Answer:
It comes from the sun unevenly heating the earth. Creating warm and cool spots.
Explanation:
Answer:
We conclude that if a flat piece of paper or blowing ball is dropped from the same height, They would hit the ground at the same time.
Thus,
option a) is correct.
Explanation:
When there is no air in the space, it is basically a vacuum.
- In other words, the vacuum can be termed as a 'space devoid of matter'.
So, when there is no air in space, it means there is no force that can act on the body.
- Therefore, no matter what objects (heavy or light) we may drop from the same height, they would all hit the ground at the same time.
As there is no force that can act on an object in the vacuum, it means every object in the vacuum seems to stand still while falling.
Therefore, we conclude that if a flat piece of paper or blowing ball is dropped from the same height, They would hit the ground at the same time.
Thus,
option a) is correct.
Explanation:
It is known that relation between velocity and height is as follows.
v = ![\sqrt{2gh}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D)
where, g = acceleration due to gravity = 9.8 ![m/s^{2}](https://tex.z-dn.net/?f=m%2Fs%5E%7B2%7D)
h = height = 0.2 m
Therefore, velocity is calculated as follows.
v = ![\sqrt{2gh}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D)
= ![\sqrt{2 \times 9.8 \times 0.2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%20%5Ctimes%209.8%20%5Ctimes%200.2%7D)
= 3.92 m/s
Also,
![m_{people}v_{people} = m_{earth}v_{earth}](https://tex.z-dn.net/?f=m_%7Bpeople%7Dv_%7Bpeople%7D%20%3D%20m_%7Bearth%7Dv_%7Bearth%7D)
![v_{earth} = \frac{m_{people}v_{people}}{m_{earth}}](https://tex.z-dn.net/?f=v_%7Bearth%7D%20%3D%20%5Cfrac%7Bm_%7Bpeople%7Dv_%7Bpeople%7D%7D%7Bm_%7Bearth%7D%7D)
Putting the given values into the above formula as follows.
![v_{earth} = \frac{m_{people}v_{people}}{m_{earth}}](https://tex.z-dn.net/?f=v_%7Bearth%7D%20%3D%20%5Cfrac%7Bm_%7Bpeople%7Dv_%7Bpeople%7D%7D%7Bm_%7Bearth%7D%7D)
= ![\frac{6 \times 10^{9} \times 74 kg/person}{6 \times 10^{24} kg}](https://tex.z-dn.net/?f=%5Cfrac%7B6%20%5Ctimes%2010%5E%7B9%7D%20%5Ctimes%2074%20kg%2Fperson%7D%7B6%20%5Ctimes%2010%5E%7B24%7D%20kg%7D)
= ![\frac{444 \times 10^{9}}{6 \times 10^{24}}](https://tex.z-dn.net/?f=%5Cfrac%7B444%20%5Ctimes%2010%5E%7B9%7D%7D%7B6%20%5Ctimes%2010%5E%7B24%7D%7D)
=
m/s
or, =
m/s
Thus, we can conclude that recoil speed of the Earth is
m/s.
Answer:
The electrical force between the given charges remains the same.
Explanation:
The expression for the electrical force is as follows as;
![F=\frac{kq_{1}q_{2}}{R^{2}}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bkq_%7B1%7Dq_%7B2%7D%7D%7BR%5E%7B2%7D%7D)
Here, k is the constant,
are the charges, F is the electrical force and R is the distance between the charges.
It is given in the problem that the magnitudes of the charges and the magnitudes of the separation between the charges are doubled.
Then, the expression of the electrical force becomes as;
![F'=\frac{k(2q_{1})(2q_{2})}{(2R)^{2}}](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7Bk%282q_%7B1%7D%29%282q_%7B2%7D%29%7D%7B%282R%29%5E%7B2%7D%7D)
![F'=\frac{k4q_{1}q_{2}}{4R^{2}}](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7Bk4q_%7B1%7Dq_%7B2%7D%7D%7B4R%5E%7B2%7D%7D)
![F'=\frac{kq_{1}q_{2}}{R^{2}}](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7Bkq_%7B1%7Dq_%7B2%7D%7D%7BR%5E%7B2%7D%7D)
![F'=F](https://tex.z-dn.net/?f=F%27%3DF)
Therefore, the electrical force between the given charges remains the same.