Answer:
yes. why do you need this answered asap? lol
The period of a pendulum is given by

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:

where the labels m and e refer to "Moon" and "Earth".
Since the gravitational acceleration on Earth is

while on the Moon is

, the ratio between the period on the Moon and on Earth is
Answer:
a
Generally from third equation of motion we have that
![v^2 = u^2 + 2a[s_i - s_f]](https://tex.z-dn.net/?f=v%5E2%20%3D%20%20u%5E2%20%2B%202a%5Bs_i%20-%20s_f%5D%20)
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
=> 
b
Explanation:
Generally from third equation of motion we have that
![v^2 = u^2 + 2a[s_i - s_f]](https://tex.z-dn.net/?f=v%5E2%20%3D%20%20u%5E2%20%2B%202a%5Bs_i%20-%20s_f%5D%20)
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
=> 
When
we have that

=> 
=>
(A) We can solve the problem by using Ohm's law, which states:

where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know

and

, therefore we can rearrange Ohm's law to find the current through the device:

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:

where

is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
They all move through empty space at the speed of light