Im not so sure but it should be the
instantaneous speed
Answer:
The answer your looking for is option 2 - Inertia
The angular speed can be solve using the formula:
w = v / r
where w is the angular speed
v is the linear velocity
r is the radius of the object
w = ( 5 m / s ) / ( 5 cm ) ( 1 m / 100 cm )
w = 100 per second
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):
![dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5](https://tex.z-dn.net/?f=dV%20%3D%20%5Cbeta%20V_%7B0%7D%20dT%5C%5CdV%20%3D%20%2869%2A10%5E%7B-6%7D%29%28%20C%29%5E%7B-1%7D%20%2A%20%281.50%2A10%5E%7B-3%7D%29m%5E%7B3%7D%20%2A%20%28298%29C%5C%5CdV%20%3D%203.0843%2A10%5E-5)
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 = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J](https://tex.z-dn.net/?f=W%20%3D%20P%20%2A%20dV%20%3D%20%281.01%20%2A%2010%5E5%29Pa%20%2A%20%283.0843%2A10%5E%7B-5%7D%29m%5E3%5C%5CW%20%3D%203.115143J)
W = 3.12 J
Hope this helps!
The answer is either C or D.