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UNO [17]
4 years ago
11

The interference pattern seen when light passes through narrow, closely spaced slits, is due to

Physics
1 answer:
Lelu [443]4 years ago
8 0
<h2>Answer: Diffraction</h2><h2 />

Diffraction  is a characteristic phenomenon that occurs in all types of waves .

In this sense,  <u>diffraction</u> happens when a wave (the light in this case) meets an obstacle or a slit .When this occurs, the light bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming <u><em>multiple patterns</em></u> with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that <u>the obstacle must be comparable in size (similar size) to the size of the wavelength. </u>

<u />

<u />

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The answer is 2. Magnetite. Hope I could help you :)
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3 years ago
Akeem is walking on the beach on a hot summer day.
creativ13 [48]

<u>Answer:</u>

<em>Sand in a beach is warmer than the water of the sea. </em>

<u>Explanation:</u>

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But water is highly reflective and can reflect off a <em>major portion of the sunlight falling on the sea</em>. The sea is also deep and thus the heat spreads through a large volume unlike in the <em>case of sand.</em> Water also has the nature of constant movement unlike sand which is stable.  

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8 0
3 years ago
Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

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