Answer:
Explanation:
If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules increases, the average force of an individual collision increases, and the average number of collisions with the wall, per unit area, per second increases.
As volume is reduced, the gas molecules come closer together, which increases the number of collisions between them and their collisions with the container walls. Also, since the distance traveled by each molecule between successive collision decreases, the molecule velocity doesn't decrease much within collisions as a result of which, the average velocity is higher compared to when the gas is stored in a larger volume. Finally, due to constant collisions, the direction of molecule travel changes rapidly owing to which the acceleration of molecules increases.
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as


= 0.42 m/s²
Answer:

So then the difference of temperature across the material would be 
Explanation:
For this case we can use the Fourier Law of heat conduction given by the following equation:
(1)
Where k = thermal conductivity = 0.2 W/ mK
A= 1m^2 represent the cross sectional area
Q= 3KW represent the rate of heat transfer
is the temperature of difference that we want to find
represent the thickness of the material
If we solve
in absolute value from the equation (1) we got:

First we convert 3KW to W and we got:

And we have everything to replace and we got:

So then the difference of temperature across the material would be 
It is a physical change, as only the dimensions changed, not the chemical structure.