Let $g_m$ be the gravitational acceleration of the moon. We know that due to the law of energy conservation, kinetic energy (and speed) of the rock when being thrown upwards from the surface and when it returns to the surface is the same. Given that $g_m$ stays constant, we can conclude that the time it takes to reach its highest point, aka 0 velocity, is the same as the time it takes to fall down from that point to the surface, which is half of the total time, or 4 / 2 = 2 seconds.

So essentially it takes 2s to decelerate from 3.2 m/s to 0. We can use this information to calculate $g_m$

$g_m = \frac{\Delta v}{\Delta t} = \frac{0 - 3.2}{2} = \frac{-3.2}{2} = -1.6 m/s^2$

So the gravitational acceleration on the Moon is 1.6 m/s2

8 0

3 years ago

A 3.0 kg ball is lifted two meters upward in two seconds, and a 6.0 kg ball is lifted one meter upward in one second? Which requ

Sav [38]

Answer:

1. They both uses same energy

2. The 6 kg ball requires more power than 3kg ball

Explanation:

Sample 1

m = 3kg

g= 10m/s^2

h = 2m

t = 2secs

W = mgh = 3 x 10 x 2 = 60J

P= w/t = 60/2 = 30watts

Sample 2

m = 6kg

g= 10m/s^2

h = 1m

t = 1sec

W = mgh = 6 x 10 x 1 = 60J

P= w/t = 60/1 = 60watts

They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball

3 0

3 years ago