Answer:
V1 = 2221.33 L
Explanation:
The system is about a ideal gas. Then you can use the equation for ideal gases for a volume V1, temperature T1 and pressure P1:
(1)
And also for the situation in which the variables T, V and P has changed:
(1)
R: constant of ideal gases = 0.082 L.atm/mol.K
For both cases (1) and (2) the number of moles are the same. Next, you solve for n in (1) and (2):
Next, you equal these equations an solve for T2:
Finally you replace the values of P2, V2, T1 and T2:
Hence, the initial volume of the gas is 2221.33 L
Answer:
It got transferred to kinetic energy
Explanation:
In the vertical direction, we know that acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity), we know our initial velocity (vi) = 15m/s and our final velocity (vf) = 0m/s if the ball stops at its max height. Using this kinematics equation: vf^2 = vi^2 + 2a*h, where h is the max height of the ball, we can find h by solving: h = (vf^2 - vi^2)/2a, now we plug in numbers h = (0^2 - 15^2)/2(-9.8) = 11.48 m.
Now we can find the time, but let's use a trick, we know the acceleration will be the same going up and down and that it will cover the same distance. Using this logic, we can know that the ball will be going the exact same speed we threw it up at when it comes back down to our hand (provided our hand is at the same height). We can then say it will take the same amount of time to reach its peak after leaving our hand as it will take to go from its peak back down to our hand. Using this, we can just get the time it takes to get to the top of its arc and then multiple it by 2. Using d = (vf + vi)*t / 2, we can solve for t, so t = 2d/(vf+vi) = 2(11.48)/(15) = 1.53 s for the trip up, doubling it for the trip down, we get a total of 3.06 s to go up and back down to our hand.
Answer:
t = 35.16 s
Explanation:
Inertia
I = (2/5)*m*r²
I = (2/5) * 8200 kg * (0.9m)²
I = 2657 kg·m²
τ = I*α
30N * 0.9m = 2657kg·m² * α
α = 0.01016 rad/s²
Θ = ½α*t²
2π rads = ½ * 0.01016rad/s² * t²
t =sqrt(4π rads/0.0106rad/s²)
t = 35.16 s
