Answer:
(a) W=217 J
(b) Tc=378K
(c) e=0.39=39%
Explanation:
For part (a)
We to calculate the mechanical work W the engine does. By knowing QC and QH can obtain the work using equation
W = IQHI — IQcl .....................eq(1)
Put given values for QH and QC into equation (1) to get
the mechanical work of the engine
W = 550 - 335
W=217 J
For part (b)
We want to determine the temperature of low temperature reservoir which means Tc
IQc|/|Qh| =TC/TH
for Tc
Tc=(IQc|/|Qh|)*TH
Now we can put values
Tc= 620K (335/
550.1)
Tc=378K
For part (c)
Here we want to find the thermal efficiency (e) of the cycle
e=1-TC/TH
e=1-(378/620)
e=0.39=39%
Answer:
v = 45.37 m/s
Explanation:
Given,
angle of inclination = 8.0°
Vertical height, H = 105 m
Initial K.E. = 0 J
Initial P.E. = m g H
Final PE = 0 J
Final KE = 
Using Conservation of energy
v = 45.37 m/s
Hence, speed of the skier at the bottom is equal to v = 45.37 m/s
1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
= 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11 AU
= 3.0.74 / 100 = 0.0374 AU
This is a trick question, isn't it? If a car is stuck in traffic, it doesn't matter how fast it can travel, because it isn't moving at all. Therefore the average speed would be zero, right?
