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Marat540 [252]
2 years ago
9

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

the top of the ride? Express your answer with the appropriate units.
Physics
1 answer:
Stells [14]2 years ago
6 0

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

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Answer: option A. a graph of the area of a circle vs. its radius r (A = πr²).



Explanation:



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The function is y =  ax² + bx + c, where x is the independent variable and y is the dependent variable.



As stated in the question, the area of a circle is given by A = πr².



In this case, A is the dependent variable and r is the independent variable.



π is assumed as the coefficient of the quadratic term, and the other coefficients are assumed 0, since there are no either terms on r or constants.



The equation a = 1/b  is an inverse relation, not a quadratic relation.



The relation of distance vs. time for a car moving at constant speed is a linear relation of the kind v = u + st.



The mass of water vs. the volume of water in a drinking glass is a direct relation, mass = density × volume



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2 years ago
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3 years ago
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1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?
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Answer: Given:

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