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Alona [7]
2 years ago
13

If the brakes are applied and the speed of the car is reduced to 12 m/s in 15 s , determine the constant deceleration of the car

.
Physics
1 answer:
zavuch27 [327]2 years ago
5 0

Given:

\triangle v = 12 m/s

Time = 15 s

To find:

Deceleration = ?

Formula used:

Deceleration is given by,

a = \frac{\triangle v}{t}

Solution:

Acceleration of the body is rate of increase of velocity. Deceleration of the body is negative acceleration i. e rate of decrease of velocity.

Thus, deceleration of the body is given by,

a = \frac{\triangle v}{t}

where a is the deceleration of the body

v is the velocity

\frac{\triangle v}{t} is rate of change of velocity

t is the time

a = \frac{12}{15}

a = 0.8 m/ s^{2}

Thus, deceleration is 0.8 m/ s^{2}.

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Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

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N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

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v = \sqrt{\mu rg}

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As  28m/s = 28m/s

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3 years ago
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Answer:

9013m/s

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6 0
2 years ago
2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential
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Answer:

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Explanation:

Given parameters:

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Unknown:

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h is the height of the building

  Insert the given parameters and solve;

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