In an experiment, a torque of a known magnitude is exerted along the edge of a rotating disk. The disk rotates about its center.
1 answer:
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Answer:
See explanation
Explanation:
The question is incomplete because the images were not attached but I will try to help you as much as possible.
Constant acceleration implies that the velocity increases uniformly with time.
The graph of constant acceleration is a straight line graph having a slope. The slope of the graph is constant at any point along the straight line.
The image attached shows a velocity-time graph depicting constant acceleration.
Answer:
B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2
C) V = 0.84m/s T = 29.92s
D) ω2 = 0.315 rad/s
Explanation:
The moment of inertia when they are standing on the edge:
where M is the mass of the merry-go-round.
I1 = 1680 kg.m^2
The moment of inertia when they are standing half way to the center:
I2 = 1120 kg.m^2
The tangencial velocity is given by:
V = ω1*R = 0.84m/s
Period of rotation:
T = 2π / ω1 = 29.92s
Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:
I1*ω1 = I2*ω2 Solving for ω2:
ω2 = I1*ω1 / I2 = 0.315 rad/s
Answer:
λ = 65.6 pm
Explanation:
Given that
λo = 65 pm
The initial energy of the electron
Now by putting the values
Eo=19.06 KeV
Given that kinetic energy KE= 0.84 KeV
Therefore the final energy
E= Eo - KE
E = 19.06 - 0.84 KeV
E= 18.22 KeV
The wavelength λ can be find as
λ = 6.56 x 10⁻¹¹ m
λ = 65.6 pm