B I think because GTOW is greater than 5 lbs than 55
Answer:
Absolute pressure , P(abs)= 433.31 KPa
Explanation:
Given that
Gauge pressure P(gauge)= 50 psi
We know that barometer reads atmospheric pressure
Atmospheric pressure P(atm) = 29.1 inches of Hg
We know that
1 psi = 6.89 KPa
So 50 psi = 6.89 x 50 KPa
P(gauge)= 50 psi =344.72 KPa
We know that
1 inch = 0.0254 m
29.1 inches = 0.739 m
Atmospheric pressure P(atm) = 0.739 m of Hg
We know that density of Hg =
P = ρ g h
P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa
P(atm) = 13.6 x 9.81 x 0.739 KPa
P(atm) =98.54 KPa
Now
Absolute pressure = Gauge pressure + Atmospheric pressure
P(abs)=P(gauge) + P(atm)
P(abs)= 344.72 KPa + 98.54 KPa
P(abs)= 433.31 KPa
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Answer:
Check the explanation
Explanation:
Question 1.
The secondary current of 250/5 amps CT when 300 amps(rated current of transmission line ) flow in TL is
(5/250 ) X 300 = 6 amps
Question 2
The correct answer to this second question is yes, when Over current relay coil will operate and relay contacts gets close, if the pickup value( Ip) of relay is set as 6 amps in relay. ( because primary current of TL is 1.2 times of CT primary)
Question 3
Tap Block figure (Fig 1) is not available/uploaded in your question.
Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft A, = 1000 rpm
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Now we know that for maximum velocity,
= 1064.1 rpm
Now we know
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Therefore moment of inertia of flywheel, I = m.
=30 X
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
=
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m