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ICE Princess25 [194]
3 years ago
12

The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun

just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn. Refer to the table of astronomical data inside the back cover.
Physics
1 answer:
Sphinxa [80]3 years ago
7 0

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

                                        I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

a)

- The intensity at the surface of Venus is calculated as:

                                       I_v = P_s / 4*pi*r^2_v

                                       I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

                                       I_v = 2,700 W / m^2

b)

- The intensity at the surface of Mars is calculated as:

                                       I_m = P_s / 4*pi*r^2_m

                                       I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

                                      I_m = 610 W / m^2

c)

- The intensity at the surface of Saturn is calculated as:

                                       I_s = P_s / 4*pi*r^2_s

                                       I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

                                      I_s = 16 W / m^2

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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
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Answer:

1.26\cdot 10^7 m/s

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Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

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Layers in Earth's atmosphere are defined as regions in which _____.
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b. different types of weather occur

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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
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Answer:

71.4583 Hz

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Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

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