Respuesta :
Answer:
The the average magnitude of the force exerted on the ball by the floor [tex]11.2 \times 10^{3}[/tex] N
Explanation:
Given:
Mass of ball [tex]m = 0.4[/tex] kg
Initial speed [tex]v_{i} = -8 \frac{m}{s}[/tex]
Rebound speed [tex]v_{f} = 6 \frac{m}{s}[/tex]
Contact time interval [tex]\Delta t = 0.5 \times 10^{-3}[/tex] sec
For finding the average magnitude of the force on the ball by the floor is given by,
[tex]F_{avg} = \frac{\Delta P}{\Delta t}[/tex]
Here [tex]\Delta P = m (v_{f}- v_{i} )[/tex]
[tex]F_{avg} = \frac{m (v_{f} -v_{i} )}{\Delta t}[/tex]
[tex]F_{avg} = \frac{0.4 \times (6 -( -8 ) )}{0.5 \times 10^{-3} }[/tex]
[tex]F_{avg} = 11.2 \times 10^{3}[/tex] N
Therefore, the the average magnitude of the force exerted on the ball by the floor [tex]11.2 \times 10^{3}[/tex] N
The average magnitude of the force released on the ball by the floor would be [tex]11.2[/tex] × [tex]10^3[/tex] N, Calculations are provided below.
Find the magnitude
Given that,
Mass carried by the ball = 0.4 Kg or 400g
Initial speed carried by the ball or [tex]v_{i}[/tex] = 8 m/s
Bounced speed = 6 m/s
The difference in time or Δt = [tex]0.5[/tex] × [tex]10^-3[/tex]
We know that,
Average Magnitude = Change in P/Change in t
Change in P = m(Final velocity - Initial velocity)
So,
Average Magnitude [tex]= [0.4 (6 - (-8)]/[0.5 * 10^-3}][/tex]
= [tex]11.2[/tex] × [tex]10^3[/tex] N
Thus, [tex]11.2[/tex] × [tex]10^3[/tex] N is the correct answer.
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