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DaniilM [7]
4 years ago
13

A tow truck exerts a force of 8000 N to move a car a distance of 4 m in 20 seconds. Calculate the power of the tow truck..

Physics
2 answers:
jenyasd209 [6]4 years ago
8 0

Answer:

Power  = 1600Watts

Explanation:

Power is given by the rate of energy expended or work done by a body (or object).

Power = \frac{Workdone}{time}

Where;

Workdone = Force x distance

=> Power = Force x distance / time   -----------------------(i)

<em>According to the question,</em>

Force = 8000N

distance = 4m

time = 20 s

<em>Substitute these values into equation (i) above</em>

=> Power = 8000 x 4 / 20

=> Power = 1600Watts

Therefore the  power of the tow truck is 1600Watts

olga_2 [115]4 years ago
6 0
3. 1600 Do the math on the calculator
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0°

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The angle between force and displacement should be 0° in order to get the maximum work done.

 

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  Ф is the angle between force and displacement

When Ф is zero, the maximum work done is attained

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A and B

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A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h
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Explanation:

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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

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