rce versus time during the collision of a 44- tennis ball with a wall. The initial velocity of the ball is 36 perpendicular to t
he wall; it rebounds with the same speed, also perpendicular to the wall. What is the maximum value of the contact force during the collision
1 answer:
Answer:
Maximum contact force = 487.4 N
Explanation:
The graph showing force vs time is attached to the solution of this question.
According to the an interpretation of Newton's second law of motion, the magmitude of the change in momentum is equal to the magnitude of the impulse experienced duting this impact.
F.t = Δ(momemtum)
Δ(momemtum) = 0.044 [36 - (-36)] = 0.044 × 72 = 3.168 kgm/s
Meaning, Impulse = 3.168 kgm/s too.
The area under the curve above is the impulse we just obtained.
If the maximum contact force is called F
The impulse = Area under this graph
= (1/2)(F)(4-0) + (F)(7-4) + (1/2)(F)(10-7)
= 2F + 3F + 1.5F = (6.5F) mN.s = (0.0065F) N.s
0.0065F = 3.168
F = (3.168/0.0065) = 487.4 N
Hope this Helps!!!
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The wind tunnel speed that must be used to test the car 1 / 5 model of a car that normally runs at 30 mph is 150 mph
Re = u L / ν
Re = Reynold's number
u = Flow speed
L = Characteristic linear dimension
ν = Kinematic viscosity
ν = 1.516 *
m² / s
ν = ν
= 1.516 *
m² / s
L = L
/ 5
u = 30 mph
Re = u
L
/ 5 * 1.516 *
Re = 30 L
/ 1.516 *
In tunnel,
Re = Re
u L
/ 5 * 1.516 *
= 30 L
/ 5 * 1.516 *
u = 30 * 5
u = 150 mph
Therefore, the wind tunnel speed that must be used to test the car is 150 mph
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