(2n+1) 5^(2n) + 2^n * 5^(n+1) + 2)

2. pipo says:

I think there is also a nice patron in:
27648 * 84672 = 48384 ^2
2763648 * 8463672 = 4836384 ^2
276363648 * 846363672 = 483636384 ^2
Where 27648 = 9 * 3072
2763648 = 9 * 307072
276363648 = 9 * 30707072
84672 = 9 * 9408
8463672 = 9 * 940408
846363672 = 9 * 94040408
Where 48384 = 9 * 5376 (or 84*576)
4836384 = 9 * 537376 (or 84*57576)
483636384 = 9 * 53737376 (or 84*5757576)
etc

3. pipo says:

Better:
27648 = 144*192
2763648 = 144*19192
276363648 = 144*1919192
84672 = 441 * 192
8463672 = 441 * 19192
846363672 = 441 * 1919192
48384 = 252 * 192
4836384 = 252 * 19192
483636384 = 252 * 1919192

pipo


text:
# Magnitude of Induced EMF MCQ [Free PDF] – Objective Question Answer for Magnitude of Induced EMF Quiz

1. Which, among the following, is the formula for induced emf?

A. e = dϕ /dt
B. e = dt/dϕ
C. e = t × ϕ
D. e = t2ϕ

The formula for induced emf is e = dϕ /dt because the induced emf is the flux linkage per unit time.

2. According to _________________ induced emf is equal to rate of change of magnetic flux.

MCQ on Virus - Examsegg Biology

A. Newton’s law
B. Lenz law
D. Coulomb’s law

According to Faraday’s law of electromagnetic induction, induced emf is equal to the rate of change of magnetic flux.

3. The emf induced in a coil having N turns is?

A. e = ϕ/t
B. e = N × ϕ/t
C. e = N × ϕ × t
D. e = N2 × ϕ × t

The emf induced in a coil having N turns is, e = N × ϕ/t. This is because, the emf in a single-coil is the flux linkage per unit time, that is, phi/t. Hence the flux induced in N turns is N × ϕ/t.

4. According to ____________________ induced emf oppose the cause due to which they are produced.

A. Newton’s law
B. Lenz law
D. Coulomb’s law

According to Lenz law, emf is induced in such a way that it opposes the cause due to which it is produced.

5. North pole induces __________

A. Clockwise current
B. Anti-clockwise current
C. Zero current
D. Infinite current

A north pole will always induce an anti-clockwise current whereas a south pole will always induce a clockwise current due to electromagnetic theory.

h({});

6. What is the principle of the transformer?

A. Gauss law
B. Coulomb’s law
C. Electromagnetic induction
D. Ampere’s law

In a transformer, flux in the secondary coil change due to the current in the primary coil, and hence current get induced in the secondary coil.

7. Voltage induced in secondary coil of transformer is given by_________

A. NP × VP/NS
B. NS × VP/NP
C. (NP/VP) × NS
D. NP/(VP × NS)

In transformer, VP/VS  = NP/NS

So, VS  = NS × VP/NP.

8. The relation between the direction of force and the direction of the magnetic field is _________

A. Same direction
B. Opposite direction
C. Perpendicular
D. Unrelated

When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor, and the magnetic field in the conductor are mutually perpendicular to each other.

9. The relation between the direction of current and the direction of the magnetic field is ________

A. Same direction
B. Opposite direction
C. Perpendicular
D. Unrelated

When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor, and the magnetic field in the conductor are mutually perpendicular to each other.

10. The relation between the direction of current and the direction of the force is _________

A. Same direction
B. Opposite direction
C. Perpendicular
D. Unrelated

When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor, and the magnetic field in the conductor are mutually perpendicular to each other.

