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masha68 [24]
2 years ago
8

A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875

Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?
Physics
2 answers:
Fed [463]2 years ago
8 0

Answer:

a

  v _{max } =  6.82 \ m/s

b

 a_{max} =  37489.5 \  m/s^2

Explanation:

From the question we are told that

   The amplitude is A =  1.24 \  mm  =  1.24 * 10^{-3} \  m

   The frequency is  f =  875 \  Hz

   

Generally the maximum speed is mathematically represented as

           v _{max } =  A *  2 *  \pi * f

=>        v _{max } =  1.24*10^{-3} * 2 * 3.142 *  875

=>        v _{max } =  6.82 \ m/s

Generally the maximum acceleration is mathematically represented as

       a_{max} =  A  * (2 *  \pi *  f)

=>      a_{max} =  1.24*10^{-3} *  (2 * 3.142 *  875 )^2

=>       a_{max} =  37489.5 \  m/s^2

mario62 [17]2 years ago
5 0

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

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GalinKa [24]

Answer:

Ceiling fan

Explanation:

Ceiling fan is a perfect and typical example of electrical energy being converted to mechanical energy.

In most systems, energy is usually transformed from one form to another. Energy is not created neither is it destroyed. We know this by virtue of law of conservation of energy.

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In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
Damm [24]

Answer:

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Explanation:

given data

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before collision velocity v2 = 0

angle =  35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ   with x axis    .............1

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so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

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form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ  

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another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

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