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Korolek [52]
2 years ago
6

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

0 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional). Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
Physics
1 answer:
Rus_ich [418]2 years ago
7 0

Answer:

6.99535\times 10^{-6}\ V/m

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

Intensity of Electric radiation

I=\frac{P}{A}\\\Rightarrow I=\frac{P}{4\pi r^2}\\\Rightarrow I=\frac{1000}{4\pi\times 35000000^2}\ W/m^2

Intensity of Electric radiation

I=\frac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\frac{2\times \frac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m

The amplitude of the electric field vector is 6.99535\times 10^{-6}\ V/m

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A conducting rod (length = 80 cm) rotates at a constant angular rate of 15 revolutions per second about a pivot at one end. A un
Studentka2010 [4]

Answer:

3.62 V

Explanation:

L = 80 cm = 0.8 m

f = 15 rps

B = 60 m T = 0.060 T

ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s

v = r ω

here, r be the radius of circular path. Here r = length of rod = L

v = 0.80 x 94.2 = 75.36 m/s

The motional emf is given by

e = B  v  L = 0.060 x 75.36 x 0.8 = 3.62 V

4 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
Bess [88]
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
2 years ago
The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
2 years ago
If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i
slavikrds [6]

Answer:

3t_{1/2}  

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}     (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em>t_{1/2}<em>: is the half-lifes.</em>

By solving equation (1) for t we have:

\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}

<u>Having that:</u>

Nt = 450

N₀ = 3150 + 450 = 3600,

The half-lifes of the isotope is:

t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

3 0
2 years ago
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