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Kamila [148]
2 years ago
7

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

Physics
2 answers:
mezya [45]2 years ago
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]2 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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Help with 1 2 and 3 please
geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. I = \frac{q}{t} Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3.  So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

\frac{p1L1}{A1}  = \frac{p2L2}{A2}\\

We are looking for L2 so we can isolate using algebra to get:

\frac{A2(\frac{P1L1}{A1}) }{P2} = L2

If you fill in those values you get 0.0205

or 2.05 cm



6 0
3 years ago
Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.
katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

 M v + m v' = 0

 5000 x v - 10 x 4000 = 0

 5000 v = 40000

    v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

F = \dfrac{m(v_i-v_f)}{\Delta t}

F = \dfrac{5000(8-0)}{10}

      F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

4 0
3 years ago
Read 2 more answers
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
PLEASE HELP, evaporation,liquids,solids, and gases involved
sukhopar [10]
Yes they are involved
5 0
3 years ago
All of the waves in the electromagnetic spectrum are _______ waves.
crimeas [40]
<span>All of the waves in the electromagnetic spectrum are transverse waves.</span>
5 0
3 years ago
Read 2 more answers
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