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4 years ago

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A 58-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of

-1 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her).

1 answer:

masha68 [24]4 years ago

5 0

Answer:

F = -72.5 N

Explanation:

You can do this excercise in two ways.

The first way is using the second law of Newton which is:

F = m*a

You should calculate the acceleration and then solve for F.

The second way is to use the formula of momentum which is:

F*t = m*V

Here, you only need to put the data of the excercise and solve for F.

Either way you do this, you'll get the same result.

I'll do it by both methods to prove it:

First method:

Assuming that the innitial speed is 0 (because is standing still) acceleration would be:

a = Vf/t

Solving for a:

a = -1/0.8 = -1.25 m/s^2

Now, let's solve for F:

F = 58 * (-1.25) = -72.5 N

Second method:

We know that F*t = m*V so:

F = m*V/t

F = 58 * (-1) / 0.8

F = -72.5 N

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3 years ago

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A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

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Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$

If the old current output was 425 kV, the ratio of current was:

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1$

Then, the ratio of the new output over the old output is:

$\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86$

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

$P_L=I^2\cdot R$

Then, the ratio of losses is (R is constant for both power losses):

$\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74$

The losses in the new line are 74% the losses in the old line.

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