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DedPeter [7]
3 years ago
7

Help please!

Physics
1 answer:
kipiarov [429]3 years ago
3 0
<h2>Answer: Diamond</h2>

Explanation:

This described situation is known as Refraction, a phenomenon in which the light bends or changes it direction when passing through a medium with a index of refraction different from the other medium.  

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.  

According to Snell’s Law:  

n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2}) (1)  

Where:  

n_{1}=1 is the first medium index of refraction  (air)

n_{2} is the second medium index of refraction (the value we want to know)

\theta_{1}=27\° is the angle of the incident ray  

\theta_{2}=11\° is the angle of the refracted ray

Now, let's find n_{2} from (1):

n_{2}=n_{1}\frac{sin(\theta_{1}}{sin(\theta_{2}} (2)  

Substituting the known values:

n_{2}=(1)\frac{sin(27\°)}{sin(11\°)}}

Finally:

n_{2}=2.379\approx 2.4

If we compare this result with the given table, the index of refraction value that is close to this number is diamond's index of refraction.

Therefore, the correct option is A: the material is diamond.

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Which of the following are found within the electromagnetic spectrum? Check all that apply. sound waves visible light X rays ult
sattari [20]

Answer:

Visible light

X rays

ultraviolet radiation

gamma rays

microwave radiation

Explanation:

Electromagnetic waves consist of oscillating electric and magnetic fields which vibrate in a direction perpendicular to the direction of motion of the wave (transverse wave). Electromagnetic waves have all same speed in a vacuum (c=3.0\cdot 10^8 m/s, known as speed of light) and are classified into 7 different types according to their frequency and wavelength. This classification is called electromagnetic spectrum.

From lowest to highest wavelength, the 7 types are:

Gamma rays

X-rays

Ultraviolet radiation

Visible light

Infrared radiation

Microwaves

Radio waves

Sound waves, on the contrary, do not belong to the electromagnetic spectrum, since they are another type of wave called mechanical waves (which consist of vibrations of the particles in a medium).

8 0
3 years ago
How does the "lollipop moment", exemplify for us the opportunity we all have to exercise leadership on a normal everyday basis?
Mrrafil [7]

Answer:

WHat do you mean by that

Explanation:

I dont get what you re asking

8 0
2 years ago
Does Ap Physics have anything to do with probablity?
padilas [110]
So I'm a junior. I am currently taking AP Calc BC and AP Physics B.

As of now, I'm not sure if I should take AP Probability and Statistics or Differential Equations/Calc III next year. Also, I'm debating between taking AP Physics C or AP Chemistry.

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5 0
3 years ago
Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

Where Ep = Gravitational potential, m = mass of the diver,h = Height.

But,

W = mg.................... Equation 2

Where W = weight of the diver.

Substitute equation 2 into equation 1

Ep = Wh

Make W the subject of the equation

W = Ep/h................... Equation 3

Given: Ep = 25000 J, h = 44 m

Substitute into equation 3

W = 25000/44

W = 568.18 N.

Hence the weight of the diver = 568.18 N

5 0
3 years ago
A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
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