Which types of interactions are negligible at standard temperature and pressure?
1 answer:
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Answer:
W = 3.12 J
Explanation:
Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:
β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C
So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):
So ΔV = 3.0843*10^-5 m^3
Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa
To get work, multiply the air pressure and the volume change.
W = 3.12 J
Hope this helps!
Answer:
Explanation:
<u>Uniform Acceleration </u>
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:
Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.
Solving for a:
Substituting: