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Alex17521 [72]
3 years ago
5

An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00

0 Hz as it approaches and then passes by the observer without slowing down. Assuming the speed of sound is 340 m/s, how much of a frequency change did the observer hear?
Physics
1 answer:
SSSSS [86.1K]3 years ago
7 0

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

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A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is
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<u>Given the following data;</u>

Radius, r = 2.6 km

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<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

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Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

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Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

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Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

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