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atroni [7]
3 years ago
13

An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.63 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the magnetic field amplitude of the signal at that point
Physics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

B_2=2.1nT

Explanation:

From the question we are told that:

Frequency F=800kHz

Distance d=4.5km

Electric field amplitude B_2=0.63V/m

Generally the equation for momentum is mathematically given by

 B=\frac{E}{C}

Therefore

 B_2=\frac{0.63}{3*10^8}

 B_2=0.21*10^{-8}

 B_2=2.1nT

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In a convex lense f=20cm, m=1,then what is u and v?​
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A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

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Answer:

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