(a) 0.249 (24.9 %)
The maximum efficiency of a heat engine is given by
where
Tc is the low-temperature reservoir
Th is the high-temperature reservoir
For the engine in this problem,
Therefore the maximum efficiency is
(b-c) 0.221 (22.1 %)
The second steam engine operates using the exhaust of the first. So we have:
is the high-temperature reservoir
is the low-temperature reservoir
If we apply again the formula of the efficiency
The maximum efficiency of the second engine is
Answer:
It can only display one record at a time
Explanation:
Form ;
1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.
2.This is the most popular method of data entry
3.It may contain images in the background.
4.This can be sorted data regardless of its source of information.
Only option C is wrong.
Therefore the answer C is correct.
Answer:
Velocity
Explanation:
<u>Velocity</u> is the rate that an object moves in certain direction.
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
