Answer:
To help wheels move in a circle
Explanation:
1.Ross fixes a dishwasher for a homeowner.
3.Cassandra fixes holes in an old road.
The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.
<h3>Ohm's law</h3>
Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as
V = IR
Where
V is the potential difference
I is the current
R is the resistance
<h3>Technician A</h3>
High resistance causes an increase in current flow
V = IR
Divide both side by I
R = V / I
Thus, technician A is wrong as high resistance suggest low current flow
<h3>Technician B</h3>
Higher than normal voltage drop could indicate high resistance
V = IR
Thus, technician B is correct as high voltage indicates high resistance
<h3>Conclusion </h3>
From the above illustration, we can see that technician B is correct
Learn more about Ohm's law:
brainly.com/question/796939
Answer:
Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero
Explanation:
On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.
For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.
Answer:
The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s
Explanation:
Q = MCp(T2 - T1)
Q (quantity of heat) = Power (P) × time (t)
Density (D) = Mass (M)/Volume (V)
M = DV
Therefore, Pt = DVCp(T2 - T1)
V/t (volume flow rate) = P/DCp(T2 - T1)
P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K
Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)
The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s
