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MArishka [77]
3 years ago
15

Which of these is a result of gravitational lensing?

Physics
1 answer:
valentina_108 [34]3 years ago
7 0

Answer:

Answer is the formation of multiple images of the same object .

Explanation:

I hope it's helpful!

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Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

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A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is​
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W = 25 J

Explanation:

Work done on an object is defined as

W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}

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The temperature of a body is from 200 to 300C.The change of temp at absolute scale is
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A car starts from rest with an acceleration of 8 m/s/s for a time of 4 s. What was the car's final velocity?
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2 years ago
A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle
Contact [7]
<h2>Answers: </h2>

1) 1.359, 1.403

2) 2.207(10)^{8}m/s,  2.138(10)^{8}m/s    

Explanation:

The described situation is known as Refraction.  

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index n is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum (c=3(10)^{8}m/s) and the speed of light v in the second medium:

n=\frac{c}{v}   (1)

On the other hand we have the Snell’s Law:  

n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})   (2)  

Where:  

n_{1} is the first medium refractive index . We are told is the air, hence n_{1}\approx 1

n_{2} is the second medium refractive index  

\theta_{1} is the angle of the incident ray  

\theta_{2} is the angle of the refracted ray  

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and \theta_{1}=57.0\º   \theta_{2}=38.1\º:

(1)sin(57.0\º)=n_{2}sin(38.1\º)  

Finding n_{2}:

n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}  

n_{2}=1.359   (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and \theta_{1}=57.0\º   \theta_{2}=36.7\º:

(1)sin(57.0\º)=n_{2}sin(36.7\º)  

Finding n_{2}:

n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}  

n_{2}=1.403   (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

n_{red}=\frac{c}{v_{red}}

v_{red}=\frac{c}{n_{red}}

Substituting (3) in this equation:

v_{red}=\frac{3(10)^{8}m/s}{1.359}

v_{red}=2.207(10)^{8}m/s >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

n_{violet}=\frac{c}{v_{violet}}

v_{violet}=\frac{c}{n_{violet}}

Substituting (4) in this equation:

v_{violet}=\frac{3(10)^{8}m/s}{1.403}

v_{red}=2.138(10)^{8}m/s >>>>Speed of violet light

3 0
3 years ago
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