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expeople1 [14]
3 years ago
8

What are the (time varying) amplitudes of the E and H fields if summer sunlight has an intensity of 1150 W/m2 in any Town?

Physics
1 answer:
Archy [21]3 years ago
3 0

Explanation:

Given that,

Intensity = 1150 W/m²

(a). We need to calculate the magnetic field

Using formula of intensity

I=\dfrac{E^2}{2\mu_{0}c}

E=\sqrt{2\times I\mu_{0}c}

Put the value into the formula

E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}

E=931.17\ N/C

Using formula of magnetic field

B = \dfrac{E}{c}

Put the value into the formula

B=\dfrac{931.17}{3\times10^{8}}

B=0.0000031039\ T

B=3.10\times10^{-6}\ T

(b). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth

We need to calculate the gravitational force

Using gravitational force

F=\dfrac{Gm_{s}M_{e}}{r^2}

Put the value into the formula

F=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}

F=3.522\times10^{22}\ N

We need to calculate the radiation force

Using formula of force

F_{R}=\dfrac{I}{c}\pi\timesR_{E}^{2}

Put the value into the formula

F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.378\times10^{6})^2

F_{R}=4.8\times10^{8}\ N

The gravitational and solar electromagnetic pressure forces of the sun on the earth

\dfrac{F_{G}}{F_{R}}=\dfrac{3.522\times10^{22}}{4.8\times10^{8}}

\dfrac{F_{G}}{F_{R}}=7.3375\times10^{13}

Hence, This is the required solution.

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Ejercicio 1: Un cuerpo gira en un círculo de 8cm de diámetro con una rapidez constante
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Answer:

Exercise 1;

The centripetal acceleration is approximately 94.52 m/s²

Explanation:

1) The given parameters are;

The diameter of the circle = 8 cm = 0.08 m

The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m

The speed of motion = 7 km/h = 1.944444 m/s

The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²

The centripetal acceleration ≈ 94.52 m/s²

8 0
3 years ago
Is rose quartz igneous sedimentary or metamorphic
Dmitriy789 [7]
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7 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
A 12-volt automotive circuit has a current of 3 amps. Technician A says the electric power in this circuit is 36 watts. Technici
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Answer:

Technician A is right.                                                    

Explanation:

Given that,

Voltage of circuit, V = 12 volt

Current in the circuit, I = 3 A

Technician A says the electric power in this circuit is 36 watts. Technician B says the electric power in this circuit is 4 watts. We need to say that which technician is correct.

The power of any circuit is given by :

P=V\times I

P=12\ V\times 3\ A

P = 36 watts

So, technician A is right. Hence, this is the required solution.

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2 years ago
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Option 2nd is the answer
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