Answer:
Explanation:
The two triangles are similar what is the length of DE
Answer:c>a>b
Explanation:
Given
All the engine extracts same amount of heat(Q) from High-temperature reservoir
For a) 400 and 500 K




For b)500 K and 600K


For c) 400 K and 600 K


So c will give the highest amount of work
c>a>b
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:

if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.
Answer: a) 10.4km
b) 0.62h
C) 17km/h
d) 9.1km/h
Explanation:
A) suppose you move in the positive direction of an x axis, from a first position of x1 = 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km.
B) time interval Dtwlk (= 0.50 h), but we don't know the driving time interval Dt/dr. However, we know that for the drive the displacement Dx/dr is 8.4 km and the average velocity vavg,dr is 70 km/h.
This average velocity is the ratio of the displacement for the drive to the time interval for the drive:
dv = dx/ dt
dt = dx/dv = 8.4/70 = 0.12h
t = ti + two
t = 0.12 + 0.50 = 0.62h
C) avg speed for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip.
Avg v = distance/time
10.4/0.62 = 16.8km/h
D) average speed is the ratio of the total distance you move to the total time interval you take to make that move. The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0.12 h + 0.50 h + 0.75 h = 1.37 h.
Avg speed = 12.4/1.37 = 9.1km/h