Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:
Explanation:
Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.
.
Now, let's use the ideal gas equation to the initial and the final state:
Let's recall that the term nR is a constant. That is why we can match these equations.
We can find a relation between the volumes of the initial and the final state.
Combining this equation with the first equation we have:
Now, we just need to solve this equation for T₂.
Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.
Here,
Finally, T2 will be:
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Just remember
Voltage = current times resistance
current = voltage over resistance
Current = 9/3 = 3