Answer:
Yes equation is valid.
Explanation:
Given:
h = (0.04 to 0.09)(D/d)^4*V^2/2*g
Using SI units to assign dimensions to every quantity as follows:
Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]
Hose diameter D = [m]
Nozzle tip diameter d = [m]
Fluid velocity in the hose V = [ m s^-1 ]
Acceleration of gravity g = [ m s^-2 ]
Using the Given Equation and plug the SI units of respective quantities:
h = (0.04 to 0.09)(D/d)^4*V^2/2*g
[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]
Simplify the equation above:
[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]
[m] = [m]
Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.
Answer:
a. The final temperature is 69.8°C
b. The final pressure is 2.67bar
c. The amount of entropy produced is 0.38568kJ/K
Explanation:
Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.
Probably low heat and a longer time. Better sad than sorry
Answer:
The condition is true when their voltage and current specifications with their impedance are matched or complementary to each other.
Explanation:
Solution
Yes it is possible or true to interface an IC with a different technology like the TTL to HCS12 ports. but the condition is that their current and voltage specifications should be matched and their impedance and power also should be matched.
What this implies is that both their voltage and current requirements should be complementary to each other so as their impedance.
Answer:
Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.
Explanation:
Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.
The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.
