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igomit [66]
4 years ago
8

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

er beam toward Earth to signal its approach. The emitted wavelength of the laser light is 697 nm . Find the light's observed wavelength on Earth.
Physics
1 answer:
bagirrra123 [75]4 years ago
8 0

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship v= 0.201c

Wavelength of laser \lambda= 697\ nm

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}

\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}

\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}

\lambda=5.685\times10^{-7}\ m

\lambda=568.5\times10^{-9}\ m

\lambda=568.5\ nm

Hence, The wavelength of observed light on earth is 568.5 nm

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Answer:

Air molecules and suspended water droplets collide as they swirl around in the clouds. Warmer air and water droplets rise, carrying charges with them. The result is an excess of positive charge near the cloud tops, and an excess of negative charge in the bottom layers of the clouds.

they are knocked off some ice and added to other ice as they crash past each other

Explanation:

6 0
3 years ago
You want to estimate the height of a cathedral ceiling. You note that a pendulum extending from the ceiling almost touches the f
barxatty [35]
<h2>The height of the room is 36 m </h2>

Explanation:

The time period of the pendulum  T = 2π\sqrt{\frac{l}{g} }

here l is the length of pendulum and g is the acceleration of gravity

Thus l = \frac{T^2 g}{4\pi^2 }

and l = \frac{(12)^2 x 10}{4 x 9.8} = 36 m approx

4 0
4 years ago
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not n
Harlamova29_29 [7]

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

6 0
3 years ago
What will Al’s charge be when it comes an ion
kolbaska11 [484]
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5 0
4 years ago
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration
stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is \mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

F=G\frac{Mm}{r^{2}}(1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

M=5.972\times10^{24}\,kg

m=7.34767309\times10^{22}\,kg

r=384400\,km

G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}

Using those values on (1)

F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}

F\approx1.98193\times10^{20}N

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}

6 0
4 years ago
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