# 0.2 Force, momentum and impulse  (Page 21/35)

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${a}_{o}=\frac{F}{{m}_{o}}$

So we substitute equation [link] into Equation  [link] , and we find that

${a}_{o}=\frac{G{m}_{e}}{{r}^{2}}$

Since it doesn't matter what ${m}_{o}$ is, this tells us that the acceleration on a body (due to the Earth's gravity) does not depend on the mass of the body. Thus all objects experience the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using $a$ we use $g$ which we call the gravitational acceleration.

## Comparative problems

Comparative problems involve calculation of something in terms of something else that we know. For example, if you weigh 490 N on Earth and the gravitational acceleration on Venus is 0,903 that of the gravitational acceleration on the Earth, then you would weigh 0,903 x 490 N = 442,5 N on Venus.

## Principles for answering comparative problems

• Write out equations and calculate all quantities for the given situation
• Write out all relationships between variable from first and second case
• Write out second case
• Substitute all first case variables into second case
• Write second case in terms of first case

A man has a mass of 70 kg. The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is 9,8 m $·$ s ${}^{-2}$ ?

1. The following has been provided:

• the mass of the man, $m$
• the mass of the planet Zirgon ( ${m}_{Z}$ ) in terms of the mass of the Earth ( ${m}_{E}$ ), ${m}_{Z}=2{m}_{E}$
• the radius of the planet Zirgon ( ${r}_{Z}$ ) in terms of the radius of the Earth ( ${r}_{E}$ ), ${r}_{Z}={r}_{E}$
2. We are required to determine the man's weight on Zirgon ( ${w}_{Z}$ ). We can do this by using:

$w=mg=G\frac{{m}_{1}·{m}_{2}}{{r}^{2}}$

to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Zirgon.

3. $\begin{array}{ccc}\hfill {w}_{E}& =& m{g}_{E}=G\frac{{m}_{E}·m}{{r}_{E}^{2}}\hfill \\ & =& \left(70\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right)\left(9,8\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\right)\hfill \\ & =& 686\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$
4. Write the equation for the gravitational force on Zirgon and then substitute the values for ${m}_{Z}$ and ${r}_{Z}$ , in terms of the values for the Earth.

$\begin{array}{ccc}\hfill {w}_{Z}=m{g}_{Z}& =& G\frac{{m}_{Z}·m}{{r}_{Z}^{2}}\hfill \\ & =& G\frac{2{m}_{E}·m}{{r}_{E}^{2}}\hfill \\ & =& 2\left(G\frac{{m}_{E}·m}{{r}_{E}^{2}}\right)\hfill \\ & =& 2{w}_{E}\hfill \\ & =& 2\left(686\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\right)\hfill \\ & =& 1\phantom{\rule{0.166667em}{0ex}}372\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$
5. The man weighs 1 372 N on Zirgon.

A man has a mass of 70 kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the Earth. Gravitational acceleration on Earth is 9,8 m $·$ s ${}^{-2}$ .

1. The following has been provided:

• the mass of the man on Earth, $m$
• the mass of the planet Beeble ( ${m}_{B}$ ) in terms of the mass of the Earth ( ${m}_{E}$ ), ${m}_{B}=\frac{1}{2}{m}_{E}$
• the radius of the planet Beeble ( ${r}_{B}$ ) in terms of the radius of the Earth ( ${r}_{E}$ ), ${r}_{B}=\frac{1}{4}{r}_{E}$
2. We are required to determine the man's weight on Beeble ( ${w}_{B}$ ). We can do this by using:

$w=mg=G\frac{{m}_{1}·{m}_{2}}{{r}^{2}}$

to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Beeble.

3. $\begin{array}{ccc}\hfill {w}_{E}& =& m{g}_{E}=G\frac{{m}_{E}·m}{{r}_{E}^{2}}\hfill \\ & =& \left(70\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right)\left(9,8\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\right)\hfill \\ & =& 686\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$
4. Write the equation for the gravitational force on Beeble and then substitute the values for ${m}_{B}$ and ${r}_{B}$ , in terms of the values for the Earth.

$\begin{array}{ccc}\hfill {w}_{B}=m{g}_{B}& =& G\frac{{m}_{B}·m}{{r}_{B}^{2}}\hfill \\ & =& G\frac{\frac{1}{2}{m}_{E}·m}{{\left(\frac{1}{4}{r}_{E}\right)}^{2}}\hfill \\ & =& 8\left(G\frac{{m}_{E}·m}{{r}_{E}^{2}}\right)\hfill \\ & =& 8{w}_{E}\hfill \\ & =& 8\left(686\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\right)\hfill \\ & =& 5\phantom{\rule{0.166667em}{0ex}}488\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \end{array}$
5. The man weighs 5 488 N on Beeble.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
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Sherica
right! what he said ⤴⤴⤴
Tamia
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yes
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I'm not good at math so would you help me
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many many of nanotubes
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
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not now but maybe in future only AgNP maybe any other nanomaterials
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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