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3 years ago

10

In optics, a diffraction ___________ is an optical component with a periodic structure, which splits and diffracts light into se

veral beams traveling in different directions

2 answers:

Grace [21]3 years ago

5 0

Answer:

<u>grating</u>

Explanation:

Diffraction is a phenomenon in which light bends at the edges of an object because of which light falls even in the shadow region.

Diffraction grating consists of thousands of slits through which the light splits. The diffracted light travels in different directions and forms a alternate dark and bright fringes.

galina1969 [7]3 years ago

4 0

In optics, a diffraction grating is an optical component with a periodic structure, which splits and diffracts light into several beams travelling in different directions.

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xercise 15.29: At a distance of 6.00×1012 m from a star, the intensity of the radiation from the star is 14.7 W/m2. Assuming tha

Fynjy0 [20]

Answer:

$6.65*10^{27}W$

Explanation:

The relationship of intensity:

$I=\frac{P}{4\pi r^{2} }$

Intensity: $I=14W/m^{2}$

radius: $r=6.00*10^{12} m$

Power is unknown

Now we can apply the formula

$I=\frac{P}{4\pi r^{2} }\\P=I(4\pi r^{2})\\P=14.7(4*\pi *(6.00*10^{12} )^{2} )\\P=6.65*10^{27} W$

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3 years ago

rutherford was the first to hypothesize that electrons orbit a positively charged nucleus a. true b.false

konstantin123 [22]

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3 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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