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Irina-Kira [14]
2 years ago
15

A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude r

eduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.
Physics
1 answer:
iren [92.7K]2 years ago
8 0

The energy becomes 0.50 times in 6.72 s.

Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.

Et=0.50E

1

k(4₂)² = (0.5) - kA²

(4₂)² = (0.5) A²

At = 0.71A

So, the amplitude of the oscillator becomes 0.71 times its initial ar

0.71A = = A(0.96)¹2

log(0.71)

log(0.96)

8.4

n=

So, the time taken for n oscillation is obtained as,

t = n (0.800 s)

= (8.4) (0.800)

= 6.72 s

learn more about oscillators brainly.com/question/15169199

#SPJ1

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Each capacitor carry the same charge 'q'.

Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:

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This voltage may also be calculated using capacitance and charge;

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Provided that the total charge is 'q', hence the total voltage can be expressed as:

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A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
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Explanation:

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(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

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