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andrew-mc [135]
3 years ago
10

Imagine that a relativistic train travels past your lecture room at a speed of 0.894 c. The passengers of the train claim that w

hen they measure the length of the train using standard meter bars found on the train, the length comes out to be 90.6 m. What would be the length of the train in the frame of the lecture room, if you measured it?
Physics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

The length of the train in the frame of the lecture room is 40.59 m.

Explanation:

Given that,

Speed = 0.894 c

Original length = 90.6 m

We need to calculate the length of the train in the frame of the lecture room

Using formula of length contraction

L=L_{0}\sqrt{1-(\dfrac{v}{c})^2}

Where, L_{0}=original length

v = speed of train

Put the value into the formula

L=90.6\sqrt{1-(\dfrac{0.894c}{c})^2}

L=90.6\sqrt{1-0.894^2}

L=40.59\ m

Hence, The length of the train in the frame of the lecture room is 40.59 m.

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When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

       F  =  mg  =  0.350 * 9.8

       F  =3.43 \ N

So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

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7 0
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Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far
olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

5 0
3 years ago
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