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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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A toy plane is flying in a horizontal circle by being attached to a 0.75 meter string. The plane has a mass of 101.7 grams and m
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Answer:

F = 2,894 N

Explanation:

For this exercise let's use Newton's second law

      F = m a

The acceleration is centripetal

     a = v² / r

Angular and linear variables are related.

     v = w r

Let's replace

     F = m w² r

The radius r and the length of the rope is related

    cos is = r / L

    r = L cos tea

Let's replace

    F = m w² L cos θ

Let's reduce the magnitudes to the SI system

     m = 101.7 g (1 kg / 1000g) = 0.1017 kg

      θ = 5 rev (2π rad / rev) = 31,416 rad

     w =  θ / t

     w = 31.416 / 5.1

     w = 6.16 rad / s

     F = 0.1017 6.16² 0.75 cos  θ

     F = 2,894 cos  θ

The maximum value of F is for  θ equal to zero

     F = 2,894 N

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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

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We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

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distance to the negative image is before the lens

B) the magnification of the lenses is given by

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        h’= - q / p h

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        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

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