The period of a simple pendulum is given by
![T= 2 \pi \sqrt{ \frac{L}{g} }](https://tex.z-dn.net/?f=T%3D%202%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg%7D%20%7D%20)
where
L is the pendulum length
g is the acceleration of gravity
If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
![T_e= 2 \pi \sqrt{ \frac{L}{g_e} }](https://tex.z-dn.net/?f=T_e%3D%202%20%5Cpi%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg_e%7D%20%7D%20)
where
![g_e](https://tex.z-dn.net/?f=g_e)
is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
![T_m= 2 \pi \sqrt{ \frac{L}{g_m} }](https://tex.z-dn.net/?f=T_m%3D%202%20%5Cpi%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg_m%7D%20%7D%20)
where
![g_m](https://tex.z-dn.net/?f=g_m)
is the acceleration of gravity on the Moon.
If we do the ratio of the two periods, we get
![\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }](https://tex.z-dn.net/?f=%20%5Cfrac%7BT_m%7D%7BT_e%7D%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7Bg_e%7D%7Bg_m%7D%20%7D%20%20)
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write
![g_e = 6 g_m](https://tex.z-dn.net/?f=g_e%20%3D%206%20g_m)
and we can rewrite the previous ratio as
![\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}](https://tex.z-dn.net/?f=%5Cfrac%7BT_m%7D%7BT_e%7D%20%3D%20%5Csqrt%7B%20%5Cfrac%7B6%20g_m%7D%7Bg_m%7D%20%7D%3D%20%20%5Csqrt%7B6%7D%20)
so the period of the pendulum on the Moon is
I think the amount of force will decrease and the amount of work will increase
Circularity system........…….......
1 pound ≈ 0.4536 kg
170 pounds ≈ 170 * 0.4536 kg
≈ 77.112 kg
Answer: c.mesosphere
Explanation: Temperatures can reach lows of -90 degrees Celsius in the mesosphere, making it the coldest layer of the earth's atmosphere. The mesosphere extends above the stratosphere and ranges from 50 km to 87 km above the surface.