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seropon [69]
3 years ago
15

A car of mass 487 kg travels around a flat, circular race track of radius 53.3 m. The coefficient of static friction between the

wheels and the track is 0.19. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track? Answer in units of m/s.
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

9.96 m/s

Explanation:

mass of car, m = 487 kg

radius of track, R = 53.3 m

coefficient of static friction, μ = 0.19

acceleration due to gravity, g = 9.8 m/s^2

let v be the maximum speed so that the car can go without flying off the track.

The formula for the maximum speed is given by

v_{max}=\sqrt{\mu Rg}

v_{max}=\sqrt{0.19\times53.3\times9.8

vmax = 9.96 m/s

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Answer:

4.32\cdot 10^5 J

Explanation:

Power is related to energy by the following relationship:

P=\frac{E}{t}

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In this problem, we know that

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E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J

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(a) The distance of the image formed by the concave mirror is 19.1 cm.

(b) The image formed is diminished and real.

<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

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The image distance is smaller than object distance, thus the image formed is diminished and real.

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As, we know that the same charges repeal each other both of the balloons with be apart from each other.

This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.

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