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3 years ago

How are lasers used to determine the distance from earth to the moon?

2 answers:

5 0

A beam of laser is directed at a reflecting surface put on the moon when the beam of laser is reflected a receiver on the each measure the time since the beam was sent till it was received. Laser is simply light so it has constant velocity in vacuum ~ air (c = 3 x 10^8 m/s)

to find the distance:

t : time measured between launching the beam and receiving it

d : distance

d = ct

Over [174]3 years ago

3 0

Explanation :

Lasers are used to determine the distance from earth to the moon. It is known as the Lunar Laser Ranging experiment. Laser is nothing but a a light source.

This can be done by passing a flash of light towards the moon. A reflector is placed on the surface of moon by the members of Apollo mission. With the help of this reflector light is reflected back to the earth.

The time taken by the light to travel the surface of moon and back to earth is calculated. It was approximately 2.55 seconds. So, for one side time is 1.275 seconds.

We know that the speed of light is c i.e. $3\times 10^8\ m/s$ .

Distance from Earth to the Moon is, d = c t

So, $d=3\times 10^8\ m/s \times 1.275\ s$

$d=3.825\times 10^8\ m$

$d=382500\ Km$

Hence, this is the required solution.

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A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no

ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, $\vec{F_{net}} = \vec{F'} - \vec{F}$

$\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N$

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

$\vec{F_{net}} = 7 N$ towards North

4 0

3 years ago

A refrigerator is being pulled up a ramp with a horizontal force P, which acts at the top corner. The refrigerator has a mass of

vaieri [72.5K]

Answer:

(a) P = 459.055 N.

(b) the refrigerator tips.

Explanation:

Given, the angle of ramp is 20°.

When the weight of refrigerator is resolved in directions parallel and perpendicular to ramp, 75×g×sin(20°) and 75×g×cos(20°).

⇒ normal contact force is 75×g×cos(20°).

⇒ frictional force is 0.3×75×g×cos(20°) = 207.414 N

so, total opposite force is 207.414 + 75×g×sin(20°) = 459.055 N.

so, the force needed is P = 459.055 N

And as the moment due to both opposite force and P force are in same direction the refrigerator tips rather than just sliding.

4 0

3 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

Which phrase best describes wave motion?

Fittoniya [83]

I think the correct answer from the choices listed above is option A. Wave motion is a movement of energy through space or a medium . Some waves are visible light waves, heat waves, sound waves and the like. Hope this answers the question.

6 0

3 years ago

What type of circuit is illustrated?

Elis [28]

A. Parallel circuit

Short circuits are circuits that is like plugging in a USB into your computer and the other end to your iPhone, so it's not D.

Series circuits can come in all kinds of different shapes, so it's not C.

Open circuits are circuits that are opened, meaning that whenever you turn on the light and it's an open circuit, it won't work cause the two wires are not connected, meaning it's not D.

Parallel circuits are circuits that are in one straight line, just like parallel lines, meaning the answer would be A. (Look at the picture and don't get confused with the other question that's the same question and has the same answers, they both have different pictures, so look at the pictures they give you whenever they ask the question so your getting the right answer)

Hope this helped.

8 0

3 years ago

Read 2 more answers