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Verizon [17]
3 years ago
6

When light reaches a barrier or the edge of an opening it diffracts, which means that the light _________.

Physics
2 answers:
katrin2010 [14]3 years ago
7 0
<span>The light slightly "bends" as it goes around the edges of the object. This bending is typically dependent upon the size of the barrier in comparison to the wavelength of the object being diffracted. The more similar in size the two are, the much more noticeable the diffraction tends to be.</span>
Reptile [31]3 years ago
3 0
Diffraction is the slight bending of light when it passes around the edge of an object. The amount of bending depends on the relative of the  wavelength of the light wave to the size of the opening. If the opening is much larger than the wavelength the bending will be almost unnoticeable but if the two are equal or closer in length then the diffraction will be noted. Therefore, the correct answer is bends. 
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Lady bird [3.3K]

Answer:

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Explanation:

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4 0
3 years ago
The small ball of mass m = 0.5 kg is attached to point A via string and is moving at constant speed in a horizontal circle of ra
babymother [125]

Answer:

d = 2.45 meters

Explanation:

Mass of the ball, m = 0.5 kg

Radius of the circle, r = 0.16 m

The angular speed of the ball around the circle is, \omega=2\ rad/s

The attached figure shows the whole scenario. Let F_t is the force acting on the ball in tangential direction. The forces will balanced each other at equilibrium.

In horizontal direction,

T\ sin\theta=F_t=mr\omega^2................(1)

In vertical direction,

T\ cos\theta=mg...............(2)

From equation (1) and (2) :

tan\theta=\dfrac{r\omega^2}{g}

Also,

tan\theta=\dfrac{r}{d}

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3 0
3 years ago
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t
stellarik [79]

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

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magnitude of magnetic field, B = 30 mT = 0.03 T

maximum value of induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

E = NBA(2πf)

f = \frac{E_{max}}{2\pi*NBA}

where;

f is the frequency of the coil

f = \frac{E_{max}}{2\pi*NBA}\\\\f = \frac{8}{2\pi(200)(0.03)(0.03)} \\\\f = 7.07 \ Hz

Therefore, the frequency of the coil is 7.07 Hz

5 0
3 years ago
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KATRIN_1 [288]
Answer:
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