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Yakvenalex [24]
3 years ago
8

A 7780 ‑kg car is travelling at 30.7 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 404 m al

ong the exit ramp, the car's speed is 13.3 m/s, and it is h = 13.1 m above the freeway. What is the magnitude of the average drag force F drag exerted on the car?
Physics
1 answer:
vredina [299]3 years ago
6 0

Answer:

4899.47 N

Explanation:

mass of the car = 7780 kg, u, initial speed =  30.7 m/s, v, final speed = 13.3 m/s

d = 404 m and h = 13.1 m

initial kinetic energy of  the car = potential energy attained by the car climbing the ramp + the work done by friction against the motion + final kinetic energy

1/2 mu² = mgh + (F × d) + 1/2mv²

3666286.1 J = 998796.4 J + ( 404 F) + 688102.1 J

404 F = 3666286.1 J - 998796.4 J - 688102.1 J

F = 1979387.6 / 404 = 4899.47 N

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If you use a force of 90 N to pick up a 10 pound bag of charcoal, what is the acceleration?
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9ms^2

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8 0
3 years ago
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

\theta=-33.42^{\circ}

The direction of the momentum with respect to east

\theta=180-33.42=146.58^{\circ}

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0
3 years ago
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