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mezya [45]
4 years ago
8

What average net force is required to stop a 4 kg bowling ball in 0.5 s if it’s initially traveling at 10 m/s?

Physics
1 answer:
Makovka662 [10]4 years ago
4 0

Answer:

-80 N

Explanation:

First of all, let's calculate the deceleration of the ball, which is given by:

a=\frac{v-u}{t}

where

v = 0 m/s is its final velocity

u = 10 m/s is its initial velocity

t = 0.5 s is the time taken

Substituting,

a=\frac{0-10 m/s}{0.5 s}=-20 m/s^2

And now we can calculate the average net force required to stop the ball by using Newton's second law: in fact, the net force is equal to the product between the ball's mass (4 kg) and the acceleration, so:

F=ma=(4 kg)(-20 m/s^2)=-80 N

Where the negative sign simply means that the force is in the opposite direction to the motion of the ball.

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hichkok12 [17]
The answer is 117.6 J

The potential energy of the object is actually its stored energy:

<span>E = m · g · h
E - the potential energy of the object,
m - the mass of the object,
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h - the height of the object.

m = 4 kg
g = 9.8 m/s</span>²
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A 48.5 kg student runs down the sidewalk and jumps with a horizontal speed of 4.25 m/s onto a stationary skateboard. The student
storchak [24]

<u>Answer:</u>

2.39 kg

<u>Explanation:</u>

There is conservation of momentum here in this problem so we will use the following problem:

m_1u_1+m_2u_2=(m_1+m_2)v

where the mass of the student m_1 is 48.5 kg,

the mass of the skateboard m_2 is m_2 kg,

the initial speed of the student u_1 is 4.25 m/s; and

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So substituting the given values in the above formula to get:

(48.5*4.25) + (m_2 * 0) = (48.5 + m_2 ) * 4.05

206.125=196.425+4.05m_2

206.125 - 196.425 = 4.05m_2

m_2=\frac{9.7}{4.05}

m_2=2.39

Therefore, the mass of the skateboard is 2.39 kg.

7 0
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Her center of mass never moved, and all of her motion produced no change
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