Pitch is related to frequency
Answer:
<h2>A. a spring & B. a well dried into an aquifer.</h2>
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
The way I do it is suddenly, in the same sort of way that magicians try to pull a table cloth off a table when there's things on the table cloth.The sudden approach acts as an impulse of force and starts to accelerate the roll. But, the piece (assuming it has perforations) is off the roll before the roll can move, due to inertia. Then the roll will acclerate, move, slow down and stop. However, in accelerating, the roll will unravel. The bigger the impulse the more it will unravel.+++++++++++++++++++++++++++++++++++++++If on the other hand, the piece of paper is held firmly, and the roll is pulled, then the impulse is presumably given to the paper and the hand whose inertia is a lot more than that of the roll. So, I think I'd actually go for choice c)+++++++++++++++++++++++++++++++++++++This assumes that the roll is free to rotate.I think that a similar idea is behind the design and use of a "ballistic galvanometer". The charge is passed through the galvanometer quickly, as a current pulse. Then the needle starts to deflect, and the deflection is arranged to depend on the total charge that has passed through in the time of the current pulse.
Answer:
Explanation:
Moment of inertia of a disc = 1/2 M R²
Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and - ω₀ .
We shall apply law of conservation of angular momentum .
initial total angular momentum
I x ω₀ - 4I x ω₀ = - 3Iω₀
Let final common angular momentum be ω
total final angular momentum = ( I + 4I ) ω
Applying law of conservation of angular momentum
( I + 4I ) ω = - 3Iω₀
ω = - 3 / 5 ω₀ .
b )
Initial total rotational K E
= 1/2 I ω₀² + 1/2 4I ω₀²
= 1/2 x5I ω₀²
Final total rotational K E
= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²
= 1/2 x 9 / 5 I ω₀²
= 9 / 10I ω₀²
change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²
(9/10 - 5/2) xI ω₀²
=( .9 - 2.5 )I ω₀²
= - 1.6 I ω₀² Ans
