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Triss [41]
4 years ago
5

Assume that a 15-kg ball moving at 8 m/s strikes a wall perpendicularly and rebounds elastically at the same speed. What is the

amount of impulse given to the wall? Be sure to show all work to support your answer.
Physics
1 answer:
bulgar [2K]4 years ago
6 0
So we are told that the ball rebounds elastically. What this means momentum and kinetic energy is conserved. So our initial condition equals our final condition.

Our initial condition is when the ball is moving at 8m/s and is 15 kg.

p_0 = (8m/s)(15kg) = 120kg*m/s

Next we our told it bounces off in the negative direction moving at 8m/s

p_f = (-8m/s)(15kg) = -120kg*m/s

Impulse, denoted as J, which is the change in momentum.

J=p_f-p_0 = -120kg*m/s - 120kg*m/s = -240kg*m/s

J = -240kg*m/s

Hope this helps! :D
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2.92 m

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Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each cha
Sloan [31]

Answer:

<em>The force on each charge = 16 N</em>

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

<em>When d = 1 m, F = 1 N,</em>

<em>Constant: 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 2,</em>

<em>q₁q₂ = 1×1²/9×10⁹ </em>

<em>q₁q₂ = 1/9×10⁹  C²</em>

<em>When d = 1/4 m, q₁q₂ = 1/9×10⁹  C² and 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1</em>

<em>F =  9×10⁹×1/9×10⁹ /(1/4)²</em>

<em>F = 1/(1/16)</em>

<em>F = 16 N</em>

<em>Therefore the force on each charge = 16 N</em>

<em />

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