Answer:
No, the claim is not reasonable for 20 W electric power consumption.
It is reasonable for 40 W electric power consumption.
Explanation:
Power = (1/2)*mass flow rate*(square of velocity)
mass flow rate = 1 kg/s
velocity = 8 m/s
square of velocity = 64 m^2 / s^2
Power = (1/2)*(1)*(64)
Power = 32 W
For a fan that consumes 20 W power it is not possible to deliver more power than 20 W but this one is delivering 32 W hence it is a false claim.
For a fan that consumes 40 W it is indeed possible to deliver 32 W considering the efficiency. Hence this claim is reasonable.
Answer:
11.2mm or 0.45in
Explanation:
The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.
Please go through the attached file for a step by step solution to this question.
Answer:
Explanation:
This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.
Answer:
I = 8.3 Amp
potential drop = 83 V
Explanation:
Power = 100 KW
V = 12,000 V
R = 10 ohms
a)
Calculate current I in each wire:
P = I*V
I = P / V
I = 100 / 12 = 8.333 A
b)
Calculate potential drop in each wire:
V = I*R
V = (8.3) * (10)
V = 83 V
