The figure(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle, starting from θ0=0 rad at t
=0s. Draw the angular-position-versus-time graph. Include an appropriate scale on both axes.
1 answer:
Answer:
As position is the product of velocity and time, the area under the velocity-time curve will be the position.
Between 2 and 4 s, the area increases linearly to 20(4) = 80
Between 4 and 6 s, the additional area is 0(2) = 0
Between 6 and 8 s the area increases linearly by -10(2) = -20
which is a decrease.
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Answer:
The system loses 90 kJ of heat
Explanation:
We can answer the question by using the 1st law of thermodynamics, which states that:
where
is the change in internal energy of the system
is the heat absorbed by the system (positive if absorbed, negative if released by the system)
is the work done by the system (positive if done by the system, negative if done by the surrounding on the system)
In this problem, we have:
is the work done (negative, because it is done by the surrounding on the system)
is the increase in internal energy
Using the equation above, we can find Q, the heat absorbed/released by the system:
And the negative sign means that the system has lost this heat.