Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.
A. very small objects behave like like particles.
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
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
