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klemol [59]
2 years ago
10

It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Ass

ume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 32.0 m/s.
Physics
1 answer:
Vinvika [58]2 years ago
5 0

The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

<h3>Acceleration of the car </h3>

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

<h3>Distance traveled when the speed is 32 m/s</h3>

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

Learn more about distance here: brainly.com/question/4931057

#SPJ1

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A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
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14.98\ \text{kg m/s}

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P_2 = Momentum of the ball after hit

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\theta = Angle the pin makes with the horizotal after getting hit by the ball

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P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

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