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Sidana [21]
3 years ago
15

A cardiac defibrillator stores 1275 J of energy when it is charged to 5.6kV What is the capacitance? O 11.3 pF 96.1 pF O 813 F

Physics
1 answer:
Angelina_Jolie [31]3 years ago
5 0

Answer: 813.13(10)^{-7}F

Explanation:

The answer is not among the given options. However, this is a good example of the relation between the energy stored in a capacitor Wand its capacitance C, which is given by the following equation:

W=\frac{1}{2}CV^{2}   (1)

Where:

W=1275J

V=5.6kV=5.6(10)^{3}V is the voltage

C is the capacitance in Farads, the value we want to find

Isolating C from (1):

C=\frac{2W}{V^{2}}   (2)

C=\frac{2(1275J)}{(5.6(10)^{3}V)^{2}}   (3)

Finally:

C=0.000081313F=813.13(10)^{-7}F This is the capacitance of the cardiac defibrillator

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Answer:

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Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

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The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

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When you run sin^{-1}(1/5) through a calculator, you get about 11.537 degrees, or θ = 11.537.

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