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VashaNatasha [74]
3 years ago
5

Which describes Michael Faraday’s work with electricity and magnetism?

Physics
1 answer:
Viefleur [7K]3 years ago
8 0
Genius right dear friend

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(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i
a_sh-v [17]

Answer:

a)<em> It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves. </em>

b) <em>The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds. </em>

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

  • Gamma rays
  • X-rays
  • Ultraviolet rays
  • Visible region
  • Infrared
  • Microwave
  • Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of 3x10^{8}m/s.  

<em>a) Find the time it took for his voice to reach the Earth via radio waves.</em>

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

c = \frac{d}{t}  (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

t = \frac{d}{c}  (2)

The distance from the Earth to the Moon is 3.85x10^{8} m, therefore.

t = \frac{3.85x10^{8} m}{3x10^{8}m/s}

t = 1.28s

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

<em>b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.</em>

The distance from the Earth to the Mars at its closest approach is 5.76x10^{10}m, therefore.

t = \frac{5.76x10^{10}m}{3x10^{8}m/s}

t = 192s

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0
3 years ago
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

g_e = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

g_e=G\frac{M}{r^2}

g=G\frac{0.9M}{r^2}

Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0
3 years ago
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th
Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0
3 years ago
Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as h
Arturiano [62]

Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.

So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle

As we know that arc length is given as

L = R\theta

length of chord is given as

L_c = 2Rsin(\frac{\theta}{2})

so here

L > L_c

R\theta > 2R sin(\frac{\theta}{2})

so we have

\frac{\theta}{2} > sin(\frac{\theta}{2})

6 0
2 years ago
PLEASE HELP! Can you do the write about it! Will thank and try to mark brainiest!
goblinko [34]
Move the objects faster to get more friction.
8 0
3 years ago
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