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melisa1 [442]
3 years ago
13

a cheetah starting from rest accelerates at a rate of 2.5m/s^2 to reach a speed of 45 m/s.how fare does the cheetah travel

Physics
1 answer:
Helen [10]3 years ago
7 0

Answer:

405 m

Explanation:

Given:

v₀ = 0 m/s

a = 2.5 m/s²

v = 45 m/s

Find: Δx

v² = v₀² + 2aΔx

(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx

Δx = 405 m

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One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 6
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Solution :

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Therefore, fundamental frequency, F = f' - f

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2 years ago
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Answer:

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So, 125=(wavelength)*(325)

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Hope this helped!

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