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MissTica
3 years ago
12

A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact ti

me between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.
Physics
1 answer:
Digiron [165]3 years ago
7 0

Answer:

The average force between the ball and bat during contact is 3006.72  N.

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the ball, u = 27 m/s

Time of contact between bat and ball is 2.5 ms, t=2.5\times 10^{-3}\ s

After striking the bat, it pops straight up to a height of 31.5 m. The final velocity of the ball is given by using third equation of motion as :

v^2-u^2=2as

a = -g

And initially, u = 0

v=\sqrt{2gs}

v=\sqrt{2\times 9.8\times 31.5}

v = -24.84 m/s (as it pops straight up)

Let F is the average force between the ball and bat during contact. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.145\times (24.84 -(-27))}{2.5\times 10^{-3}}

F = 3006.72  N

So, the average force between the ball and bat during contact is 3006.72  N. Hence, this is the required solution.                                  

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To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

\rho = \frac{m}{V}

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The mass of the neutron star is 1.4times to that of the mass of the sun

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V = \frac{4}{3}\pi R^3

Replacing at the equation we have that

\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}

\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}

\rho = 5.75*10^{17}kg/m^3

Therefore the density of a neutron star is 5.75*10^{17}kg/m^3

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3 years ago
What concerns people in regards to<br> the federal government?
inysia [295]

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Mulitple Changes.

Explanation:

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which of the following helps to explain why you can lift heavy objects more easily when they are in water
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3 years ago
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca
4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

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Now

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s= 33.9 * 17.38

⇒ s = 589.3 m

3 0
3 years ago
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