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alex41 [277]
3 years ago
14

A truck moving at 45 m/s passes a police car moving at 36 m/s in the opposite direction. if the frequency of the siren is 500 Hz

relative to the police car, what is the frequency heard by an observer in the truck after the police car passes the truck
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer: observed frequency (f') = 511.97Hz

Explanation: when there is a relative motion between an observer and a sound source, the frequency of sound wave perceived by the observer is different from the frequency of originally sent out by the sound source.

This is called Doppler effect and given mathematically below as

f' = (v + v') /(v- vs) * f

f' = observed frequency

v = speed of sound in air = 340m/s

v' = velocity of observer= 45m/s

vs = velocity of source relative to observer = - 36m/s ( the negative sign came as a result of the fact that the velocity of the source is in opposite direction to the velocity of the observer)

f = original frequency of sound source = 500Hz

f' = (340 + 45)/{340 -(-36)} * 500

f' = 385/ (340 + 36) * 500

f' = 385/ 376 * 500

f' = 1.0239 * 500

f' = 511.97Hz

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The force is F = 1041.7N

Explanation:

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               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

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3 years ago
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