A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum.
1 answer:
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The first law of thermodynamics can be written as
where
is the variation of internal energy of the system
is the amount of heat absorbed by the system
is the work done by the system on the surrounding.
Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)
where
Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)
Answer:
(a) 1.939 m/h
(b) 0.926 m/h
(c) -0.315 m/h
(d) -1.21 m/h
Explanation:
Here, we have the water depth given by the function of time;
D(t) = 7 + 5·cos[0.503(t-6.75)]
Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;
D'(t) =
= 5×(-sin(0.503(t-6.75))×0.503
= -2.515×(-sin(0.503(t-6.75))
= -2.515×(-sin(0.503×t-3.395))
Therefore we have;
(a) at 5:00 AM = 5 - 0:00 = 5
D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h
(b) at 6:00 AM = 6 - 0:00 = 6
D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h
(c) at 7:00 AM = 7 - 0:00 = 7
D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h
(d) at Noon 12:00 PM = 12 - 0:00 = 12
D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.