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joja [24]
2 years ago
8

Please subscribe wonder melodies please on you tube channel​

Physics
2 answers:
Anuta_ua [19.1K]2 years ago
8 0

Sure , I'm subscribing!!!

Reika [66]2 years ago
3 0

Answer:

WTH

i thought this should be full with smarts ones/people

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Alenkasestr [34]
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)

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3 years ago
Does anyone know what uranium is???
schepotkina [342]
The answer to your question is "Yes".  Great numbers of people ... chemists,
physicists, and common folks alike ... know what Uranium is.

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