Answer:
35psig
Explanation:
Pressure is a force that acts on a surface area. The psig measure the pressure pounds per square inch gauge. It measure the difference in pressure between supply tank and outside air. This pressure is relative to atmospheric pressure. When charging an R-410A system, the vapor pressure should be at least 35 psig before switching to liquid charging.
Answer:
The velocity the air will enter the room through an opening is 20.9
Explanation:
We have to consider 2 points
Point 1 = Being away from the laboratory door
Point 2 = Gab between the floor and the laboratory door
Using the Bernoulli equation
Assuming
p1 = patm
p2 = patm + plab
patm will be cancel and both points will be at the ground level
= - у H2O * hH2O
Substituting the results
V2 = 20.9
Answer:
R = 1 kΩ
i_L = 2.2 mA
Explanation:
The complete question is given in attachment:
Given:
- The gain i_L / i_i = 11
- Given circuit Attachment
Find:
- Find the required value of R
- If the amplifier is fed with a current source having a current of 0.2 mA and a source resistance of 10 kΩ, find iL
Solution:
- Use KVL on right most loop (load loop):
v_x = 0 - i_i*10 = -10i_i
- Using ohm's law:
i = (0 - v_x) / R
i = 10*i_i / R
- Use KCL at the node:
i_L = i + i_i
- Substitute the the two i:
i_L = 10*i_i / R + i_i
i_L / i_i = ( 10/ R + 1 )
11 = ( 10/ R + 1 )
10 / R = 10
R = 1 kΩ
- The input resistance = 0 because there is a virtual ground at input. Then the value of source resistance will have no effect on resulting i_l, hence:
i_L = i_L / i_i * i_i
i_L = 11*0.2
i_L = 2.2 mA
Answer:
a fracture will occur, because the Kc value is greater than the KIC (48.9901 MPa > 40 MPa)
Explanation:
the solution is in the attached Word file