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Kay [80]
3 years ago
13

Which material produces static charge when rubbed?

Physics
1 answer:
natita [175]3 years ago
5 0
Many materials produce static charge

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Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is
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For a body to be in SHM, the restoring force must be __________ to displacement from equilibrium.
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Proportional

Explanation:

The conditions that must be met to produce SHM are;

-The restoring force must be proportional to the displacement and act opposite to the direction of motion with no drag forces or friction.

- The frequency of oscillation does not depend on the amplitude.

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How do you calculate average speed?
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Divide
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by
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Although all sports require a person to be in shape to some degree, some sports demand more from the body. Which sport does not
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I  think golf does not demand high level of physical fitness. A golfer just need be skillful and able to play under pressure. In terms of physical exercise i think he/she just need to do some muscle stretches.

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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
enot [183]

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.

7 0
3 years ago
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