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Arte-miy333 [17]
3 years ago
8

Has anyone done the science project: modeling waves/properties of light?

Physics
2 answers:
LekaFEV [45]3 years ago
8 0

Answer:

Here the other guy is right I’m answering so he can be marked brainliest

Explanation:

igor_vitrenko [27]3 years ago
7 0
<h2>MARK BRAINLIEST</h2>

For this assignment, you will develop several models that show how light waves and mechanical waves are reflected, absorbed, or transmitted through various materials. For each model, you will write a brief description of the interaction between the wave and the material. You will also compose two <u><em>typewritten</em></u> paragraphs. The first will compare and contrast light waves interacting with different materials. The second will explain why materials with certain properties are well suited for particular functions.

<h2><u>Background Information</u></h2>

A wave is any disturbance that carries energy from one place to another. There are two different types of waves: mechanical and electromagnetic. A mechanical wave carries energy through matter. Energy is transferred through vibrating particles of matter. Examples of mechanical waves include ocean waves, sound waves, and seismic waves. Like a mechanical wave, an electromagnetic wave can also carry energy through matter. However, unlike a mechanical wave, an electromagnetic wave does not need particles of matter to carry energy. Examples of electromagnetic waves include microwaves, visible light, X-rays, and radiation from the Sun.

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A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate
PilotLPTM [1.2K]

Answer:

T(1)=21

Explanation:

The equation of the position in kinematics is given:

x(t)=x_{0}+v_{0}t+0.5at^{2}

  • x(0) is the initial position, in this it is 0
  • v(0) is the initial velocity (20 m/s)
  • a is the acceleration (2 m/s²)

So the equation will be:

x(t)=20t+0.5*2*t^{2}

x(t)=20t+t^{2}    

Now, the Taylor polynomial equation is:

f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

f(t)=x(t)=20t+t^{2}

f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t

f''(t)=\frac{dv(t)}{dt}=a(t)=2

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}

f(0)=x(0)=0

f'(0)=v(0)=20

f''(0)=a(0)=2

T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}

T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}

T(t)=20t+t^{2}

Let's put t=1 so find the how far the car moves in the next second:

T(1)=20*1+1^{2}

T(1)=21

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!

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Answer:

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