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4 days ago

A 600 kg car is at rest, and then accelerates to 5 m/s.

Physics

1 answer:

ValentinkaMS [1.1K]4 days ago

3 0

Answer:

7500J

Explanation:

Provided details:

Weight of car = 600kg

Speed = 5m/s

Unknown variables:

Initial kinetic energy =?

Final kinetic energy =?

Work performed =?

Solution method:

Kinetic energy relates to the energy associated with a body's motion.

This can be expressed mathematically as follows:

K.E = $\frac{1}{2}$ m v²

where m denotes mass

and v signifies velocity

Initial kinetic energy:

Since the car began at rest, v = 0, so K.E = 0

Final kinetic energy:

K.E = $\frac{1}{2}$ x 600 x 5² = 7500J

Work done:

Work done = Final K.E - Initial K.E = 7500 - 0 = 7500J

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Answer:

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b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

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This gives us our initial equation.

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$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

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Utilizing $v_{o}$ in our first equation (1)

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$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

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