Answer:
Explanation:
We are given:
m = 1.06Kg
T = 22kj
Therefore we need to find coefficient performance or the cycle
= 5
For the amount of heat absorbed:
= 5 × 22 = 110KJ
For the amount of heat rejected:
= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;
= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at 
= 12.4°c we have 442KPa
Answer:
See explanation for step by step procedure to get answer.
Explanation:
Given that:
The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.
See the attachments for complete steps to get answer.
Answer:
Mass of cement used is 62.5 lb
Mass of gravel used is 225 lb
Explanation:
The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6
So, for 150 lb of sand
C : S : G = 1 : 2.4 : 3.6
Mass of cement used is 62.5 lb
Mass of gravel used is 225 lb
