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MAXImum [283]
1 year ago
13

Amy, standing near a straight road, records the sound of the horn of a car traveling at a constant speed. The recorded frequency

as a function of the car's position along the road for trials at different car speeds is accurately graphed below.
True or false:
1. The speed in trial U is ..... that in Q.
2. The speed in trial T is ..... that in other trials.
3. The distance of Amy from the path in R is ..... in S.
4. The speed in trial U is ..... that in other trials.
5. When the car is far away, the average of the frequencies moving in and out is ..... the frequency at rest.
6. The speed in trial R is ..... to that in S.

7. Using the data on the figure, calculate the speed of the car in trial T. Use 340 m/s for the speed of sound.

Physics
1 answer:
myrzilka [38]1 year ago
6 0

(1) The speed in trial U is less than that in Q.

(2) The speed in trial T is greater than that in other trials.

(3) The distance of Amy from the path in R is equal to distance in S.

(4) The speed in trial U is less than that in other trials.

(5) When the car is far away, the average of the frequencies moving in and out is equal to the frequency at rest.

(6) The speed in trial R is equal  to that in S.

(7) The speed of the car in trial T is 58.6 m/s.

<h3>Speed of wave</h3>

The speed of a wave is directly proportional to its frequency and wavelength.

v = fλ

where;

  • v is speed of the wave
  • f is frequency of the wave
  • λ is wave length

The speed in trial U is less than that in Q, due to greater frequency observed in trial Q.

The speed in trial T is greater than that in other trials, due to greater frequency observed in trial T.

The distance of Amy from the path in R is equal to distance in S.

The speed in trial U is less than that in other trials.

<h3>Average of the frequencies when the car is far away</h3>

f(avg) = (1,055 + 1,020 + 986 + 985 + 940 + 862 + 830 + 828 + 808 + 730)/10

= 904.3 ≈ 900 Hz

When the car is far away, the average of the frequencies moving in and out is equal to the frequency at rest.

The speed in trial R is equal  to that in S.

<h3>Speed of the car in trial T</h3>

f = f₀(v / v - vs)

where;

  • f is frequency of the car at zero position
  • f₀ is the original frequency
  • vs is the speed of the car
  • v is speed of sound

900 = 1055(340 / 340 - vs)

900/1055 = 340/340 - vs

0.853 = 340/340 - vs

340 - vs = 340/0.853

340 - vs = 398.6

vs = 398.6 - 340

vs = 58.6 m/s

Learn more about speed of wave here: brainly.com/question/2142871

#SPJ1

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To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

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\theta= Angle between the difraccion paterns and the source of light

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d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

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e is the electron charge
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Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
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