Answer:
The acceleration of each ball while in motion is acceleration due to gravity, that is 9.81 m/s²
Explanation:
Here one child simply drops a ball. At the same time, the second child throws a ball downward so that it has an initial speed of 10 m/s.
In first case acceleration on the ball is acceleration due to gravity,
In second case also acceleration on the ball is acceleration due to gravity,
Only gravitational force is acting on both cases so their acceleration is acceleration due to gravity.
The acceleration of each ball while in motion is acceleration due to gravity, that is 9.81 m/s²
Answer:
yes
Explanation:
sound can only vibrate of particles.
Answer:
![\frac{a_{r,earth}}{a_{r,mars}} = 2.325](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7Br%2Cearth%7D%7D%7Ba_%7Br%2Cmars%7D%7D%20%3D%202.325)
Explanation:
The distance of Earth from the Sun is
and of Mars from the Sun is
. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:
![a_{r} = \frac{v^{2}}{R}](https://tex.z-dn.net/?f=a_%7Br%7D%20%3D%20%5Cfrac%7Bv%5E%7B2%7D%7D%7BR%7D)
Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:
![v=\frac{2\pi\cdot r}{\Delta t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%5Cpi%5Ccdot%20r%7D%7B%5CDelta%20t%7D)
Earth:
![v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}](https://tex.z-dn.net/?f=v_%7Bearth%7D%20%3D%20%5Cfrac%7B2%5Cpi%5Ccdot%20%28149.6%5Ctimes%2010%5E%7B9%7D%5C%2Cm%29%7D%7B%28365%5C%2Cdays%29%5Ccdot%28%5Cfrac%7B24%5C%2Chours%7D%7B1%5C%2Cday%7D%20%29%5Ccdot%28%5Cfrac%7B3600%5C%2Cs%7D%7B1%5C%2Ch%7D%20%29%7D)
![v_{earth}=29806.079\,\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bearth%7D%3D29806.079%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Mars:
![v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}](https://tex.z-dn.net/?f=v_%7Bmars%7D%20%3D%20%5Cfrac%7B2%5Cpi%5Ccdot%20%28227.9%5Ctimes%2010%5E%7B9%7D%5C%2Cm%29%7D%7B%28687%5C%2Cdays%29%5Ccdot%28%5Cfrac%7B24%5C%2Chours%7D%7B1%5C%2Cday%7D%20%29%5Ccdot%28%5Cfrac%7B3600%5C%2Cs%7D%7B1%5C%2Ch%7D%20%29%7D)
![v_{mars}=24124.244\,\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bmars%7D%3D24124.244%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Now, centripetal accelarations can be found:
Earth:
![a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}](https://tex.z-dn.net/?f=a_%7Br%2Cearth%7D%20%3D%20%5Cfrac%7B%2829806.079%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%7D%7B149.6%5Ctimes%2010%5E%7B9%7D%5C%2Cm%7D)
![a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}](https://tex.z-dn.net/?f=a_%7Br%2Cearth%7D%20%3D%205.939%5Ctimes%2010%5E%7B-3%7D%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
Mars:
![a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}](https://tex.z-dn.net/?f=a_%7Br%2Cmars%7D%20%3D%20%5Cfrac%7B%2824124.244%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%7D%7B227.9%5Ctimes%2010%5E%7B9%7D%5C%2Cm%7D)
![a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}](https://tex.z-dn.net/?f=a_%7Br%2Cmars%7D%20%3D%202.554%5Ctimes%2010%5E%7B-3%7D%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:
![\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7Br%2Cearth%7D%7D%7Ba_%7Br%2Cmars%7D%7D%20%3D%20%5Cfrac%7B5.939%7D%7B2.554%7D)
![\frac{a_{r,earth}}{a_{r,mars}} = 2.325](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7Br%2Cearth%7D%7D%7Ba_%7Br%2Cmars%7D%7D%20%3D%202.325)
Answer:
a)![E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BQ%7D%7B%5Cvarepsilon%20_o%5Ctimes%204%5Cpi%20r%5E2%7D%5C%20N%2FC)
b)E=0
c)![E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BQ%7D%7B%5Cvarepsilon%20_o%5Ctimes%204%5Cpi%20r%5E2%7D%5C%20N%2FC)
Explanation:
Given that
A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .
a) r < a
At a radius r ,from gauss theorem
![E.ds=\dfrac{q_i}{\varepsilon _o}](https://tex.z-dn.net/?f=E.ds%3D%5Cdfrac%7Bq_i%7D%7B%5Cvarepsilon%20_o%7D)
![E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}](https://tex.z-dn.net/?f=E%5Ctimes%204%5Cpi%20r%5E2%3D%5Cdfrac%7BQ%7D%7B%5Cvarepsilon%20_o%7D)
![E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BQ%7D%7B%5Cvarepsilon%20_o%5Ctimes%204%5Cpi%20r%5E2%7D%5C%20N%2FC)
b) a < r < b
![E.ds=\dfrac{q_i}{\varepsilon _o}](https://tex.z-dn.net/?f=E.ds%3D%5Cdfrac%7Bq_i%7D%7B%5Cvarepsilon%20_o%7D)
The total induce in this surface = - Q+ Q =0
![E.ds=\dfrac{0}{\varepsilon _o}](https://tex.z-dn.net/?f=E.ds%3D%5Cdfrac%7B0%7D%7B%5Cvarepsilon%20_o%7D)
E = 0
c) r > b
![E.ds=\dfrac{q_i}{\varepsilon _o}](https://tex.z-dn.net/?f=E.ds%3D%5Cdfrac%7Bq_i%7D%7B%5Cvarepsilon%20_o%7D)
![E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}](https://tex.z-dn.net/?f=E%5Ctimes%204%5Cpi%20r%5E2%3D%5Cdfrac%7BQ%7D%7B%5Cvarepsilon%20_o%7D)
![E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BQ%7D%7B%5Cvarepsilon%20_o%5Ctimes%204%5Cpi%20r%5E2%7D%5C%20N%2FC)