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Marat540 [252]
3 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]3 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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2 years ago
A 210 Ohm resistor uses 9.28 W of
IgorLugansk [536]
  • Resistance=R=210Ohm
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Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

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\\ \sf\longmapsto I\approx\sqrt{0.04}

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4 0
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Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript
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Answer:

Part a)

F_A = 4.59 N

Part B)

F_B = 1.28 N

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

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as we know that

a = 0.554 m/s^2

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now we will have

F_A + F_B = 10.6(0.554)

F_A + F_B = 5.87 N

Now two forces are in opposite direction then we have

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F_A - F_B = 3.32 N

Part A)

Now we will have from above two equation

F_A = 4.59 N

Part B)

Similarly for other force we have

F_B = 1.28 N

5 0
3 years ago
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