Physics (Electricity)

Physics

2 answers:

Georgia [21]3 years ago

7 0

Here are the 4 parts a-d.

Elanso [62]3 years ago

5 0

8(a) charges at A and B are 4q(of which its initially q and later 3q is added) and q respectively.

answer is $4*4*10^{-5}$

You might be interested in

The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively, and their centers are separated by 3.84×108m.

bija089 [108]

Answer:

C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.

Explanation:

At first we assume that both Earth and Moon can be treated as particles, the center of mass of the Earth-Moon system is obtained by using this formula:

$r_{CM} = \frac{r_{E}\cdot m_{E}+r_{M}\cdot m_{M}}{m_{E}+m_{M}}$ (Eq. 1)

Where:

$r_{E}$ - Location of the center of the Earth, measured in kilometers.

$r_{M}$ - Location of the Moon, measured in kilometers.

$r_{CM}$ - Location of the center of mass, measured in kilometers.

$m_{E}$ - Mass of the Earth, measured in kilograms.

$m_{M}$ - Mass of the Moon, measured in kilograms.

If we know that $r_{E} = 0\,km$ , $r_{M} = 3.84\times 10^{8}\,m$ , $m_{E} = 5.98\times 10^{24}\,kg$ and $m_{M} = 7.35\times 10^{22}\,kg$ , the location of the center of mass respect to the Earth is:

$r_{CM} = \frac{(0\,km)\cdot (5.98\times 10^{22}\,kg)+(3.84\times 10^{8}\,m)\cdot (7.35\times 10^{22}\,kg)}{5.98\times 10^{24}\,kg+7.35\times 10^{22}\,kg}$

$r_{CM} = 4.662\times 10^{6}\,m$

The Earth has a radius of $6.371\times 10^{6}$ meters, we notice that center of mass in located inside the Earth and the radius of rotation of the Earth around the center of mass is much greater than the radius of rotation of the Moon around the center of mass. That center of mass follows an orbit around the sun.