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Rainbow [258]
3 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]3 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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2 years ago
A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

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Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

8 0
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7 0
3 years ago
A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
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To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

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Where,

m= Mass

a = Acceleration

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Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

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By condition of Balance the friction force must be equal to the total net force, that is to say

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Re-arrange to find acceleration,

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3 years ago
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