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dlinn [17]
3 years ago
8

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 .

At the window, the electric field of the wave has an rms value 0.0600 V/m .
How much energy does this wave carry through the window during a 30.0-s commercial? Express your answer with the appropriate units.
Physics
1 answer:
vesna_86 [32]3 years ago
3 0

Answer:

The energy of the wave is 1.435 x 10⁻⁴ J

Explanation:

Given;

area of the window, A = 0.5 m²

the rms value of the field, E = 0.06 V/m

The peak value of electric field is given by;

E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m

The average intensity of the wave is given by;

I_{avg} = \frac{c \epsilon_o E_o^2 }{2}\\\\I_{avg} =  \frac{(3*10^8)( 8.85*10^{-12}) (0.0849)^2 }{2}\\\\I_{avg} = 9.569*10^{-6} \ W/m^2

The average power of the wave is given by;

P = I x A

P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)

P = 4.784 x 10⁻⁶ W

The energy of the wave is given by;

E = P x t

E = (4.784 x 10⁻⁶ W)(30 s)

E = 1.435 x 10⁻⁴ J

Therefore, the energy of the wave is 1.435 x 10⁻⁴ J

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6. In an integrated circuit, each wafer is cut into sections, which
Katen [24]

Answer:

B. carry a single circuit and are placed in individual cases.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

Similarly, an integrated circuit (IC) also referred to as microchip can be defined as a semiconductor-based electronic component that comprises of many other tiny electronic components such as capacitors, resistors, transistors, and inductors.

Integrated circuits (ICs) are often used in virtually all modern electronic devices to carry out specific tasks or functions such as amplification, timer, oscillation, computer memory, microprocessor, etc.

A wafer can be defined as a thin slice of crystalline semiconductor such as silicon and germanium used typically for the construction of an integrated circuit.

In an integrated circuit, each wafer is cut into sections, which generally comprises of a single circuit that are placed in individual cases.

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7 0
3 years ago
Please help will be marked most brainlist !!!
Neko [114]

Answer:

d

Explanation:

all of the rest are true

6 0
2 years ago
Layla shot a balloon straight up in the air and it is in the air round trip for 6.80 s.
Katena32 [7]

Answer:

24mph

Explanation:

it really depends how high but the average speed for that quick will be atleast 24mph if not try 42mph if it is wrong

5 0
3 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
Anarel [89]

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

3 0
4 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
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