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format: plain


text:
• ### On the Production of an Isotensor Dibaryon in the $pp \to pp\pi^+\pi^-$ Reaction(1803.03192)

April 18, 2018 hep-ex, nucl-ex
Exclusive measurements of the quasi-free $pp \to pp\pi^+\pi^-$ reaction have been performed by means of $pd$ collisions at $T_p$ = 1.2 GeV using the WASA detector setup at COSY. Total and differential cross sections have been obtained covering the energy region $T_p = 1.08 - 1.36$ GeV ($\sqrt s$ = 2.35 - 2.46 GeV), which includes the regions of $N^*(1440)$ and $\Delta(1232)\Delta(1232)$ resonance excitations. Calculations describing these excitations by $t$-channel meson exchange are at variance with experimental differential cross sections and underpredict substantially the measured total cross section. An isotensor $\Delta N$ dibaryon resonance with $I(J^P) = 2(1^+)$ produced associatedly with a pion is able to overcome these deficiencies. Such a dibaryon was predicted by Dyson and Xuong and more recently calculated by Gal and Garcilazo.
• ### An Isotensor Dibaryon in the $pp \to pp\pi^+\pi^-$ Reaction?(1803.03193)

April 18, 2018 hep-ex, nucl-ex
Exclusive measurements of the quasi-free $pp \to pp\pi^+\pi^-$ reaction have been carried out at WASA@COSY by means of $pd$ collisions at $T_p$ = 1.2 GeV. Total and differential cross sections have been extracted covering the energy region $T_p = 1.08 - 1.36$ GeV, which is the region of $N^*(1440)$ and $\Delta(1232)\Delta(1232)$ resonance excitations. Calculations describing these excitations by $t$-channel meson exchange are at variance with the measured differential cross sections and underpredict substantially the experimental total cross section. An isotensor $\Delta N$ dibaryon resonance with $I(J^P) = 2(1^+)$ produced associatedly with a pion is able to overcome these deficiencies.
• ### Measurement of the $np \to np\pi^0\pi^0$ Reaction in Search for the Recently Observed $d^*(2380)$ Resonance(1409.2659)

Sept. 9, 2014 nucl-ex
Exclusive measurements of the quasi-free $np \to np\pi^0\pi^0$ reaction have been performed by means of $dp$ collisions at $T_d$ = 2.27 GeV using the WASA detector setup at COSY. Total and differential cross sections have been obtained covering the energy region $\sqrt s$ = (2.35 - 2.46) GeV, which includes the region of the ABC effect and its associated $d^*(2380)$ resonance. Adding the $d^*$ resonance amplitude to that for the conventional processes leads to a reasonable description of the data. The observed resonance effect in the total cross section is in agreement with the predictions of F\"aldt and Wilkin as well Albadajedo and Oset. The ABC effect, {\it i.e.} the low-mass enhancement in the $\pi^0\pi^0$-invariant mass spectrum, is found to be very modest - if present at all, which might pose a problem to some of its interpretations.
• ### Neutron-Proton Scattering in the Context of the $d^*$(2380) Resonance(1408.4928)

Aug. 21, 2014 hep-ex, nucl-ex
New data on quasifree polarized neutron-proton scattering, in the region of the recently observed $d^*$ resonance structure, have been obtained by exclusive and kinematically complete high-statistics measurements with WASA at COSY. This paper details the determination of the beam polarization, checks of the quasifree character of the scattering process, on all obtained $A_y$ angular distributions and on the new partial-wave analysis, which includes the new data producing a resonance pole in the $^3D_3$-$^3G_3$ coupled partial waves at ($2380\pm10 - i40\pm5$) MeV -- in accordance with the $d^*$ dibaryon resonance hypothesis. The effect of the new partial-wave solution on the description of total and differential cross section data as well as specific combinations of spin-correlation and spin-transfer observables available from COSY-ANKE measurements at $T_d$ = 2.27 GeV is discussed.
• ### Cross section ratio and angular distributions of the reaction p + d -> 3He + eta at 48.8 MeV and 59.8 MeV excess energy(1402.3469)

