The relationship between frequency and wavelength for an electromagnetic wave is
where
f is the frequency
is the wavelength
is the speed of light.
For the light in our problem, the frequency is
, so its wavelength is (re-arranging the previous formula)
At the bottom of the tank :
P = ρgH
P = (1000 kg/m³)(10 m/s²)(1 m)
P = 10000 N/m²
F = P • A
F = (10000 N/m²)(1 m²)
F = 10000 N
At the side of the tank :
Pav = ½ρgH
Pav = ½(1000 kg/m³)(10 m/s²)(1 m)
Pav = 5000 N/m²
F = P • A
F = (5000 N/m²)(1 m²)
F = 5000 N
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the 
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Answer:
Quantity of charge = 80 Coulombs
Explanation:
Given the following data;
Current = 2 A
Time = 40 seconds
To find the amount of charge flowing through the light bulb;
Mathematically, the quantity of charge passing through a conductor is given by the formula;
Quantity of charge = current * time
Substituting into the formula, we have;
Quantity of charge = 2 * 40
Quantity of charge = 80 Coulombs
Ur answer is 3 and i'm sure of it
