According to Keplar's second law of planetary motion, "The areas swept by the radial vector connecting the star (sun) and the planet orbiting it are equal for equal intervals of time
Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
Radiation
Explanation:
Radiation is the type of heat transfer through the electromagnetic waves. Transfer of heat by radiation can occur in space/vacuum as in between the sun and the earth
On the other hand:
Conduction is the transfer of heat through touching on objects while convection is the transfer of heat as a result of moving currents in liquids or gases.
Hope this helps :)
