(1) The wavelength of the wave is 1.164 m.
(2) The velocity of the wave is 23.7 m/s.
(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.
<h3>
Wavelength of the wave</h3>
A general wave equation is given as;
y(x, t) = A sin(Kx - ωt)
<h3>Velocity of the wave</h3>
v = ω/K
From the given wave equation, we have,
y(x, t) = 0.048 sin(5.4x - 128t)
v = ω/K
where;
- ω corresponds to 128
- k corresponds to 5.4
v = 128/5.4
v = 23.7 m/s
<h3>Wavelength of the wave</h3>
λ = 2π/K
λ = (2π)/(5.4)
λ = 1.164 m
<h3>Maximum speed of the wave</h3>
v(max) = Aω
where;
- A is amplitude of the wave
- ω is angular speed of the wave
v(max) = (0.048)(128)
v(max) = 6.14 m/s
Thus, the wavelength of the wave is 1.164 m.
The velocity of the wave is 23.7 m/s.
The maximum speed in the y-direction of any piece of the string is 6.14 m/s.
Learn more about wavelength here: brainly.com/question/10728818
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Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer:
I = I₀ + M(L/2)²
Explanation:
Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.
The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.
The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀
The distance between the two axes is L/2 (total length of the rod divided by 2
From the parallel axis theorem we have
I = I₀ + M(L/2)²
Answer:19.32 m/s
Explanation:
Given
initial speed of car(u)=4.92 m/s
acceleration(a)=
Speed of car after 4.5 s
using equation of motion
v=u+at
v=19.32 m/s
Displacement of the car after 4.5 s
s=54.54 m