June 23, 2014 nucl-ex
We present new data for angular distributions and on the cross section ratio of the p + d -> 3He + eta reaction at excess energies of Q = 48.8 MeV and Q = 59.8 MeV. The data have been obtained at the WASA-at-COSY experiment (Forschungszentrum J\"ulich) using a proton beam and a deuterium pellet target. While the shape of obtained angular distributions show only a slow variation with the energy, the new results indicate a distinct and unexpected total cross section fluctuation between Q = 20 MeV and Q = 60 MeV, which might indicate the variation of the production mechanism within this energy interval.
• ### Investigation of the dd -> 3He n \pi 0 reaction with the FZ J\"ulich WASA-at-COSY facility(1304.3561)

March 15, 2014 nucl-ex
An exclusive measurement of the dd -> 3He n \pi 0 reaction was carried at a beam momentum of p = 1.2 GeV/c using the WASA-at-COSY facility. For the first time data on the total cross section as well as differential distributions were obtained. The data are described with a phenomenological approach based on a combination of a quasi-free model and a partial wave expansion for three-body reaction. The total cross section is found to be \sigma(tot) = (2.89 +- 0.01(stat) +- 0.06(sys) +- 0.29(norm)) \mu b. The contribution of the quasi-free processes (with the neutron being target or beam spectator) accounts for 38% of the total cross section and dominates the differential distributions in specific regions of the phase space. The remaining part of the cross section can be described within a partial wave decomposition indicating the significance of p-wave contributions in the final state.
• ### \pi^0 \pi^0 Production in Proton-Proton Collisions at Tp=1.4 GeV(1107.0879)

Feb. 14, 2014 hep-ex
The reaction pp->pppi0pi0 has been investigated at a beam energy of 1.4 GeV using the WASA-at-COSY facility. The total cross section is found to be (324 +- 21_systematic +- 58_normalization) mub. In order to to study the production mechanism, differential kinematical distributions have been evaluated. The differential distributions indicate that both initial state protons are excited into intermediate Delta(1232) resonances, each decaying into a proton and a single pion, thereby producing the pion pair in the final state. No significant contribution of the Roper resonance N*(1440) via its decay into a proton and two pions is found


text:
# Homework Help: Equations of motion for an orbit

1. Mar 8, 2007

### fasterthanjoao

Only reason I'm posting here is that i'll get more views than in the cosmology thread, I'm afraid..

(Basically, I'm working through a couple of different models and after some work I'm a bit stuck: http://trond.hjorteland.com/thesis/node21.html

I'm basically trying to integrate equation 3.33 to $$t_0$$ - without much success. Now, cosmologists will know that $$a(t)=(\frac{t}{t_0})^\frac{2}{3}$$ which I feel should be substituted, getting rid of the $$a^\frac{1}{2}da$$? Any comments on this appreciated.

2. Mar 9, 2007

### dextercioby

This is to help you find the antiderivative: Make the substitutions

$$a=\Omega_{0} , \ b=1-\Omega_{0}, \ a=y^{3/2}$$

You get the equation

$$H_{0}\int{}dt=\frac{2}{3}\int \frac{dy}{\sqrt{by^2 +a}}$$

Can you take it from here ?

3. Mar 9, 2007

### fasterthanjoao

Seems to me like it's at least definitely a log, I think roughly coming out to something like :

$$H_0t=\frac{2}{3}\log(2by+2\sqrt{a+by^2})$$

Which I'm doesn't seem right, and infact I'm almost certain the final form should be independent of y, since in a limit it should simplify to

$$H_0t_0=\frac{2}{3}$$

So it seems I need to at least have a definite integral, ill just say y from 0 to 1 since i've already normalised for y to be 1 at present.

Last edited: Mar 9, 2007
4. Mar 9, 2007

### fasterthanjoao

Integrating from 0 to 1 seems to be correct now, from Mathematica I see that it's an inverse hyperbolic sine (through I suppose theres an equivalent form in Log). Could obviously have used Mathematica at the beginning, but I'd like to be able to work through it since it's bugging me now!

If you can shed light on how this is done, that would be great. Thanks.

