a large cargo trucks needs to cross a bridge the truck is 30m long, and 3.2 wide, the cargo exerts a force of 54,000 N the bridg
1 answer:
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Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Answer:
31.66 m/s
Explanation:
mass of player, M = 75 kg
mass of ball, m = 0.45 kg
initial velocity of player, U = + 4 m/s
initial velocity of ball, u = - 24 m/s
Let the final speed of player is V and the ball is v.
use conservation of momentum
Momentum before collision = momentum after collision
75 x 4 - 0.45 x 24 = 75 x V + 0.45 x v
289.2 = 75 V + 0.45 v .... (1)
As the collision is perfectly elastic, coefficient of restitution,e = 1
So,
V - v = u - U
V - v = -24 - 4 = - 28
V = v - 28, put this value in equation (1), we get
289.2 = 75 (v - 28) + 045 v
289.2 = 75 v - 2100 + 0.45 v
2389.2 = 75.45 v
v = 31.66 m/s
Thus, the velocity of ball after collision is 31.66 m/.