To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is




The amplitude of electric field at the receiver is


The amplitude of induced emf by this signal between the ends of the receiving antenna is


Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,


Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
Answer:
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Explanation:
Answer:
1777.92 m/s
Explanation:
R = Radius of asteroid = 545 km
M = Mass of planet
g = Acceleration due to gravity = 2.9 m/s²
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
Acceleration due to gravity is given by

The expression of escape velocity is given by

The escape speed is 1777.92 m/s
<span>Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.</span>
Answer:
The force due to air resistance is 256 N.
Explanation:
Given;
mass of the plane, m = 5 kg
applied force on the plane, Fa = 706 N
the net force on the plane, ∑F= 450 N
Let the force due to air resistance = Fr
The net force on the plane is given as;
Net force = applied force - force due to air resistance
∑F = Fa - Fr
Fr = Fa - ∑F
Fr = 706 - 450
Fr = 256 N.
Therefore, the force due to air resistance is 256 N.