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2 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]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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What does a lunar eclipse and a solar eclipse have in common

liraira [26]

During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".

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3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

2 years ago

How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s

solong [7]

Answer:

71 rpm

Explanation:

Given that:

Angular momentum (L) = 0.26

Diameter = 25cm = 0.25 cm

Radius, r = (d/2) = 0.125m

Mass = 5.6 kg

Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

= 0.175

= 0.175 / 5

= 0.035 kgm²

Angular speed (w) ;

w = L / I

w = 0.26 / 0.035

= 7.4285714

= 7.429 rad/s

w = (7.429 * 60/2π)

w = 445.74 / 2π rpm

w = 70.941724

Angular speed = 70.94 rpm

= 71 rpm

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2 years ago

How much have jellyfish populated since 2015?

My name is Ann [436]

About a mil sience 2014-2015

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3 years ago

Read 2 more answers

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago