Answer:

Explanation:
Moment of inertia of given shell
where
M represent sphere mass
R -sphere radius
we know linear speed is given as 
translational 
rotational 
total kinetic energy will be


fraction of rotaional to total K.E

Answer:For example, standard atmospheric pressure (or 1 atm) is defined as 101.325 kPa. The millibar, a unit of air pressure often used in meteorology, is equal to 100 Pa. (For comparison, one pound per square inch equals 6.895 kPa.)
Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.
I hope this helps.... I'm sorry if it doesn't
No force is necessary to keep a moving object moving (in a straight line at a constant speed).
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
Learn more about work done here: brainly.com/question/25573309
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Answer:
993.52 Hz
Explanation:
The frequency of sound emitted by the stationery train is 1057 Hz.
The car travels away from the train at 20.6 m/s.
The frequency the observer hears is given by the formula:

where v = velocity of sound = 343 m/s
vo = velocity of observer
f = frequency from source
This phenomenon is known as Doppler's effect.
Therefore:

The frequency heard by the observer is 993.52 Hz.