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3 years ago

5

in a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance th

ey travel by a multiple of

1 answer:

blagie [28]3 years ago

3 0

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

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A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

a

$l_o =52 \ m$

b

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

So

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

So

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

3 years ago

A divot is created _____.

STatiana [176]

I think it is when shot scrapes off the top of the turf

4 0

3 years ago

Read 2 more answers

Shayla was late for school. She ran the last half mile in 4 minutes. What was her average speed in miles per hour?

Shtirlitz [24]

Answer:

7.5 miles per hour

Explanation:

6 0

3 years ago

An observer sees a full moon in the night sky. What will the Moon's phase be 10 days later?

algol [13]

I'm pretty sure the moon would be a crescent within the 10 day period because it takes about 28 days for the moon to go through the different stages.

5 0

3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago