Answer:
(ω₁ / ω₂) = 1.9079
Explanation:
Given
R₁ = 3.59 cm
R₂ = 7.22 cm
m₁ = m₂ = m
K₁ = K₂
We know that
K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²
if
v₁ = ω₁*R₁
and
I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²
∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² <em>(I)</em>
then
K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²
if
v₂ = ω₂*R₂
and
I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²
∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² <em>(II)</em>
<em>∵ </em>K₁ = K₂
⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²
⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²
⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²
⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)
⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²
⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)
⇒ (ω₁ / ω₂) = 1.9079
Answer:
C) three
Explanation:
Let gram of gold required be m . Let temperature change in both be Δ t .
heat absorbed = mass x specific heat x change in temperature
for copper
heat absorbed = 1 x .385 x Δt
for gold
heat absorbed = m x .129 x Δt
So
m x .129 x Δt = 1 x .385 x Δt
m = 2.98
= 3 g approximately .
Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
Answer: R=24.2Ω
Explanation: <u>Power</u> is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:
P=V.i
P=R.i²
The resistance of the system is:
R = 24.2Ω
<u>For the device, resistance is 24.2Ω.</u>
