Answer: Yes both gases would have the same entropy.
Explanation:
The formula for the change in the entropy is as follows,
Here, \Delta S is the change in the entropy, Q is the heat transfer and T is the temperature.
If the temperature of the system increases then there will be increase in the entropy as the randomness of the system increases.
In the given problem, if the both gases were initially at the same absolute temperature. Then there will be same entropy change in both gases.
Therefore, yes both gases would have the same entropy.
You do 0.5 x 2kg x 4 squared so
0.5x2x16
So the answer is 16
The amount of current flowing through the wire
The number of coils in the wire
The material in the core
The answer is D) The outcome of the experiment will be non observable
Answer:
p = 60.6N*s
Explanation:
v_f = v_0+a*t
a = (v_f-v_0)/t
a = (1.8m/s)/2.35s
a = 0.77m/s²
F = m*a
F = (25kg+8.5kg)*0.77m/s²
F = 25.8N
^p = F*t
p = 25.8N*2.35s
p = 60.6N*s