Answer:
The required cross-sectional area of the copper wires used to connect the source to the mine lights is 0.029mm²
Explanation:
Given.
Copper Resistivity = 1.69 *10^-8Ωm
The mine lights use a total of 5 kW and operate at 120 V dc.
So,
Power = 5kW
Power Required = 5% of 5kW
Power Required = 0.05 of 5000W
Power Required = 250W
Calculating the Resistance
The power lost in the wires is given by 120² / R, where R is the resistance.
250 = 120²/R
R = 120²/250
R = 57.6 Ω
This is small amount over 100 m.
Calculating the Cross-sectional area.
The resistance of the wires is given by:
R = 1.69 * 10^(-8) *100 / A
R = 1.69 * 10^-6/A, where A is the cross-sectional area.
1.69 * 10^(-6)/A = 57.6
A = 1.69 * 10^-6/57.6
A = 2.9340277777777E−8m²
A = 0.029mm²
Answer:
C which is square millimeters which is right.
Answer:
depends on the size
Explanation:
Length x width x depth x 7.5 = volume (in gallons)
Length times width gives the surface area of the pool. Multiplying that by the depth gives the volume in cubic feet. Since there are 7.5 gallons in each cubic foot, multiply the cubic feet of the pool by 7.5 to arrive at the volume of the pool, expressed in gallons.
Answer:
The electrical power is 96.5 W/m^2
Explanation:
The energy balance is:
Ein-Eout=0
if:
Gsky=oTsky^4
Eb=oTs^4
qc=h(Ts-Tα)
if Gl≈El(l,5800)
lt= 2*5800=11600 um-K, at this value, F=0.941
The hemispherical emissivity is equal to:
lt=2*333=666 K, at this value, F=0
The hemispherical absorptivity is equal to: