The 9-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point Clocated on the vertica
1 answer:
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In this case the rubber raft has horizontal and vertical motion.
Considering vertical motion first.
We have displacement , u = Initial velocity, t = time taken, a = acceleration.
In vertical motion
s = 1960 m, u = 0 m/s, a = 9.81
So raft will take 20 seconds to reach ground.
Now considering horizontal motion of raft
u = 109 m/s, t = 20 s, a = 0
So
So shipwreck was 2180 meter far away from the plane when the raft was dropped.
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.