5. Mar 9, 2007

### fasterthanjoao

Think I have it ok now, was just a little rusty on my integration. Thanks for the help dextercioby.


text:
Problem: The density of osmium (the densest metal) is 22.57 g/cm3. If a 1.00-kg rectangular block of osmium has two dimensions of 4.00 cm x 4.00 cm, calculate the third dimension of the block.

FREE Expert Solution
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Problem Details

The density of osmium (the densest metal) is 22.57 g/cm3. If a 1.00-kg rectangular block of osmium has two dimensions of 4.00 cm x 4.00 cm, calculate the third dimension of the block.


text:
Monday, October 30, 2017

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text:
# 16.6. Summary¶

The bias-variance trade-off allows us to more precisely describe the modeling phenomena that we have seen in this chapter: under-fitting relates to model bias; over-fitting results in model variance. In Figure 16.4, the $$x$$-axis measures model complexity and the $$y$$-axis measures the components of risk - model bias squared and model variance. Notice how as model complexity increases, model bias decreases and model variance increases. Thinking in terms of test error, we have seen this error first decrease and then increase as the model variance outweighs the decrease in model bias. To select a useful model, we must strike a balance between model bias and variance.

Fig. 16.4 Bias-variance Trade-off Diagram. As model complexity increases, model variance increases and model bias decreases. In the other direction, model variance decreases and model bias increases as model complexity decreases.

Collecting more observations reduces bias if the model can fit the population process exactly. If the model is inherently incapable of modeling the population (as in the example above), even infinite data cannot get rid of model bias. In terms of variance, collecting more data reduces variance. One recent trend is to select a model with low bias and high intrinsic variance (such as a neural network) but collect many data points so that the model variance is low enough to make accurate predictions. While effective in practice, collecting enough data for these models tends to require large amounts of time and money.

Creating more features, whether useful or not, typically increases model variance. Models with many parameters have many possible combinations of parameters and therefore have higher variance than models with few parameters. On the other hand, adding a useful feature to the data, such as a quadratic feature when the underlying process is quadratic, reduces bias, but even adding a useless feature rarely increases bias.

Being aware of the bias-variance tradeoff can help… And techniques of train-test, cv, and reg are designed to ameliorate this issue.


text:
## Constants & Units

### Physical Constants

$\quad\rlap{a_0}$ $\qquad\rlap{\rm Bohr~radius}$ $\hskip150pt 5.29177 \times 10^{-9} {\rm ~cm}$
$\quad\rlap{c}$ $\qquad\rlap{\rm speed~of~light~in~vacuum}$ $\hskip150pt 2.99792 \times 10^{10}{\rm ~cm~s}^{-1}$
$\quad\rlap{e}$ $\qquad\rlap{\rm electron~charge}$ $\hskip140pt -4.80325 \times 10^{-10} {\rm ~statcoulomb~(=esu)}$
$\quad\rlap{\rm eV}$ $\qquad\rlap{\rm electron~volt}$ $\hskip150pt 1.60218 \times 10^{-12} {\rm ~erg}$
$\quad\rlap{G}$ $\qquad\rlap{\rm gravitational~constant}$ $\hskip150pt 6.67428 \times 10^{-8} {\rm ~dyne~cm}^2 {\rm ~g}^{-2}$
$\quad\rlap{h}$ $\qquad\rlap{\rm Planck's~constant}$ $\hskip150pt 6.62607 \times 10^{-27} {\rm ~erg~s}$
$\quad\rlap{k}$ $\qquad\rlap{\rm Boltzmann's~constant}$ $\hskip150pt 1.38065 \times 10^{-16} {\rm ~erg~K}^{-1}$
$\quad\rlap{m_{\rm e}}$ $\qquad\rlap{\rm electron~mass}$ $\hskip150pt 9.10938 \times 10^{-28} {\rm ~g}$
$\quad\rlap{m_{\rm p}}$ $\qquad\rlap{\rm proton~mass}$ $\hskip150pt 1.67262 \times 10^{-24} {\rm ~g}$
$\quad\rlap{R_\infty}$ $\qquad\rlap{\rm Rydberg~constant}$ $\hskip150pt 1.09737 \times 10^5 {\rm ~cm}^{-1}$
$\quad\rlap{R_\infty c}$ $\qquad\rlap{\rm Rydberg~frequency}$ $\hskip150pt 3.28984 \times 10^{15} {\rm ~s}^{-1}$
$\quad\rlap{\sigma}$ $\qquad\rlap{\rm Stefan-Boltzmann~constant}$ $\hskip150pt 5.67040 \times 10^{-5} {\rm ~erg~cm}^{-2}{\rm ~s}^{-1} {\rm ~K}^{-4}$
$\quad\rlap{\sigma_{\rm T}}$ $\qquad\rlap{\rm Thomson~cross~section}$ $\hskip150pt 6.65245 \times 10^{-25} {\rm ~cm}^2$
$\quad\rlap{u}$ $\qquad\rlap{\rm atomic~mass~unit}$ $\hskip150pt 1.66054 \times 10^{-24} {\rm ~g}$

