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3 years ago

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Consider a 145 gram baseball being thrown by a pitcher. The ball approaches the batter with a speed of 44 m/s. The batter swings

and sends the ball flying back at a speed of 90 m/s into the outfield.
If the ball was in contact with the bat for a time of 3 ms, what is the average force exerted by the bat during the collision?

1 answer:

larisa [96]3 years ago

4 0

For the development of this problem it is necessary to apply the concepts related to Force, momentum and time.

By definition we know that momentum can be expressed as

$P = F*t$

Where

P = Momentum

F = Force

t = Time

At the same time, another way of expressing the momentum is through mass and speed, that is,

P = mv

Where

m = Mass

v = Velocity

Equating both equations

F*t = mv

$F = \frac{mv}{t}$

Our values are given as

$m = 145*10^{-3}Kg$

$t = 3*10^{-3}s$

$v = 90-44 = 46m/s \rightarrow$ Net velocity

Replacing we have then,

$F = \frac{mv}{t}$

$F = \frac{3*10^{-3}*46}{3*10^{-3}}$

$F = 2223.33N$

Therefore the average force exerted by the bat during the collision is 2223.3N

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The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

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# A cheetah can start from rest and attain the velocity 72km/h in 2 seconds. Calculate the acceleration of cheetah

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<h2> $10 \:$ m/s ^2</h2><h2 />

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One major difference between plant and animal cells is that plant cells are the only ones with A) cellulose. B) lysosomes. C) nu

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An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

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3 years ago

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

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3 years ago