A movable piston having a mass of 8.00 kg and a cross-sectional area of 5.00 cm2 traps 0.430 moles of an ideal gas in a vertical
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Answer:
Part a)
Part b)
Part c)
Part d)
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Explanation:
Part a)
When elevator is ascending with constant speed then we will have
So it will read same as that of the mass
Part b)
When elevator is decending with constant speed then we will have
So it will read same as that of the mass
Part c)
When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have
Reading is given as
Part d)
Here the speed of the elevator is constant initially
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass