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.
Answer:
i) O m/s² as velocity is constant
ii) 20m/s
iii)Hint:Calculate the area of the graph,that is;below the line, for the first 15 seconds only
Hope this helps.
It is A I have done this in my science.Try Quizlet for more info.<span />
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
The displacement covered by the worker in the whole trip is 30km.
The easiest way to solve this problem is using the displacement equation:
Δx = x₂ - x₁
A worker covers a distance of 40 km from his house to his place of
work, and 10 km towards his house back. So:
Δx = x₂ - x₁ = 40km - 10km = 30km