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Ostrovityanka [42]
2 years ago
8

A siren on the top of the police car emits sound at frequency of 823 Hz. The car is approaching a bystander on a sidewalk at spe

ed 39.5 m/s. What is the freguency the bystander hears? Assume the speed of sound in air to be 340 m/s.
Physics
1 answer:
statuscvo [17]2 years ago
3 0

Answer:

The bystander will hear a frequency of 931.18 Hz.

Explanation:

let V = 39.5 m/s be velocity of the police car and Fs be the frequency a siren of the police car emits. let v = 0 m/s be the velocity of the bystander and Vs = 340 m/s is the velocity of sound in air.

then, we know that the bystander has to hear a higher frequency than the one emitted and the frequency that the bystander hears is given by the doppler relation:

Fo = [(Vs)/(Vs - V)]×Fs

      = [(Vs)/(Vs - V)]×Fs

      = [(340)/(340 - 39.5)]×(823)

      = 931.18 Hz

Therefore, the bystander will hear a frequency of 931.18 Hz.

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Answer:

The induced emf between two end is 34.02 \times 10^{-5} V

Explanation:

Given:

Length of rod l = 1 m

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For finding induced emf,

  \epsilon = Blv

Where v = velocity of rod,

For finding the velocity of rod.

From kinematics equation,

 v^{2} = v_{o} ^{2} + 2gh

Where v_{o} = initial velocity, g = 9.8 \frac{m}{s^{2} }

   v = \sqrt{2gh}

   v = \sqrt{2 \times 9.8 \times 1.23}

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Put the velocity in above equation,

   \epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91

   \epsilon = 34.02 \times 10^{-5} V

Therefore, the induced emf between two end is 34.02 \times 10^{-5} V

6 0
2 years ago
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.
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Answer: The mean value <u>= 9.85m/s².</u>

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Number of measurements =9

Sum of measurements =  88.69

Mean = \dfrac{88.69}{9}=9.85444444\approx9.85

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Answer:

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Answer:

1 point is earned for stating that the conservation of energy should be applied to this situation.

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