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galina1969 [7]
3 years ago
11

Can someone help me with b, d and e? Please need this today

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

Explanation:

a) the free body diagram of the lower pulley would have two force vectors of equal magnitude acting upward and the block weight acting downward. These must balance so 2F = Wt

                                          F = 262/2 = 131 N

b) as the force needed is half of the block weight, the mechanical advantage is 262 / 131 = 2

c) with a mechanical advantage of 2, he would have to pull twice the distance of block motion or 2(3.00) = 6.00 m

d) W = Fd = 131 N(6.00 m) = 786 J

which is the same as the increase in potential energy of the block

e) W = mgh = 262(3.00) = 786 J

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To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
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Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
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Explanation:

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Explanation:

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