Answer:
The Romans adopted the Greek's use of realistic features when making human art.
Explanation:
i took the test
Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
Answer
TL/TH- TL
Because we know that power coefficient is. = QL/QH-QL
=so using this for performance we have
=>Perf= TL/(TH-TL)
Answer:
The pressure exerted by this man on ground
(a) if he stands on both feet is 8.17 KPa
(b) if he stands on one foot is 16.33 KPa
Explanation:
(a)
When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:
Area = A = 2 x 480 cm²
A = 960 cm²
A = 0.096 m²
The force exerted by man on his area will be equal to his weight.
Force = F = Weight
F = mg
F = (80 kg)(9.8 m/s²)
F = 784 N
Now, the pressure exerted by man on ground will be:
Pressure = P = F/A
P = 784 N/0.096 m²
<u>P = 8166.67 Pa = 8.17 KPa</u>
(b)
When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:
Area = A = 480 cm²
A = 0.048 m²
The force exerted by man on his area will be equal to his weight, in this case, as well.
Force = F = Weight
F = mg
F = (80 kg)(9.8 m/s²)
F = 784 N
Now, the pressure exerted by man on ground will be:
Pressure = P = F/A
P = 784 N/0.048 m²
<u>P = 16333.33 Pa = 16.33 KPa</u>
