. Draw the sketches of two waves A and B such that wave A has twice the wavelength and half the amplitude of wave B.
1 answer:
Wave B will, therefore, be considered more powerful than wave A. Waves with higher frequencies are more energetic than those with lower frequencies. The equation below demonstrates this. The higher the amplitude also, the higher the energy of the wave.
v = f λ
Where,
v = velocity of the wave,
f = frequency of the wave,
λ = wavelength.
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Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ =
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k