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viva [34]
3 years ago
8

What percentage of the earth's mass is the moon's mass?

Physics
1 answer:
Schach [20]3 years ago
5 0

Answer:

ssssssssss

Explanation:

hshjdjjfjjjjhhddd

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Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to
Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x (\frac{Rc}{Rc + Ri})

where

  • Rc = resistance of copper = \frac{ρl}{a}  (l = length , a = area, ρ = resistivity of copper)
  • Ri = resistance of iron = \frac{ρ₀l}{a}  (l = length , a = area, ρ₀ = resistivity of copper)

Vc =  V x (\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}})

Vc = V x (\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})})

Vc = V x (\frac{ρ}{ρ + ρ₀})

where

  • ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
  • ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x (\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}})

Vc = 2.41 v

5 0
3 years ago
PLEASE HELP NO LINKS NEED HELP FAST
MrRa [10]

Answer:

The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you

Explanation:

7 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
Which statements accurately describes density?
lawyer [7]
<span>the degree of compactness of a substance.

</span>
4 0
3 years ago
Which of the following CANNOT happen when an EM wave strikes a surface? A. Transmission B. Absorption C. Multiplication D. Refle
kobusy [5.1K]
D because there no way it will reflect making the way come back
3 0
3 years ago
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