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user100 [1]
3 years ago
6

Imagine you are riding on a yacht in the ocean and traveling at 20 mph. You then hit a golf ball at 100 mph from the deck of the

yacht. You see the ball move away from you at 100mph, while a person standing on a near by beach would observe your golf ball traveling at 120 mph (20 mph + 100 mph). Now imagine you are aboard the Hermes spacecraft traveling at 0.1c (1/10 the speed of light) past Mars and shine a laser from the front of the ship. You would see the light traveling at c (the speed of light) away from your ship. According to Einstein’s special relativity, how fast will a person on Mars observe the light to be traveling?
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
A naive guess would be 1.1c, according to classical mechanics. But Special relativity posited a new special law of adding velocities; so when one is travelling at 5m/s on a platform moving 5m/s, the result is very very close to 10m/s, but it is not exact. This deviation is more visible in higher speeds. There is a specific formula that gives us the speed of an object when it moves in a frame of reference, but in this case the answer is simple. The speed of light is an absolute barrier to the speed of any object and it is preserved in all frames of reference. Thus, a person will also measure a velocity of c for the light.
Whitepunk [10]3 years ago
6 0

Before solving this question, first we have to understand the special theory of relative.

As per classical mechanics, the velocity of light will be different in different frame of reference. The light moves in the ether medium which exists every where in the entire universe.

Let us consider a body which moves with a velocity v. Let light is coming along the direction of the body. As per classical mechanics,the velocity of light with respect to the body will be [ c-v].

Let us consider that light is coming from opposite direction. Hence, the velocity of light with respect to the observer will be c+v.

From above we see that velocity of light is different in both the cases which is wrong.

As per Einstein's special theory of relativity, the velocity of light will be same in every frame of reference i.e c=300000 km/s.

As per the question ,the space craft is moving with a velocity 0.1 c.

We are asked to calculate the velocity of the light with respect to an observer present in Mars.

Considering Einstein's theory of relativity, the velocity of light will be c [300000 km/s] with respect to the person in Mars.

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A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nu
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Answer:

First Question

    E   =   1.065*10^{-12} \  J

Second  Question

   The  wavelength is for an X-ray  

Explanation:

From the question we are told that

     The  width of the wall is  w =  10\ fm =  10*10^{-15 }\ m

     The  first excited state is  n_1  =  2

     The  ground state is   n_0 = 1

Gnerally the  energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

          E   =   \frac{h^2 }{ 8 * m  *  l^2 [ n_1^2 - n_0 ^2 ] }

Here  h is the Planck's constant with value  h =  6.62607015 * 10^{-34} J \cdot s

         m is the mass of proton with value m  = 1.67 * 10^{-27} \   kg

So    

          E  =   \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27})  *  (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }

=>        E   =   1.065*10^{-12} \  J

Generally the energy of the photon emitted is also mathematically represented as

             E  =  \frac{h * c }{ \lambda }

=>          \lambda  =  \frac{h * c }{E }

=>          \lambda  =  \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }

=>         \lambda  =  1.87*10^{-10} \  m

Generally the range of wavelength of X-ray is  10^{-8} \to  1)^{-12}

So this wavelength is for an X-ray.

     

8 0
3 years ago
A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje
Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

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eduard

Explanation:

Sorry but I don't Understand question

3 0
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Which graph shows an object that is dropped?
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Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0
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