63 142
Assignments Done
99,1%
Successfully Done
In July 2018

Answer to Question #4888 in Mechanics | Relativity for lizzeth

Question #4888
A cheerleader lifts his 38.6 kg partner straight up off the ground a distance of 0.596 m before releasing her. 2
The acceleration of gravity is 9.8 m/s .
If he does this 28 times, how much work has he done?
Expert's answer
The work of doing that 1 time is:
A1 = F*h = mgh.
Then the total work
is:
A = 28*A1 = 28mgh = 28*38.6*9.8*0.596 = 6312,74 J.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions