The period of a pendulum is given by
![T=2 \pi \sqrt{ \frac{L}{g} }](https://tex.z-dn.net/?f=T%3D2%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg%7D%20%7D%20)
where L is the pendulum length and g is the gravitational acceleration.
We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
![\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_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%5Cfrac%7B2%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg_m%7D%20%7D%20%7D%7B2%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg_e%7D%20%7D%20%7D%3D%20%20%20%20%5Csqrt%7B%20%5Cfrac%7Bg_e%7D%7Bg_m%7D%20%7D%20)
where the labels m and e refer to "Moon" and "Earth".
Since the gravitational acceleration on Earth is
![g_e = 9.81 m/s^2](https://tex.z-dn.net/?f=g_e%20%3D%209.81%20m%2Fs%5E2)
while on the Moon is
![g_m=1.63 m/s^2](https://tex.z-dn.net/?f=g_m%3D1.63%20m%2Fs%5E2)
, the ratio between the period on the Moon and on Earth is
Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
Therefore, the relationship between force and motion is that it takes force to change the speed or direction of any object in motion.
V=IR
The more V, the more I
I need a picture plz I don’t know what to answer.