Answer:
1.312 in
Explanation:
Data provided in the question:
Weight of the compressor, W = 227 pound
Number of legs = 4
Maximum pressure = 42 psi
Now,
Let F be the force taken by the legs
Therefore,
W = 4F
or
227 pound = 4F
or
F = 56.75 pounds
Also,
Force = Pressure × Area
or
56.75 pounds = 42 psi × πr² [ r is the diameter of one leg]
or
r² = 0.4301
or
r = 0.656
therefore,
diameter = 2r = 2 × 0.656
= 1.312 in
Answer:
quivalent force-couple system at D, (b) the resultant of the loading and its line of action.Explanation:
Answer:
D. Overrule any other laws and traffic control devices.
Explanation:
Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.
Also, in the case of instructions given by construction flaggers, these instructions have priority over those from controlling devices. This is because during construction traffic controls are redirected from the norms. Therefore, the flaggers such be given more importance.
Answer: to be exact you need 28mm of tubing for that
Explanation:
When the election
The total amount of daily heat transfer is 1382.38 M w.
The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.
<u>Explanation:</u>
Given data,
= 10° C
= 250 w/ 
k
Pipe length = 20 m
Inner diameter 
= 6 cm, 
= 3 cm
Outer diameter 
= 8 cm, 
= 4 cm
The thickness of insulation is 4 cm.
= + 4
= 4+4
= 8 cm
is the heat transfer coefficient of convection inside, 
is the heat transfer coefficient of convection outside.
The heat transfer rate between ambient and steam is
watt
= 
watt
= 
watt
q = 15999.86 watt
The total amount of daily heat transfer = 15999.86 × 86400
= 1382.387904 watt
= 1382.38 M w
The total amount of daily heat transfer is 1382.38 M w.
b) The temperature on the outside surface of the gypsum plaster insulation.
q = 
15999.86 
- 10 = 7.96
= 17.96 ° C.
