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Oduvanchick [21]
3 years ago
10

21 of 35 Review A friend of yours is loudly singing a single note at 401 HzHz while racing toward you at 23.0 m/sm/s on a day wh

en the speed of sound is 343 m/sm/s . Part A What frequency do you hear
Physics
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

The frequency of the sound note as heard = 429 Hz

Explanation:

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = ?

f₀ = real frequency of sound = 401 Hz

v₀ = velocity at which the sound source is moving towards the reference point = 23.0 m/s

v = velocity of sound waves = 343 m/s

+f = 401/[1 - (23/343)]

+f = 429 Hz

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Answer:

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Explanation:

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alekssr [168]

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s

The motion on the X axis is a constant velocity motion so:

t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s

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We know that:

Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

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6 0
3 years ago
You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s
Jobisdone [24]
Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
Since the speed of sound and light remains constant in a particular medium, we can use
      Speed = Distance/Time

For light,
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                t = d/(3 x 10^8)   -1 

For sound,
           339 = d/(t + 30)       -2

Putting value from 1 in 2.
               d = 10^4 m(approx)
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factual evidence of customer-service levels.

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evidence-based determination of process improvement priorities.

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pav-90 [236]

Answer: This is False.

Explanation:

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