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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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Which of the following describes gamma rays?
Anarel [89]

Answer: short wavelength, high frequency

Explanation:

Gamma rays are highly energetic electromagnetic waves. High energy implies high frequency.

E = h ν

h is the Planck's constant, ν is the frequency.

For electromagnetic radiation, frequency is inversely proportional to wavelength. Thus, gamma rays have high frequency but short wavelength.

The frequency of gamma rays is greater than 10¹⁹ Hz and wavelength is less 10⁻¹² m.

3 0
3 years ago
Read 2 more answers
C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F. Calculate the equivalent capacitance
grin007 [14]

The equivalent capacitance between A and B points is 2.5F.

<h3>What is parallel plate capacitor?</h3>

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is given by

C = ε₀ A/d

From the given circuit C1, C2 and C3, C4 are in parallel  C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F

C1, C2 = 4 +4 =8F

C3, C4 = 2 +4 =6F

Now , all capacitors are in series.

Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2

Ceq = 2.5 F

Thus,  the equivalent capacitance between A and B points is 2.5F.

Learn more about parallel plate capacitor.

brainly.com/question/12733413

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6 0
2 years ago
A wave has a frequency of 775 Hz. What is its period?
Leto [7]

The period is simply the inverse of the frequency, therefore:

T = 1 / f

T = 1 / 775 Hz

T = 0.001290 s    → possible answer

T = 1.29 × 10⁻³ s   → possible answer

T = 1.29 ms          → possible answer

4 0
2 years ago
A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in
sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]            ----------------------    (1)

in which R₁ is the resistance at temperature T₁

R_{T(2) is the resistance at temperature T₂

Given that V= IR

R = \frac{V}{I}

Therefore, the resistance at temperature 28°C is;

R_{28}= \frac{120V}{1.36A}

= 88.24Ω

R_{T(2) = \frac{120V}{1.23A}

= 97.56Ω

From (1) above;

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]      

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

\frac{0.1056}{4.5*10^{-4}}= T-28^0C

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

P= I^2_2R_{T^0C

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0
3 years ago
A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

F = q(v \times B)

F = qvB \: sin \:  θ

F = 1.6 \times 10 ^{ - 19}  \times 5.02 \times 10 ^{6}  \times 0.180 \times  \: sin \: 60°

= 1.25 \times 10 ^{ - 13}  \: N

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

#SPJ4

4 0
1 year ago
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