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2 years ago

What do diffraction and refraction have in common? pick only one

Physics

2 answers:

evablogger [386]2 years ago

8 0

B. they both involve wave interaction.

xeze [42]2 years ago

5 0

<h3><u>Answer;</u></h3>

They both involve wave interactions

<h3><u>Explanation;</u></h3>

<em><u>Refraction is a property of a wave that refers to the bending of a wave as it moves from one medium to another.</u></em>
<em><u>Diffraction on the other hand is a property of a wave such that an object causes a wave to change direction and bend towards it.</u></em>
<u><em>Diffraction and refraction differ in that, refraction occurs when waves pass through an object while diffraction occurs when waves pass around an object.</em></u>
<u><em>The similarity between refraction and diffraction is that they both involve interactions of waves and causes waves to bend thus changing their direction.</em></u>

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What are the evidences of molecular theory of magnetism

Assoli18 [71]

Answer:

i hope it helps you please mark me as brainliest

Explanation:

Molecular theory of magnetism states, "If molecular magnets align in a row, then the substance exhibits magnetic property. If they are kept haphazardly, they do not exhibit magnetic property." This is the molecular theory of magnetism. If molecular magnets align in a row, then the substance exhibits magnetic property.

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2 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

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11 months ago

Is a cinder cone volcanoe constructive or deconstructive

Ksivusya [100]

Cinder cone volcanoes can be associated with either constructive or destructive margins.<span>
</span>

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2 years ago

If it takes you 20 joules of work to move a couch 10 meters in 10 seconds, what is the power?

Debora [2.8K]

Answer:

Power = 2 j/s

Explanation:

Power = Work / Time

= 20/10

= 2 j/s

4 0

1 year ago

U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.

ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2, the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

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1 year ago