Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Answer:
The further implementations of nuclear power are limited because although nuclear energy does not produce CO2 the way fossil fuels do, there is still a toxic byproduct produced from uranium-fueled nuclear cycles: radioactive fission waste.
Explanation:
Answer:
If you are using a machine that has a mechanical advantage of 2.5,you will have increased your EFFORT DISTANCE by 2.5 times. This allows you to REDUCE the EFFORT FORCE needed by 2.5 times.
Answer:

Explanation:
Given:
- diameter of pipe,

- diameter of throat,

- velocity of flow,

<u>Pressure in the pipe is twice the atmospheric pressure:</u>

<u>Now the force of flow of water:</u>

<u>now we find cross sectional area of the pipe:</u>



<u>Therefore,</u>


<u>Now the area at throat:</u>



<u>Therefore pressure at throat:</u>


