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lisov135 [29]
2 years ago
12

Water at the top of Horseshoe Falls (part of Niagara Falls) is moving horizontally at 9.2 m/s as it goes off the edge and plunge

s 53 m to the pool below.
Part A
If you ignore air resistance, at what angle with respect to the vertical is the falling water moving as it enters the pool?
Physics
1 answer:
dangina [55]2 years ago
4 0

Water at the top of Horseshoe Falls (part of Niagara Falls) is moving horizontally at 9.2 m/s, the angle with respect to the vertical  is mathematically given as

\theta=15.93

<h3>What angle with respect to the vertical is the falling water moving as it enters the pool?</h3>

Generally, the equation for velocity is mathematically given as

V^2=u^2+2as

Therefore

V^{2}=0+2(9.81)(53)

V=32.2m/s

In conclusion, The angle

\theta =tan^{-1}\frac{(9.20m/s)}{(32.23m/s)}

\theta=15.93

Read more about Speed

brainly.com/question/4931057

#SPJ1

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Explanation:

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Hence, this is the required solution.

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1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

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Using binomial theorem in the above eqn:

We know that:

(1 + x)^{a} = 1 + ax

Thus eqn (1) becomes:

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Answer:

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<u>Given the following the details;</u>

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Mathematically, the quantity of charge with respect to electric potential is given by the formula;

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Therefore, the quantity of charge must be <em>5 Coulombs.</em>

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