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Vera_Pavlovna [14]
4 years ago
7

If the total dissipated power is to be reduced by 10%, how much should the voltage be reduced to maintain the same leakage curre

nt? note: power is defined as the product of voltage and current.
Physics
1 answer:
harkovskaia [24]4 years ago
6 0
<span>Since P = V x I, a 10% reduction of power would lead to a 10% reduction in the product of voltage and current. What is left is 90% of the original power: .9P = .9(V x I). If we assume that current must be the same, then we can regroup the terms on the right-hand side as follows: .9P = (.9V) x I In this case, voltage is also reduced by 10% (100% - 90% = 10%).</span>
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A girl lifts a 160-N load to a height of 1 min 0.5 s. How much power is used to lift the load?
Aleonysh [2.5K]

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Explanation:

P = W/t

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t = 0.5 s

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8 0
2 years ago
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How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
solong [7]

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71 rpm

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5 0
3 years ago
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anastassius [24]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
\rho =  \frac{AR}{L}
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\rho is the resistivity
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R is the resistance
L is the wire length

the cross-sectional area is given by
A=\pi r^2
where r is the radius of the wire. Substituting in the previous equation ,we find
\rho =  \frac{\pi r^2 R}{L}

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
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5 0
3 years ago
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