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3 years ago

Maureen takes notes in class. Wave Interactions

Physics

2 answers:

Monica [59]3 years ago

7 0

A. <span>I .................
</span>

m_a_m_a [10]3 years ago

5 0

Answer;

Explanation;

Wave interactions results from the interaction of waves and matter, these interactions occur when waves pass from one medium to another. The types of interactions includes; reflection, refraction, interference and diffraction.

Reflection takes place when waves bounce back from a surface that they cannot pass through.
Refraction takes place when waves bend as they enter a new medium at an angle and start traveling at a different speed.
Diffraction is another wave interaction that occur when waves spread out as they travel around obstacles or through openings in obstacles.
Interference occurs when two or more waves meet at the same place a the same time.

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Which point lies on the line with point slope equation : y+5=2(x+8)

insens350 [35]

Answer:

A. (-8,-5)

Explanation:

If you fill in A you get

-5 + 5 = 2(-8+ 8)

0 = 0

So this point is on the line

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4 years ago

Read 2 more answers

there is a skier at the top of the ski slope the skier has potential energy what gives the skier her potential energy

scoundrel [369]

Her height on the mountain gives her the PE that she has.

She got it from the work that some person or machine did to lift her up there.

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3 years ago

A thin flake of mica ( n = 1.58 ) is used to cover one slit ofa double slit interference arrangement. The central point on thevi

ELEN [110]

Answer:

the thickness of the mica is 6.64μm

Explanation:

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

Where

L = Thickness

n = Index of refraction of each material

$\lambda = Wavelength$

Our values are given as

$\Phi = 7(2\pi)L=tn_1 = 1.58n_2 = 1\lambda = 550nm$

Replacing our values at the previous equation we have

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))$

$t = \frac{7*550}{1.58-1}\\t = 6637.931nm \approx 6.64\mu m$

the thickness of the mica is 6.64μm

5 0

3 years ago

10. Find the mass density of a ball with 22.0 g and a radius of 0.875 cm.

Ede4ka [16]

Answer:

Mass (m) = 22.0 g

Radius (r) = 0.875 cm

Density (d) = ?

First find ( volume) ;

$v = \frac{4}{3} \pi \: {r}^{3} \\ \\ v = \frac{4}{3} \times 3.14 \times ( {0.875)}^{3} \\ \\ v = \frac{4}{3} \times 3.14 \times 0.66992187 \: {cm}^{3} \\ \\ v = 2.8 \: {cm}^{3}$