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d1i1m1o1n [39]
3 years ago
5

A race car, traveling at constant speed, makes one lap around a circular track of radius 200 m. When the car has traveled halfwa

y around the track, what is the magnitude of the displacement from the starting point?
Physics
1 answer:
mario62 [17]3 years ago
8 0

The magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

Answer:

Explanation:

Since the race track is a circular track, the distance for one lap will be equal to the circumference of the circular track. And the circumference will be equal to the circumference of the circle.

Since the radius of the track is given as 200 m, then the circumference of the circular track will be

Circumference = 2πr = 2 × 3.14 × 200

So the circumference of the circular track = 1256 m.

So the starting point or position of the track is considered as zero and if the car has traveled half way means, the car has covered half of the circumference of the track.

As the circumference = 1256 m, then half of the circumference of the circle = 1256/2 = 256 m.

So the displacement is the measure of difference between the final position and initial position. As here the initial position is zero and the final position is the halfway around the track which is equal to 256 m.

Then Displacement = Final-Initial = 256-0= 256 m.

So the magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

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3 years ago
If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy
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Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

                                    P.E = mgh

                                     K.E = 0

At height 'h/4', the total energy is

                                     P.E + K.E = mgh

                                     P.E = mgh/4

                                     K.E = 1/2 mv²

Therefore,

                                   mgh/4 + 1/2 mv² = mgh

                                    gh/4 + v²/2 = gh

Hence,

                                      v² = 3gh/2

Substituting in the K.E equation

                               K.E = 1/2 mv²

                                      = 1/2 m (3gh/2)

                                       = 3/4 mgh

                                        = 3/4 x 40000

                                         = 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

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