Answer:
N₂=20.05 rpm
Explanation:
Given that
R= 19 cm
I=0.13 kg.m²
N₁ = 24.2 rpm
ω₁= 2.5 rad/s
m= 173 g = 0.173 kg
v=1.2 m
Initial angular momentum L₁
L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)
Final linear momentum L₂
L₂= I₂ ω₂
I₂ = I + m r²
The is no any external torque that is why angular momentum will be conserve
L₁ = L₂
Iω₁ - m v r = I₂ ω₂
Iω₁ - m v r = ( I + m r²) ω₂
Now by putting the all values
Iω₁ - m v r = ( I + m r²) ω₂
0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂
0.325 - 0.0394 = 0.136 ω₂
ω₂ = 2.1 rad/s
N₂=20.05 rpm
<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
Hey I hope this helps in any way.
As the front<span> moves through, cool, fair </span>weather<span> is likely to follow. Warm </span>front<span> Forms when a moist, warm air mass slides up and over a cold air mass. As the warm air mass rises, it condenses into a broad area of clouds. A warm </span>front<span> brings gentle rain or light snow, followed by warmer, milder </span>weather<span>.</span>
Answer:
Explanation:
means initial angular velocity, which is 0 rev/min
means final angular velocity, which is 
t means time t= 3.20 s
one revolution is equivalent to 2πrad so the final angular velocity is:
= (2π/60) *2.513*10^{4} rad/s
= 2628.5 rad/s
so the angular acceleration, α will be:
α = 2628.5 rad/s / 3.20 s
so the rotational motion about a fixed axis is:
+ 2αΔTita where ΔTita is the angle in radians
so now find the ΔTita the subject of the formula
ΔTita = 
Force is all the same, but pressure will be higher on 3cm
