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Oliga [24]
3 years ago
15

If the average velocity during the athlete's walk back

Physics
1 answer:
goblinko [34]3 years ago
4 0

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

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Answer:

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Explanation:

lets assume speed of sound is 350 m/s.

frequencies of a standing wave modes of an open-close tube of length L

fm = m(v/4L)

where m is 1,3,5,7......

and fm = mf1

where f1 = fundamental frequency

so therefore: f1 = 350 x 4 / 1.5

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f2 = 3 x 58.3

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5 0
4 years ago
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Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

The linear expansion coefficient of aluminium \alpha =23.6\times 10^{-6}/^{\circ}C

We know that change in length is given by \Delta L=L\alpha \Delta T

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\Delta T=36.32^{\circ}C

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3 years ago
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Explanation:

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Given that the vertical displacement = 5.0 m, we have

Total work done = Potential energy gained = Mass, m × Acceleration due to gravity, g × Vertical height, h

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Also, time, t = 5.0 s

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P = 2501.55 J/(5 s) = 500.31 J/s = 500.31 W ≈ 500 W = 0.5 kW.

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Answer:

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