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Orlov [11]
3 years ago
8

A small model car with mass m travels at a constant speed on the inside of a track that is a vertical circle radius 5.00 m. If t

he normal force exerted by the track on the car when it is at the bottom of the track (point A) is equal to 2.50 mg, how much time does it take the car to complete one revolution around the track?
Physics
1 answer:
dolphi86 [110]3 years ago
6 0
One point to another
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A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
Hey what is you answering streak and what fact dose it say
Valentin [98]

Answer:

It’s 53:) Or - quit

Explanation:

That’s my strong streak

5 0
3 years ago
Read 2 more answers
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