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3 years ago

A 3.0-kilogram toy dump truck moving with a speed of 4.0 m/s starts up a ramp. How high does the truck roll before it stops?

1 answer:

7 0

Before it reaches the ramp, the toy's kinetic energy is

(1/2) (mass) (speed²)

= (1/2) (3 kg) (16 m²/s²)

= 24 Joules .

On the way up the ramp, its kinetic energy will change to
gravitational potential energy, and it'll stop when it's height
gives it 24 joules of potential energy.

Potential energy = (mass) (gravity) (height)

24 joules = (3 kg) (9.8 m/s²) (height)

Divide each side by 29.4 kg-m/s² (that's the same as 29.4 newtons):

Height = 24 joules / 29.4 newtons

= 0.816 meter higher than the bottom of the ramp.

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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

Initial momentum

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

Impulse

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$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

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Answer:

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Explanation:

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