Answer:
(A) –14m/s
(B) –42.0m
Explanation:
The complete solution can be found in the attachment below.
This involves the knowledge of motion under the action of gravity.
Check below for the full solution to the problem.
Explanation:
Below is an attachment containing the solution.
(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

Ionic bonds are formed between a cation (metal) and an anion (nonmetal)