### Astronomical Constants

$\quad\rlap{\rm au}$ $\qquad\rlap{\rm astronomical~unit}$ $\hskip150pt 1.49598 \times 10^{13} {\rm ~cm}$
$\quad\rlap{H_0}$ $\qquad\rlap{\rm Hubble~constant}$ $\hskip150pt 72 {\rm ~km~s}^{-1}{\rm ~Mpc}^{-1}$
$\quad\rlap{\rm kpc}$ $\qquad\rlap{\rm kiloparsec}$ $\hskip150pt 10^3 {\rm ~pc}$
$\quad\rlap{L_\odot}$ $\qquad\rlap{\rm solar~bolometric~luminosity}$ $\hskip150pt 3.826 \hphantom{01~} \times 10^{33} {\rm ~erg~s}^{-1}$

$\quad\rlap{\rm ly}$ $\qquad\rlap{\rm light~year}$ $\hskip150pt 9.4605 \hphantom{0~}\times 10^{17} {\rm ~cm}$
$\quad\rlap{M_\odot}$ $\qquad\rlap{\rm solar~mass}$ $\hskip150pt 1.989 \hphantom{01~} \times 10^{33} {\rm ~g}$
$\quad\rlap{\rm Mpc}$ $\qquad\rlap{\rm megaparsec}$ $\hskip150pt 10^6 {\rm ~pc}$
$\quad\rlap{\rm pc}$ $\qquad\rlap{\rm parsec}$ $\hskip150pt 3.0856 \hphantom{0~} \times 10^{18} {\rm ~cm}$
$\quad\rlap{R_\odot}$ $\qquad\rlap{\rm solar~radius}$ $\hskip150pt 6.9598 \hphantom{0~} \times 10^{10} {\rm ~cm}$
$\quad\rlap{\rm yr}$ $\qquad\rlap{\rm year}$ $\hskip150pt 3.156 \times 10^7 \approx 10^{7.5} {\rm ~s}$

### MKS and Gaussian CGS Units

$\quad\rlap {\rm Type}$ $\hskip60pt \rlap{\rm mks~unit}$ $\hskip50pt \rlap{\rm cgs~unit}$ $\hskip60pt {\rm conversion}$

