The catalyst lowers the activation energy.
Answer:
a) 0 J
b) W = nRTln(Vf/Vi)
c) ΔQ = nRTln(Vf/Vi)
d) ΔQ = W
Explanation:
a) To find the change in the internal energy you use the 1st law of thermodynamics:
Q: heat transfer
W: work done by the gas
The gas is compressed isothermally, then, there is no change in the internal energy and you have
ΔU = 0 J
b) The work is done by the gas, not over the gas.
The work is given by the following formula:
n: moles
R: ideal gas constant
T: constant temperature
Vf: final volume
Vi: initial volume
Vf < Vi, then W < 0 and the work is done on the gas
c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.
The amount of heat is equal to the work done W
d)
Answer:
The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."
Explanation:
The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame
take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2
relative velocity of the object to the S' frame would be
Vrel = v2- v
This means the Vrel of the object with respect to the S' frame is less than the Vrel of the object with respect to the S frame
However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.
This would mean the second option is the answer, the relative speed of the object depends on the actual values.
Answer:
244mm
Explanation:
I₁ = 3.35A
I₂ = 6.99A
μ₀ = 4π*10^-7
force per unit length (F/L) = 6.03*10⁻⁵N/m
B = (μ₀ I₁ I₂ )/ 2πr .........equation i
B = F / L ..........equation ii
equating equation i & ii,
F / L = (μ₀ I₁ I₂ )/ 2πr
Note F/L = B = F
F = (μ₀ I₁ I₂ ) / 2πr
2πr*F = (μ₀ I₁ I₂ )
r = (μ₀ I₁ I₂ ) / 2πF
r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵
r = 1.4713*10⁻⁵ / 6.03*10⁻⁵
r = 0.244m = 244mm
The distance between the wires is 244m
