Calculate the volume of 10g of helium ( M= 4kg/kmol) at 25C and 600 mmHg
1 answer:
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When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
Answer:
1) 883 kgm2
2) 532 kgm2
3) 2.99 rad/s
4) 944 J
5) 6.87 m/s2
6) 1.8 rad/s
Explanation:
1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:
If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:
2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):
3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:
4)Kinetic energy before:
Kinetic energy after:
So the change in kinetic energy is: 2374 - 1430 = 944 J
5)
6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.