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Answer:
a) I_LED= 1/6 A b) Vf= 2.5V
Explanation:
Consider circuit in the attachment.
a) We will simplify current source in paraller with resistor to a voltage source in series with a resistor(see attachment 2)
Solving the circuit in attachment 2 using mesh analysis
-9+2I1+4(I1-I2)-4+2I1=0
8I1 - 4I2= 13 ............... eq 1
4+4(I2-I1)+ I2 + 2=0
4I1- 5I2 = 6 ............ eq 2
I1= 41/24 ; I2 = 1/6; I2= I_LED
b) Solving the circuit in attachment 2 again, this time I2=0
8I1 - 4I2= 13
8I1- 4(0)=13
I1= 13/8
Vf= 4(I1- I2) -4
I2=I_LED=0
Vf= 2.5 V
Answer:
0.0432 m^3/s
Explanation:
Internal diameter of smaller pipes = 2.5 cm = 0.025 m
pipa wall thickness = 3 mm = 0.003 m
internal diameter of larger pipes = 8 cm = 0.8 m
velocity of region between smaller and larger pipe = 10 m/s
Calculate discharge in m^3/s
First we calculate the area of the smaller pipe
A = =
= 0.00023571 m^2
next we calculate area of fluid between the smaller pipes and larger pipe
A =
=
= [ 0.00502857 - 0.00070713 ]
= 0.00432144 m^2
hence the discharge in m^3/s
Q = AV
= 0.00432144 * 10
= 0.0432 m^3/s