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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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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Answer:

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

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Answers

Answer:

Explanation:

It is still using F = m*a -- just in a disguised form.

Givens

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t = 40 second

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Answers

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Option B

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Answers

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Answers

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Answers

Hey mate
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Answers

Answer:

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