Which star has a similar surface temperature to our sun?
2 answers:
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Answer:
104
Explanation:
U = Energy stored in the solenoid = 6.00 μJ = 6.00 x 10⁻⁶ J
i = current flowing through the solenoid = 0.4 A
L = inductance of the solenoid
Energy stored in the solenoid is given as
U = (0.5) L i²
6.00 x 10⁻⁶ = (0.5) L (0.4)²
L = 75 x 10⁻⁶
Inductance is given as
l = length of the solenoid = 0.7 m
N = number of turns
r = radius = 5.00 cm = 0.05 m
Area of cross-section is given as
A = πr²
A = (3.14) (0.05)²
A = 0.00785 m²
Inductance is given as
N = 73
Winding density is given as
density = n =
n =
n = 104
Answer: Heat current through the insulator=196W
Electric power= 196W
Explanation: Given: Kglass = 0.040W/m
Temperature of inside glassTi=175°C
Temperature of outside glass To= 35°C
Area=1.4m^2 , L= 4×10^-2
Heat current(H)= K ×A× (Ti - To)/L
Substituting the values into the equation
H = 0.04 × K × 1.4 ×(175-35)/4×10^-2
H= 196W.
The electric power = Heat current =196W
The electric power is the magnitude of heat current
