HELP this need to be turned it tomorrow

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 108 m. Find

Goshia [24]

Answer:

It took 1.28 seconds to his voice to reach the Earth via radio waves.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, 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, the highest being that of gamma rays, followed by X-rays, ultraviolet rays and the visible region , and those of lower frequencies, which correspond to infrared, microwave and radio waves.

Light propagates as electromagnetic wave in vacuum with a speed of $3x10^{8}m/s$ . Therefore, radio waves will have in vacuum the same speed.

Then, to know the time that it took for its voice, the next equation can be used:

$c = \frac{d}{t}$ (1)

Where c 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)

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

$t = 1.28s$

Hence, it took 1.28 seconds to his voice to reach the Earth via radio waves.

7 0

3 years ago

How do forest fires affect the Earth's atmosphere?

netineya [11]

It is B because the other ones are good.

5 0

3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

Why are different constellations<br> of stars seen during different<br> seasons?

slamgirl [31]

Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ? Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.

5 0

3 years ago

Read 2 more answers

How can you show that pressure increases with the increase of force?

Julli [10]

Answer:

...........................

7 0

3 years ago

Read 2 more answers