$\quad\rlap {\rm mass}$ $\hskip60pt \rlap{\rm kg}$ $\hskip50pt \rlap{\rm g}$ $\hskip60pt 10^3 {\rm ~g} = 1 {\rm ~kg}$
$\quad\rlap {\rm length}$ $\hskip60pt \rlap{\rm m}$ $\hskip50pt \rlap{\rm cm}$ $\hskip60pt 10^2 {\rm ~cm} = 1 {\rm ~m}$
$\quad\rlap {\rm time}$ $\hskip60pt \rlap{\rm s}$ $\hskip50pt \rlap{\rm s}$
$\quad\rlap {\rm frequency}$ $\hskip60pt \rlap{\rm Hz}$ $\hskip50pt \rlap{\rm Hz}$ $\hskip60pt 1 {\rm ~Hz} = 1 {\rm ~s}^{-1}$
$\quad\rlap {\rm charge}$ $\hskip60pt \rlap{\rm coulomb}$ $\hskip50pt \rlap{\rm statcoulomb}$ $\hskip60pt 3 \times 10^9 {\rm ~statcouloumb} = 1 {\rm ~coulomb}$
$\quad\rlap {\hphantom{\rm charge}}$ $\hskip60pt \rlap{\hphantom{\rm coulomb}}$ $\hskip50pt \rlap{\hphantom{\rm statcoulomb}}$ $\hskip60pt \hskip15pt (1 {\rm ~statcoulomb} = 1 {\rm ~esu})$
$\quad\rlap {\rm current}$ $\hskip60pt \rlap{\rm ampere}$ $\hskip50pt \rlap{\rm statampere}$ $\hskip60pt 3 \times 10^9 {\rm ~statamp} = 1 {\rm ~amp}$
$\hskip60pt \hskip15pt \hskip120pt$ $(1 {\rm ~amp} = 1 {\rm ~coulomb~s}^{-1})$

$\quad\rlap {\rm electric~field}$ $\hskip60pt \rlap{{\rm v~m}^{-1}}$ $\hskip50pt \rlap{{\rm statvolt~cm}^{-1}}$ $\hskip60pt (1/3) \times 10^{-4}{\rm ~statvolt~cm}^{-1} = 1 {\rm ~v~m}^{-1}$
$\quad\rlap {\rm energy}$ $\hskip60pt \rlap{\rm joule}$ $\hskip50pt \rlap{\rm erg}$ $\hskip60pt 10^7 {\rm ~erg} = 1 {\rm ~joule}$
$\quad\rlap {\rm force}$ $\hskip60pt \rlap{\rm newton}$ $\hskip50pt \rlap{\rm dyne}$ $\hskip60pt 10^5 {\rm ~dyne} = 1 {\rm ~newton}$

$\quad\rlap {\rm magnetic~field}$ $\hskip60pt \rlap{\rm tesla}$ $\hskip50pt \rlap{\rm gauss}$ $\hskip60pt 10^4 {\rm ~gauss} = 1 {\rm ~tesla}$
$\quad\rlap{\rm resistance}$ $\hskip60pt \rlap{\rm Ohm}$ $\hskip50pt \rlap{{\rm sec~cm}^{-1}}$ $\hskip60pt (1/9) \times 10^{-11} {\rm ~s~cm}^{-1} = 1 {\rm ~Ohm}$
$\quad\rlap{\rm temperature}$ $\hskip60pt \rlap{\rm Kelvin}$ $\hskip50pt {\rm Kelvin}$

### Other Constants and Units

$\quad\rlap {\rm arcmin}$ $\hskip60pt 1/60 {\rm ~deg}$
$\quad\rlap {\rm arcsec}$ $\hskip60pt 1/60 {\rm ~arcmin}$
$\quad\rlap {\rm Angstrom}$ $\hskip60pt 10^{-10} {\rm ~m}$
$\quad\rlap {\rm dB}$ $\hskip60pt 0.1\log_{10}(P_1/P_2)$

$\quad\rlap {\rm e}$ $\hskip60pt 2.71828\dots$
$\quad\rlap {\rm GHz}$ $\hskip60pt 10^9 {\rm ~Hz}$
$\quad\rlap {\rm Jy}$ $\hskip60pt 10^{-26} {\rm ~W~m}^{-2} {\rm ~Hz}^{-1} \quad {\rm (1~mJy} = 10^{-3} {\rm ~Jy,~1~}\mu{\rm Jy} = 10^{-6} {\rm ~Jy)}$
$\quad\rlap {\rm MHz}$ $\hskip60pt 10^6 {\rm ~Hz}$
$\quad\rlap {\mu{\rm m}}$ $\hskip60pt 10^{-6} {\rm ~m}$
$\quad\rlap \pi$ $\hskip60pt 3.14159\dots$
$\quad\rlap {\rm radian}$ $\hskip60pt (180/\pi) {\rm ~deg} \approx 206264.8 {\rm ~arcsec}$

