Answer:
kinetic energy + potential energy
Answer:
E= 4.35*10^6 N/C
Explanation:
Let's find the area charge density of the plate
α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2
Now we can calculate the electric field just of the plate
E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C
Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
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!
I would have to say B failed because I think I read something about it being only 2law not 3