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Rainbow [258]
2 years ago
11

The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t

he oscillator is lost in each cycle?
Physics
1 answer:
frozen [14]2 years ago
4 0

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

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Answer:

Explanation:

Given

q_1=8.01\ nC

q_2=4.50\ nC

distance between the charges is given by d=1.87\ m

Electrostatic force between the charges is given by

F_{12}=F_{21}=\frac{kq_1q_2}{d^2}

where k=constant =9\times 10^{9}

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F_{12}=F_{21}=92.769\times 10^{-9}\ N

as both the charges are of similar nature therefore they repel each other

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Answer:

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Explanation:

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A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force
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Answer:

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Explanation:

Given that,

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