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Alenkinab [10]
2 years ago
7

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

equency of 346 Hz. (a) What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding
Physics
1 answer:
aleksandr82 [10.1K]2 years ago
4 0

Answer:

Explanation:

We shall apply Doppler's effect to solve the problem .

Formula for apparent frequency for a source of sound approaching an observer is as follows .

f₁ = f₀ V / (V - v )

where f₁ and f₀ are apparent and real frequency of source , V and v is velocity of sound and velocity of approaching source respectively .

Putting the given values and knowing that speed of sound is 340 m /s

f₁ =346x 340 / (340 - 39.6 )

f₁ = 391.6 Hz

In case of receding train , the formula is

f₂ = f₀ V / (V + v )

Putting the values

f₂ = 346x 340 / (340 + 39.6 )

= 309.9 Hz

Change in frequency =  391.6 - 309.9

= 81.7 Hz .

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Hi Pupil Here's Your answer :::





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An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.





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Does frequency affect wavelength?
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6 0
2 years ago
Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice
OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

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3 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
2 years ago
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