A tennis ball launcher works by using a spring to

propel the ball forward.

One model of tennis ball launcher has

advertised that it is capable of imparting

15,000 J of kinetic energy to a 0.5 kg ball

initially at rest. A student carefully measures the

compression of the spring to be 0.10 m.

What is the value of the spring constant for the

tennis ball launcher according to the data

provided?

Answers

Answer 1

Answer:

3MN/m

Explanation:

Step one:

given data

Kinetic Energy KE=15,000J  

mass m= 0.5kg

compression x= 0.1m

Step two:

The expression for the energy stored in spring is

[tex]KE= 1/2ke^2\\\\15000=1/2 *k*0.1^2\\\\15000=1/2*k*0.01\\\\15000=0.005k\\\\[/tex]

divide both sides by

0.005

[tex]k= 15000/0.005\\\\k=3000000\\\\k= 3MN/m[/tex]

Answer 2

From conservation of energy, the value of the spring constant for the tennis ball launcher according to the data is 3 x [tex]10^{6}[/tex] Nm

The work done by a  tennis ball launcher by using a spring to propel the ball forward is equal to the kinetic energy which is 15,000 Joule. That is,

W = K.E

The given parameters are:

K.E = 15,000 J

Mass m = 0.5 kg

compression x = 0.1 m

The elastic potential energy on the spring will be equal to the work done by the tennis ball launcher. That is,

E = W

since W = K.E

E = K.E

1/2k[tex]x^{2}[/tex] = 1/2m[tex]v^{2}[/tex] = 15,000

1/2k[tex]x^{2}[/tex] = 15000

Substitute for 0.1 for x

1/2 K ([tex]0.1^{2}[/tex]) = 15000

[tex]0.1^{2}[/tex] K = 30000

K = 30000/0.01

K = 3 x [tex]10^{6}[/tex] Nm

Therefore,  the value of the spring constant for the tennis ball launcher according to the data is 3 x [tex]10^{6}[/tex] Nm

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Answers

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Explanation:

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Answers

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Answers

Answer:

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Answers

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Answers

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Between ball B and ball C, ball ____ has LESS potential energy because ___
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Answers

Answer: b

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brainly.com/question/13584911

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Answers

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Answers

Answer:

1800J

Explanation:

Step one:

given data

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initial velocity u = 400m/s

final velocity v= 200 m/s

Step two:

1.The bullet's lost kinetic energy went inside the tree.

2. The energy transferred is computed as

= initial KE- KE final

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Answers

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

[tex]h=\frac{u^2sin^2\theta}{2g}[/tex]

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[tex]For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\[/tex][tex]\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}[/tex]

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