Answer:
W / n = - 9133 J / mol, W / n = 3653 J / mol
, e = 0.600
Explanation:
The Carnot cycle is described by
In this case they indicate that the final volume is
V = 3V₀
In the part of the heat absorption cycle from the source is an isothermal expansion
W = n RT ln (V₀ / V)
W / n = 8.314 1000 ln (1/3)
W / n = - 9133 J / mol
During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times
W / n = 8.314 400 (3)
W / n = 3653 J / mol
The efficiency of the cycle is
e = 1- 400/1000
e = 0.600
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
The letter that answers this question correctly is E .
Using the formula: ΔY = V₀y * t + (1/2) * ay * t²
Solve for time and get: 1.968s
Then use: v = d/t in the x-direction and get: d = 3.936
On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
