Answer:
25 V
Explanation:
It is convenient to use Kirchoff's current law (KCL), which tells you the sum of currents into a node is zero. The node of interest is the top left node.
The currents into it are ...
20 mA + (-5 -Vo)/(2kΩ) -(Vo/(5kΩ)) = 0
20 mA -2.5 mA = Vo(1/(2kΩ) +1/(5kΩ)) . . . . add the opposite of Vo terms
(17.5 mA)(10/7 kΩ) = Vo = 25 . . . volts . . . . divide by the coefficient of Vo
_____
You will notice that the equation resolves to what you would get if you drew the Norton equivalent of the voltage source with its 2k impedance. You have two current sources, one of +20 mA, and one of -2.5 mA supplying current to a load of 2k║5k = (10/7)kΩ. KCL tells you the total current into the node is equal to the current through that load (out of the node).
Answer:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature 
degree celcius
Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8
Q = 125.538 W
False you have to do the work and perricapate
Answer:
The requirement does not meet the feasibility standard.
Explanation:
because the people can just put on the blackbelts
