Answer:
4.43 kW
Explanation:
Since Intensity I = P/A = E²/2cμ₀ where P = Power, A = Area = 4πr² where r = distance from source = 61 m and E = electric field amplitude = 8.45 V/m.
P = E²A/2cμ₀ = E²4πr²/2cμ₀ = 2πE²r²/cμ₀
= 2π(8.45 V/m)²(61 m)²/3 × 10⁸ m/s × 4π × 10⁻⁷ Tm/A
= 4428.1 W
= 4.4281 kW ≅ 4.43 kW
Answer:
Self inductance, 
Explanation:
It is given that,
Length of the coil, l = 5 cm = 0.05 m
Area of cross section of the coil, 
Number of turns in the coil, N = 130
The self inductance relates the magnetic flux linkage to the current through the coil and it is given by :
L = 0.000127 Henry
or
So, the self inductance of the coil is 127 microhenry. Hence, this is the required solution.
Answer:
1.31Kgms-1
Explanation:
∆p = m∆v
Where:
∆p= change in momentum
m= mass of the ball
∆v= change in velocity of the ball= (5.3-2.4)= 2.9ms-1
Therefore, substituting appropriately with the values above:
∆p= 0.45×2.9= 1.31Kgms-1
Answer:
To understand all the different kind of elements and how they work. Nature is a beautiful thing and it is important we understand how they survive and how we can help. The solar system is a mandatory nature. Learning about other planets also helps us feel safe and more knowledgable.
Answer:
(of the moon) have a progressively smaller part of its visible surface illuminated, so that it appears to decrease in size.
Explanation:
