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ladessa [460]
4 years ago
12

Tap on the photo. For each diagram, explain why the light behaves in the way that it does.

Physics
1 answer:
dem82 [27]4 years ago
3 0

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

n_1 sin \theta_1 = n_2 sin \theta_2

where n_1,n_2 are the refractive index of the 1st and 2nd medium

\theta_1, \theta_2 are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (n_1), this means that \theta_2, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (n_1>n_2), so it bends away from the normal (\theta_2>\theta_1).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

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adoni [48]

Answer:

(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

(b)\ F_{No} = 771.125N

Explanation:

Given

d_D = 6000ft ---- Altitude of container in Denver

A = 0.0155m^2 -- Surface Area of the container lid

P_D = 79000Pa --- Air pressure in Denver

P_{No} = 100250Pa --- Air pressure in New Orleans

<em>See comment for complete question</em>

Solving (a): The expression for F_{No

Force is calculated as:

F = \triangle P * A

The force in New Orleans is:

F_{No} = \triangle P * A

Since the inside pressure is half the pressure at sea level, then:

\triangle P = P_{No} - \frac{P_{area}}{2}

Where

P_{area} = 101000Pa --- Standard Pressure

Recall that:

F_{No} = \triangle P * A

This gives:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Solving (b): The value of F_{No

In (a), we have:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Where

A = 0.0155m^2

P_{No} = 100250Pa

P_{area} = 101000Pa

So, we have:

F_{No} = [100250 - \frac{101000}{2}] * 0.0155

F_{No} = [100250 - 50500] * 0.0155

F_{No} = 49750* 0.0155

F_{No} = 771.125N

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3 years ago
Find the missing spaces in the charts below! Please!
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Answer:

Zinc- Zn

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atomic number 17

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symbol Br

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atomic number 14

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3 years ago
A golfer hits a golf ball at an angle of 25 degrees to the ground at a speed of 76 m/s. If the gold ball covers a horizontal dis
Lady bird [3.3K]

The ball's horizontal position <em>x</em> and vertical position <em>y</em> at time <em>t</em> are given by

<em>x</em> = (76 m/s) cos(25º) <em>t</em>

<em>y</em> = (76 m/s) sin(25º) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball's initial vertical velocity is (76 m/s) sin(25º) ≈ 32.12 m/s.

Its initial horizontal velocity is (76 m/s) cos(25º) ≈ 68.88 m/s.

The ball stays in the air for as long as <em>y</em> > 0. Solve <em>y</em> = 0 for <em>t</em> :

(76 m/s) sin(25º) <em>t</em> - 1/2 <em>g</em> <em>t</em>² = 0

<em>t</em> ((76 m/s) sin(25º) - 1/2 <em>g</em> <em>t </em>) = 0

<em>t</em> = 0   or   (76 m/s) sin(25º) - 1/2 <em>g</em> <em>t</em> = 0

Ignore the first solution.

(76 m/s) sin(25º) - 1/2 <em>g</em> <em>t</em> = 0

(76 m/s) sin(25º) = (4.90 m/s²) <em>t</em>

<em>t</em> = (76 m/s) sin(25º) / (4.90 m/s²)

<em>t</em> ≈ 6.55 s

Recall that

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>y</em>

where <em>u</em> and <em>v</em> denote initial and final velocities, <em>a</em> is acceleration, and ∆<em>y</em> is displacement. At maximum height, the ball has zero vertical velocity, and taking the ball's starting position on the ground to be the origin, ∆<em>y</em> refers to the maximum height. So we have

0² - ((76 m/s) sin(25º))² = 2 (-<em>g</em>) ∆<em>y</em>

∆<em>y</em> = ((76 m/s) sin(25º))² / (2<em>g</em>)

∆<em>y</em> ≈ 52.6 m

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