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3 years ago

11

I would appreciate some help.

1 answer:

motikmotik3 years ago

4 0

You oughtta be able to do this one.

The "efficiency" is just the portion of the input work that
comes out in a useful form.

If the efficiency is 70%, that tells you that however much work
you put INto the machine, the machine will do 70% of that much
work for you at the output side.

Put 20,000 J in ... out comes (0.70) x (20,000 J) = 14,000 J .

What happens to the other 30% of the work you put into it ?
It turns into HEAT. That's why machines always have to be
cooled somehow while they're running.

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From the question we are told that

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$\varphi_2= 0 N m^2/C$

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$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

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