Answer:
the active region is bound by cutoff region and saturation or power dissipation region.
Explanation:
Answer:
Cost of 1000 kilowatt hour = 6000 cents
Explanation:
Given that
Electricity cost is 6 cents per kilowatt hour.
And we have to found out the cost for one megawatt hour
We know that
1 kilowatt = 1000 watt
1 megawatt = = 1000000 watt
1 megawatt = 1000 kilowatt
1 megawatt hour = 1000 kilowatt hour
Given that cost of 1 kilowatt hour = 6 cents
So the cost of 1000 kilowatt hour = 6 x 1000 cents
Cost of 1000 kilowatt hour = 6000 cents
<span>3.36x10^5 Pascals
The ideal gas law is
PV=nRT
where
P = Pressure
V = Volume
n = number of moles of gas particles
R = Ideal gas constant
T = Absolute temperature
Since n and R will remain constant, let's divide both sides of the equation by T, getting
PV=nRT
PV/T=nR
Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation
P1V1/T1 = P2V2/T2
where
P1, V1, T1 = Initial pressure, volume, temperature
P2, V2, T2 = Final pressure, volume, temperature
Now convert the temperatures to absolute temperature by adding 273.15 to both of them.
T1 = 27 + 273.15 = 300.15
T2 = 157 + 273.15 = 430.15
Substitute the known values into the equation
1.5E5*0.75/300.15 = P2*0.48/430.15
And solve for P2
1.5E5*0.75/300.15 = P2*0.48/430.15
430.15 * 1.5E5*0.75/300.15 = P2*0.48
64522500*0.75/300.15 = P2*0.48
48391875/300.15 = P2*0.48
161225.6372 = P2*0.48
161225.6372/0.48 = P2
335886.7441 = P2
Rounding to 3 significant figures gives 3.36x10^5 Pascals.
(technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>
Answer:
Pressure of the gas = 12669 (Pa) and height of the oil is 1,24 meters
Explanation:
First, we can use the following sketch for an easy understanding, in the attached image we can see the two pressure gauges the one with mercury to the right and the other one with oil to left. We have all the information needed in the mercury pressure gauge, so we can determine the pressure inside the vessel because the fluid is a gas it will have the same pressure distributed inside the vessel (P1).
Since P1 = Pgas, we can use the same formula, but this time we need to determine the height of the column of oil in the pressure gauge.
The result is that the height of the oil column is higher than the height of the one that uses mercury, this is due to the higher density of mercury compared to oil.
Note: the information given in the units of the fluids is not correct because the density is always expressed in units of (mass /volume)