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3 years ago

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A flat screen is located 0.55 m away from a single slit. Light with a wavelength of 544 nm (in vacuum) shines through the slit a

nd produces a diffraction pattern. The width of the central bright fringe on the screen is 0.052 m. What is the width of the slit?

1 answer:

kobusy [5.1K]3 years ago

5 0

Answer:

Explanation:

The diffraction pattern is given as

Sinθ = mλ/ω

Where m=1,2,3,4....

Now, when m=1

Sinθ = λ/ω

Then,

ω = λ/Sinθ

The width of the central bright fringe is given as

y=2Ltanθ. From trigonometric

Then,

θ=arctan(y/2L)

Given that,

y=0.052m

L=0.55m

θ=arctan(0.052/2×0.55)

θ=arctan(0.0473)

θ=2.71°

Substituting this into

ω = λ/Sinθ

Since λ=544nm=544×10^-9m

Then,

ω = 544×10^-9/Si.2.71

ω = 1.15×10^-5m.

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A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

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Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

$v = \lambda f$

here we know that the wavelength of the wave is

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now speed of the wave is given as

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Now time taken by the wave to reach 5 m distance is

$t = \frac{L}{v}$

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3 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

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orbital speeds = 220 km/s

we know that

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Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

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f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

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$m = 2.23 \times 10^{-32} kg$

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3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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2 years ago

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Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

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Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

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= 0 + .5 x 2 x 10²

= 100 m

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s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0

3 years ago