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Sunny_sXe [5.5K]
3 years ago
9

"A capacitor charged to 1.5 V stores 2.0 mJ of energy. Determine the energy stored in the capacitor if it is charged to 3.0 V."

Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

To solve this problem we will apply the concepts related to electric potential energy. This relationship shows us that energy is equivalent to half the product between the capacitance of the body by the squared voltage. Mathematically it can be expressed as,

E = \frac{1}{2} CV^2

Here,

C = Capacitance

V = Voltage

E = Energy

The initial capacitance would then be given by

2mJ = \frac{1}{2} C (1.5)^2

C = \frac{4*10^{-3}}{(1.5)^2}

C = 1.78*10^{-3}F

Energy stored in the capacitor with 3V is

E = \frac{1}{2} (1.78*10^{-3})(3)^2

E = 8mJ

Therefore the energy stored in the capacitor if it is charged to 3V is 8mJ

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While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

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3 years ago
What is the work energy transfer equation?
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The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

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