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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 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m,
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Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

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What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude
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Answer:

The fraction of kinetic energy to the total energy is \frac{K}{T}=\frac{8}{9}.

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)

The total energy is

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Divide (1) by (2)

\frac{K}{T}=\frac{8}{9}

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