Answer:
a) 0.162 V/m
b) 0.54 nT
c) 22 kW
Explanation:
Given
Distance of flight, d = 20 km
Intensity of signal, I = 35.1 μW/m²
Magnitude of the electric component is gotten using the formula
E(m) = √(2Iμc), where
E(m) = √(2 * 35*10^-6 * 4*3.142*10^-7 * 3*10^8)
E(m) = √(2 * 35*10^-6 * 1.257*10^-6 * 3*10^8)
E(m) = √0.0264
E(m) = 0.162 V/m
Magnitude of magnetic component can be gotten by using the relation
B(m) = E(m) / c
B(m) = 0.162 / 3*10^8
B(m) = 5.4*10^-10
B(m) = 0.54 nT
Transmission power, P = IA
where A = 1/2 * 4πr²
P = 2Iπr²
P = 2 * 35*10^-6 * 3.142 * (10000)²
P = 2 * 35*10^-6 * 3.142 * 1*10^8
P = 21994 W
Thus, the transmission power is 22 kW
Answer:
Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges.
Answer: (A) "X: May use LF, HF, and UHF waves
Y: Travels at the speed of light
Z: May use MF and VHF waves"
Answer:
(a)
907.11 rad/s²
(b)
4082 rad
Explanation:
(a)
t = time taken = 3.00 s
w₀ = initial angular velocity of the drill = 0 rad/s
w = final angular velocity of the drill = 26000 rev/min = 26000 
= 2721.33 rad/s
α = angular acceleration of the drill
t = time interval = 3.00 s
using the equation
w = w₀ + α t
2721.33 = 0 + α (3)
α = 907.11 rad/s²
(b)
θ = angle
using the equation
θ = w₀ t + (0.5) α t²
θ = (0) (3.00) + (0.5) (907.11) (3.00)²
θ = 4082 rad
