A softball pitcher rotates a 0.250 kg ball around a vertical circular path of radius 0.5 m before releasing it. The pitcher exerts a 33.0 N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 13.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release

Answers

Answer 1

Answer:

The value is  [tex]u = 27.93 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass of the ball is [tex]m = 0.250 \ kg[/tex]

    The radius of the circle is  [tex]r = 0.5 \ m[/tex]

    The force exerted is  [tex]F = 33 \ N[/tex]

    The  speed of the ball on top of the circle is  [tex]v = 13.0 \ m/s[/tex]

Gnerally from the law of energy conservation we have that

     [tex]PE + W = \Delta KE[/tex]

Here PE  is the potential energy of the pitcher which is mathematically represented as

      [tex]PE = m * g * (2 r )[/tex]

=>   [tex]PE = 0.250 * 9.8 * (2 * 0.5 )[/tex]  

=>   [tex]PE = 24.5\ J[/tex]

And  W  is the workdone by the force applied which is mathematically represented as

       [tex]W = F * \pi * r[/tex]

=>    [tex]W = 33 * 3.142 * 0.5[/tex]

=>    [tex]W = 51.84 \ J[/tex]

And   [tex]\Delta KE[/tex] is the change in kinetic energy which is mathematically represented as

       [tex]\Delta KE = \frac{1}{2} * m * (v^2 - u^2 )[/tex]

=>    [tex]\Delta KE = \frac{1}{2} * 0.250 * ( 13^2 - u^2 )[/tex]

=>    [tex]\Delta KE = 0.125 * ( 13^2 - u^2 )[/tex]

=>    [tex]\Delta KE = 21.125 - 0.125u^2[/tex]

So

      [tex]24.5 + 51.84 = 21.125 - 0.125u^2[/tex]

=>    [tex]u = 27.93 \ m/s[/tex]

Answer 2

The velocity of the ball released at the bottom is required.

The velocity of the ball released at the bottom is 24.56 m/s

F = Force = 33 N

r = Radius of loop = 0.5 m

m = Mass of ball = 0.25 kg

u = Initial velocity of ball = 13 m/s

v = Final velocity

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]F\pi r+mg2r=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow v=\sqrt{\dfrac{2}{m}(F\pi r+mg2r)+u^2}\\\Rightarrow v=\sqrt{\dfrac{2}{0.25}(33\times \pi\times 0.5+0.25\times 9.81\times 2\times 0.5)+13^2}\\\Rightarrow v=24.56\ \text{m/s}[/tex]

The velocity of the ball released at the bottom is 24.56 m/s.

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Answers

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Explanation:

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a = (vf - vi)//t

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b????

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The development of the current atomic theory has changed over time as a result of different proposed models and experiments as per the chronological order of the development of the atomic models, the correct answer is option B

What are atomic models?

There are some models that are used to explain the arrangements of subatomic particles inside the atom based on the atomic theory of atom are known as the atomic models.

As given in the problem statement The development of the current atomic theory has changed over time as a result of different proposed models and experiments. The chronological order of the atomic model is given as follows

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Erwin Schrodinger's model explains the arrangement of the subatomic particles as the electron Cloud Model

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Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete but the missing figure is in the attachment below.

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The answer is true I think I had this question too

The statement "TNT is an example of a high explosive." is true.

What is TNT?

TNT, short for trinitrotoluene, is a highly explosive chemical compound that is widely used in military and industrial applications. It is a pale yellow crystalline solid that is highly sensitive to heat, shock, and friction. TNT is made by the nitration of toluene, a colorless liquid hydrocarbon that is derived from petroleum.

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