Explanation :
In transverse waves the particles are oscillating perpendicular to the direction of propagation of waves.
The uppermost part of the wave is crests and the lowermost part is troughs.
Wavelength of a transverse wave is defined as the distance between two consecutive crests or troughs.
Amplitude is the maximum distance or displacement covered by a wave.
So, crest, amplitude, trough and wavelength identifies the parts of a transverse wave.
Explanation and answer:
This type of question can be clarified and sometimes solved by drawing a proper diagram or sketch. (see below)
Solution:
Since we do not know the reaction of the support, we can take moments about the support (thereby eliminating its involvement).
CCW moment = 0.900(5.00/2 - x) kg-m
CW moment = 0.300*(5.00/2-2.00)) = 0.150 kg-m
At equilibrium, CCW moment = CW moment, so
0.900(2.50-x) = 0.150
Expand and solve
2.25 - 0.900x = 0.150
0.900x = 2.25-0.15 = 2.10
x = 2.10 / 0.900 = 2.33 m (to nearest cm)
Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.
From this question, the plane is still up in the air.
We have wind blowing in [W 60° N ]
To solve the problem we have to make use of the sine rule
We put the values in the equation, we have:
50/Sinθ = 200/sin60°
The next step is to cross multiply
50 x sin60° = 200Sinθ
50 x 0.8660 = 200sinθ
We make Sin θ the subject
Sine θ = 43.30/200
sine θ = 0.2165
we find the value of θ
θ = sine⁻¹(0.2165)
θ = 12.50
So Lindsay has to fly this plane towards this direction
[W 12.5° S]
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