Answer:
1) joule
2) 
3) 
Explanation:
1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second (
), which is equal to Watts (
).
This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.
Therefore, if we want to calculate luminosity the Joule as a unit will be used.
2) Work is expressed as force 
multiplied by the distane 
:
Where force has units of 
and distance units of 
.
If we input the units we will have:
This is 1Joule (
) in the SI system, which is also equal to 
3) The formula to calculate the percent error is:
Where:
is the experimental value
is the accepted value
This is the percent error
Slide with her left foot. hope this is helpful
Answer:
114.26
Explanation:
a)Formula for per unit impedance for change of base is
Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)
Zpu2: New per unit impedance
Zpu1: given per unit impedance
kV1: give base voltage
kV2: New bas votlage
kVA1: given bas power
kVA2: new base power
In the question
Zpu2=??
Zpu1= 0.3
kV2=24kV
kV1= 13.8 kV
kVA2= 1MVA ×1000= 1000 kVA
kVA1=500kVA
Zpu2= 0.3(13.8/24)²×(1000/500)
Zpu2= 0.198
b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,
Zbase= kV²/MVA
Zbase= 13.8²/(500/1000)
Zbase=380.88
Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:
Zpu=Zactual/Zbase
0.3= Zactual/380.88
Zactual= 114.26 ohms
