The statement which is true of a wave that’s propagating along the pavement and girders of a suspension bridge is A. The wave is mechanical, with particles vibrating in a direction that is parallel to that of the wave, forming compressions and rarefactions.
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Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
Answer: 6.47m/s
Explanation:
The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.
The time taken for 1 revolution is, 1/3.33 = 0.30s
velocity of the wheel = d/t
Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t
V = 2π * 0.309 / 0.3
V = 1.94/0.3
V = 6.47m/s
The tangential speed of the tack is 6.47m/s
