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amid [387]
2 years ago
6

Water at the top of Horseshoe Falls (part of Niagara Falls)

Physics
1 answer:
ale4655 [162]2 years ago
8 0

Answer:

\theta= (-74.42)^{\circ} C

Explanation:

Horizontal speed of water, v_{xf}=9\ m/s

Height, h = -53 (below pool)

We can find firstly the final vertical speed of the water using third equation of kinematics. So

v^2_{yf}=u^2_{yi}+2(-g)h\\\\v^2_{yf}=2\times -9.8\times -53\\\\v_{yf}=32.23\ m/s

Let \theta is the angle where the falling water moving as it enters the  pool. So,

\tan\theta=\dfrac{v_{yf}}{v_{xf}}\\\\=\dfrac{-32.3}{9}\\\\=-74.42^{\circ} C

Hence, the angle is (-74.42)°C.

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sertanlavr [38]

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 v ’= 21.44 m / s

Explanation:

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we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

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                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

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2 years ago
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