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almond37 [142]
2 years ago
14

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

on with a speed of 0.800 c relative to the mother ship. As measured on the Earth, the spaceship is 0.200 ly from the Earth when the landing craft is launched.(b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens?
Physics
1 answer:
Thepotemich [5.8K]2 years ago
5 0

The distance to the Earth at the moment of the landing craft's launch as measured by the aliens is 0.16 light years.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the distance to the Earth at the moment of the landing craft's launch as measured by the aliens.

  • It is given that; the spaceship is 0.200 ly from the Earth when the landing craft is launched.
  • By Lorentz transformation equation, the value of the distance to the Earth at the moment of the landing craft's launch as measured by the aliens will be,

        L=L_0\sqrt{1-\frac{V^2}{c^2} } =0.2ly\sqrt{1-\frac{(0.6c)^2}{c^2} }=0.2ly\sqrt{1-0.36}=  0.16ly

Thus, we can conclude that, the distance to the Earth at the moment of the landing craft's launch as measured by the aliens is 0.16 light years.

Learn more about the frame of reference here:

brainly.com/question/20897534

SPJ4

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You have not provided the diagram, therefore, I cannot provide an exact answer.
However, I will try to help by explaining how to solve this problem.

When light moves from air to glass:
1- part of the light is reflected back into the air where the angle of incidence is equal to the angle of reflection
2- part of the light enters the water and refracts. The angle of refraction can be calculated using Snell's law.

In a diagram, the reflected ray would be the one getting back into air while the refracted ray would be the one entering the water.

You can check the attached diagram for further illustrations.

Hope this helps :)

6 0
3 years ago
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A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

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How does resistant affect the current in a cirtuit
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What does the zeroth law of thermodynamics allow us to define?
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4 years ago
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A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with
romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = 30.1 \ m/s

The velocity of red car = 45.4 \ m/s

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

t=\frac{45-30}{2.7}                                                              (∵ time=\frac{distance}{time})

t=5.56 \ sec

then,

The distance covered by trooper,

t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2

   =208.33 \ m

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= 45\times 5.56

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Maximum distance = 250.2-208.33

                                = 41.87 \ m                                                        

6 0
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