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Genrish500 [490]
3 years ago
13

baseball player hits a line drive estimated to have traveled 145 meters the ball leaves the bat with a horizontal velocity of 40

.0 m/s . how much time was the ball in the air
Physics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

3.63 s

Explanation:

Since distance d = vt where v = velocity and t = time taken to cover the distance.

Now, the baseball traveled a distance of d = 145 m with a horizontal velocity of 40.0 m/s.

So, making t subject of the formula,

t = d/v

Substituting the values of d and v we have

t = 145 m/40.0 m/s

= 3.625 s

≅ 3.63 s

So the time the ball was in the air is 3.63 s

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A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .
Usimov [2.4K]

The body moves at a velocity of 1.62m/s after the bullet emerges.

<h3>Given:</h3>

Mass of bullet, m_1 = 22g

                               = 0.022 kg

Mass of the block, m_2 = 1.9 kg

Velocity of bullet , v_1 = 265 m/s

v_2 = 0

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

v^{'}_1 =125 m/s

v^{'}_2=?

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

   m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s

Therefore, It moves with a velocity of 1.62 m/s.

Learn more about momentum here:

brainly.com/question/25121535

#SPJ1

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