Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe
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2.57 joule energy lose in the bounce .
<u>Explanation</u>:
when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground
Formula for gravitational potential energy given by
Potential Energy = mgh
Where ,
m = mass
g = acceleration due to gravity
h = height
Potential energy when ball hits the ground
m= 0.375 kg
h = 1.37 m
g = 9.8 m/s²
Potential Energy = 5.03 joule
Potential energy when ball bounces up again
h= 0.67 m
Potential Energy = 2.46 joule
Energy loss = 5.03 - 2.46 = 2.57 joule
2.57 joule energy lose in the bounce
The equilibrant force of the two given forces is 14.14 N.
<h3 /><h3 /><h3>What is equilibrant force?</h3>- This is a single force that balances other given forces.
The given parameters:
- First force, F₁ = 10 N
- Second force, F₂ = 10 N
- Angle between the forces, θ = 90⁰
The equilibrant force of the two given forces is calculated as follows;
Thus, the equilibrant force of the two given forces is 14.14 N.
Learn more about equilibrant force here: brainly.com/question/8045102