Engineers and physicists prefer mks (meter, kilogram, second) units, so most radio astronomers use mks units to describe their equipment and the results of their observations.  Most astrophysicists prefer Gaussian cgs (centimeter, gram, second) units to describe astronomical sources.  Thus you have to deal with both systems of units and be able to convert between them in order to do astronomy.  J. D. Jackson's Classical Electrodynamics has an appendix explaining the different systems in detail.


text:
A good way of resetting the simulation is to make use of the /gazebo/reset_world service provided by gazebo_ros_packages.
Also I see you are using the SITL plugin.. maybe you could find some useful code related to the erlecopter in gym-gazebo. Check folders gym_gazebo/envs/erlecopter and gym-gazebo/examples/erlecopter/. Note that this is not under maintenance anymore, but you can just take a look.


text:
# Count points on elliptic curve secp256k1 with another p

Example:

secp192k1':

_ECData( p=2192 - 232 - 212 - 28 - 27 - 26 - 2**3 - 1, a=0, b=3,
Gx=0xDB4FF10EC057E9AE26B07D0280B7F4341DA5D1B1EAE06C7D,
Gy=0x9B2F2F6D9C5628A7844163D015BE86344082AA88D95E2F9D,
n=0xFFFFFFFFFFFFFFFFFFFFFFFE26F2FC170F69466A74DEFD8D

I have another curve (with another p(!), similar base point and equation ) parameters.

I need to calculate n

How can I do it with code in sage, python or another software please?

Br

edit retag close merge delete

It seems there are formatting problems in the way your question was entered, can you fix them?

( 2020-07-30 22:48:09 +0200 )edit

I edited the lines in the question, marking the many letters that may be part of a pseudo-code as code, but we still have no well defined question. To have a well defined question, we need the "other" p and the parameters for the "new" curve. For small values of p, sage can compute via E.order() the number of $\Bbb F_p$-rational points, e.g.

sage: E = EllipticCurve(GF(7), [3,4])
sage: E.order()
10

but for greater values, we have to invest mathematical effort to find the "order". (Of $E(\Bbb F_p)$.) Please fill in all needed defining data. And best give references / links.

( 2020-08-03 17:57:58 +0200 )edit

Possibly related

But for anything to happen here, can you please edit your post into a better formed question?

( 2020-08-03 18:19:22 +0200 )edit


text:
# How do you divide 7 4/5 div 1 1/4?

Oct 26, 2016

$= 6 \frac{6}{25}$

#### Explanation:

so first we convert to improper fractions,

$7 \frac{4}{5} \div 1 \frac{1}{4}$

$= \frac{39}{5} \div \frac{5}{4}$

dividing by a fraction is the same as multiplying by the inverse of the fraction so,

$= \frac{39}{5} \cdot \frac{4}{5}$

so simplifying,

$= \frac{39 \cdot 4}{25}$

$= \frac{156}{25}$

turning back into a mixed fraction,

$= 6 \frac{6}{25}$


text:
# Question about linear momentum and angular

Tags:
1. Nov 22, 2014

### Queren Suriano

1. The problem statement, all variables and given/known data
If an analysis of momentum and impulse, a rigid body has initial linear and angular momentum zero, does this mean that the vector sum of the impulses and the vector of LINEAR momentum final, will be in the same direction?

2. Relevant equations
Systema moment 1 + System Extern Impulses (1-->2) = Sistem Momenta 2

3. The attempt at a solution
I supposed that, because they are vectors, right??

2. Nov 25, 2014

### OldEngr63

If we let L denote the linear momentum and I the linear impulse (both vector quantities), then
L2 - L1 = I
This is a vector equation. If (for your problem) L1 = 0, then this reduces to
L2 = I
which says that both vectors must have the same direction.

Angular momentum is not involved at all here. Why did you mention it in the original question?

3. Nov 27, 2014

### Queren Suriano

Ok, thank you!! I was talking about angular momentum becasue it's included in an analysis for solve a problem where a rigid body is in translation and rotation.