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Lyrx [107]
3 years ago
5

A train with proper length L has clocks at the front and back. A photon is fired from the front to the back. Working in the trai

n frame, we can easily say that if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c. Now consider this setup in the ground frame, where the train travels by at speed v. Rederive the above frame-independent result (namely, if the photon leaves the front of the train when a clock there reads zero, then it arrives at the back when a clock there reads L/c) by working only in the ground frame.
Physics
1 answer:
tamaranim1 [39]3 years ago
7 0

Explanation:

In train's rest frame, the speed of photon is c and the proper length of the train is L. The time taken by the photon to cross the train is t=\frac{L}{c}

In ground frame, the speed of the photon is given as follows:

v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}

=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c

The speed of light or photon remains same in every frame of reference.

Now, the speed of train is very less as compared to the speed of photon so that v So that, \frac{v}{c} \ll 1

The length contraction in the ground frame is given as follows:

L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}

=L

Time taken by the photon to travel the length of the train in ground frame is .

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You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb
mafiozo [28]

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

7 0
3 years ago
What type of electromagnetic radiation is being used in the picture?
pashok25 [27]

Answer: b i think

Explanation:

6 0
3 years ago
A 2kg Book is sitting on a table.a 10n Force is pulling to the right.a 3n Force is pulling to the left what is the net force act
salantis [7]


The net force = sum of all forces acting on the body



If we take left side as -ve and right side as +ve,
then,

The net force here would be equal to,
10N + (- 3N)
= 7N.


Therefore, a net force of +7N ( + indicates it's moving towards right) is acting on the book of mass 2kg.


4 0
3 years ago
Which of the following is strenuous?
Arada [10]

Answer:

I think it c

Explanation:

7 0
3 years ago
Read 2 more answers
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
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