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oksano4ka [1.4K]
3 years ago
5

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m . If the track is completely flat and the race car is traveling at a constant 26.5 m/s (about 59 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Physics
1 answer:
Svetach [21]3 years ago
7 0

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

a_c=\dfrac{V^2}{R}

Now by putting the values in the above equation we get

a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

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Determine the change in thermal energy of 100 g of copper (M = 63,5, Debye 348K) if it is cooled from
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Answer:

given,

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The students changed the release distance of the sphere during their investigation. Which statements best describe the purpose f
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Since unbalanced forces acts on the spheres, the aim of increasing the release distance of the spheres is to observe the effect of unbalanced forces acting on the spheres.

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An experiment is a study whose objective is to study cause and effect relationships. In this case, the students had to increase the distance at which the spheres was released.

The aim of increasing the release distance of the spheres is to observe the effect of unbalanced forces acting on the spheres.

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A baseball is thrown horizontally at 55m/s. The ball slows down at a rate of -10 m/s2. How long is the ball in the air before co
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The time of motion of the ball before coming to rest is  determined as 5.5 s.

<h3>Time of motion of the ball</h3>

The time of motion of the ball before coming to rest is calculated as follows;

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