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Taya2010 [7]
3 years ago
13

Is the facceleration due to gravity the same on Earth and Earth's moon?

Physics
1 answer:
Alchen [17]3 years ago
3 0

Answer:

Earth's average surface gravity is about 9.8 meters per second per second. ... The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass but inversely proportional to the square of its radius.

Explanation:

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12. AABC is a right triangle. If AB = 3 and AC = 7, find BC. Leave your answer in simplest radical form.
Dennis_Churaev [7]

Answer: A 2 square root 3

Explanation:

4 0
3 years ago
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
You decide to impress Grandpa by showing him how fast sound travels. You have a piece of plastic pipe with an adjustable closed
fredd [130]

Answer:

336.96m/s

Explanation:

answer is in photo above

3 0
3 years ago
What is the equivalent resistance in this circuit?<br><br> What is the current in this circuit?
kramer

All these resistors are in series so we can take the sum of them by:

Rtotal = R1 + R2 + R3......

So...

Rtotal = 2 + 3 + 4 + 6

Rtotal = 15

So now the total resistance in the circuit is 15 ohms and the potential difference applied to the circuit is 45 volts

Now we can use:

V = IR

Isolate for I

V/R = I

45/15 = I

I = 3 amps (A)

3 0
3 years ago
Read 2 more answers
How much work will a 500 watt motor do in 10 seconds?
Viktor [21]

Answer:

50j

Explanation:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

4 0
2 years ago
Read 2 more answers
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