From the law of conservation of momentum
m1u1+ m2u2= m1v1+ m2v2
110*8+ 110*-10= 110*-10 + 110* v2
v2= 8 m/sec
Answer:
21.35 cm^3
Explanation:
let the volume at the surface of fresh water is V.
The volume at a depth of 100 m is V' = 2 cm^3
temperature remains constant.
density of water, d = 1000 kg/m^3
Pressure at the surface of fresh water is atmospheric pressure,
P = Po = 1.013 x 10^5 N/m^2
The pressure at depth 100 m is P' = Po + hdg
P' = 
P' = 10.813 x 10^5 N/m^2
Use the Boyle's law
P V = P' V'
V = 21.35 cm^3
Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.
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
