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2 years ago

7

As a student is performing a double slit experiment to determine the wavelength of a light source, she realizes that the nodal l

ines are too close together to be accurately measured. To increase the distance between the nodal lines, she could:

1 answer:

aleksandr82 [10.1K]2 years ago

6 0

To increase the distance between the nodal lines as it may be difficult to determine the wavelength of the light source because the nodal lines are too close, the student should spread the fringe pattern by having to be able to decrease the separation of the slit.

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Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

$\lambda = 0.186 m$

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

$f' = f_0 (\frac{v + v_o}{v - v_s})$

here we know that

$f_0 = 1750 Hz$

$v_o = 15 m/s$

$v_s = 15 m/s$

now we will have

$f' = (1750)(\frac{340 + 15}{340 - 15})$

$f' = 1911.5 Hz$

Part b)

Apparent wavelength is given by the formula

$\lambda = \frac{v_{relative}}{f_{app}}$

here we will have

$\lambda = \frac{340 + 15}{1911.5}$

$\lambda = 0.186 m$

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2 years ago

A planet of mass M and radius R has no atmosphere. The escape velocity at its surface is ve. An object of mass m is at rest a di

zubka84 [21]

Answer:

Explanation:

Expression for escape velocity

ve = $\sqrt{\frac{2GM}{R} }$

ve² R / 2 = GM

M is mass of the planet , R is radius of the planet .

At distance r >> R , potential energy of object

= $\frac{-GMm}{r}$

Since the object is at rest at that point , kinetic energy will be zero .

Total mechanical energy = $\frac{-GMm}{r}$ + 0 = $\frac{-GMm}{r}$

Putting the value of GM = ve² R / 2

Total mechanical energy = ve² Rm / 2 r

This mechanical energy will be conserved while falling down on the earth due to law of conservation of mechanical energy . So at surface of the earth , total mechanical energy

= ve² Rm / 2 r

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2 years ago

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

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2 years ago

You are provided with three polarizers with filters making angles of (A) 90 ∘ , (B) 180 ∘ and (C) −45 ∘ with respect

irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) $90^{\circ}$ :

The light intensity after passing through the first polarizer is $I_{o}$ and this intensity does not depend on the angle of the polarizer.

Consider $90^{\circ}$ with the vertical, the intensity is given by:

$I = I_{o}cos^{2}90^{\circ}$

$I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}$

(B) $180^{\circ}$ :

Suppose the second polarizer is $45^{\circ}$ with the vertical.

Now, intensity through the second polarizer:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)$

$I' = \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}$

Now, if we consider the second polarizer to be $180^{\circ}$ ,

$I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0$

(C) $- 45^{\circ}$ :

Now,

Intensity through the third polarizer, if it is $180^{\circ}$ with the vertical:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))$

$I' = \frac{I_{o}}{8}$

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2 years ago

An astronomer observes that the wavelength of light from a distant star is shifted toward the red part of the visible spectrum.

balandron [24]

Answer:

The distance between the earth and the star is increasing.

Explanation:

When we observe an object and its electromagnetic radiation has been displaced to blue, it means that it is getting closer to us, causing the light waves it emits to get closer together and its wavelength to decrease towards blue, this is knowm as blueshift.

On the contrary, when an object is rapidly moving away from us, the light waves or electromagnetic radiation it emits have been stretched from their normal wavelength to a longer wavelength, towards the red part of the spectrum. This is known as redshift.

This phenomenon of changes in wavelength and frequency due to movement (whether the source approaches or moves away) is described by the Doppler effect.

So for this case because the light we perceive from the star has moved to the red part of the visible spectrum, we can conclude that it is moving away from the earth, and that the distance between the star and the earth is increasing.

7 0

2 years ago