How much force did the gymnast use to climb the rope?
Answer:
-39.2m/s
Explanation:
Given that :
t = 4secs
g = -9.8m/s^2
v = ?
u = 0m/s ( since it was at rest )
V = u +at............. 1
Where v is the final velocity
a = -g = -9.8m/s^2 since the ball was dropped from a height which will eventually make it move against gravity
t = 4secs
Substitute the values into 1
v = 0 - 9.8×4
v = -39.2m/s
Answer:
The speed of the banana just before it hits the water is:
√(2 · g · h) = v
Explanation:
Hi there!
Before Emily throws the banana, its potential energy is:
PE = m · g · h
Where:
PE = potential energy.
m = mass of the banana.
g = acceleration of the banana due to gravity.
h = height of the bridge (distance from the bridge to the ground).
When the banana reaches the water, all its potential energy will have converted to kinetic energy. The equation for kinetic energy is as follows:
KE = 1/2 · m · v²
Where:
KE = kinetic energy.
m = mass of the banana.
v = speed.
Then, when the banana hits the water:
m · g · h = 1/2 · m · v²
multiply by 2 and divide by m both sides of the equation:
2 · g · h = v²
√(2 · g · h) = v
<span>when it returns to its original level after encountering air resistance, its kinetic energy is
decreased.
In fact, part of the energy has been dissipated due to the air resistance.
The mechanical energy of the ball as it starts the motion is:
</span>

<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>