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pochemuha
3 years ago
6

If the light strikes the plastic (from the water) at an angle θw, at what angle θa does it emerge from the plastic (into the air

)? Express your answer in terms of nw, np, na, and θw. Remember that the inverse sine of a number x should be entered as asin(x) in the answer box.
Physics
2 answers:
Natasha2012 [34]3 years ago
6 0

Answer:

\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})

Explanation:

According to Snell's law the angle of incidence and angle of refraction are related by:

\frac{sin(\theta _w)}{sin(\theta _a)}=\frac{n_a}{n_w}\\\\\therefore sin(\theta _a)=\frac{sin(\theta _w)\times n_w}{n_a}\\\\\theta _a=sin^{-1}(\frac{sin(\theta _w)\times n_w}{n_a})\\\\\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})

umka2103 [35]3 years ago
6 0

Answer:

θa = asin(nwsin(θw)/na)

Explanation:

We are going to look at the two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface). In order to ascertain their relationship, draw a picture with the path the light follows in the plastic and the normals to both surfaces. Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normals are parallel lines. An important theorem from geometry will give you the relationship between the angles.

Using Snell's Law,

θa = asin(nwsin(θw)/na)

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Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
kifflom [539]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = \sqrt[4]{P/Ae}

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

4 0
3 years ago
a mass of 10kg is placed on a horizontal table with coefficient of friction is 0.5. if the mass is static, determine weight of t
zloy xaker [14]

Answer:

10

Explanation:

5 0
3 years ago
Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te
zlopas [31]

Answer: C

Explanation: Side post terminals need to be removed to inspect them for corrosion.

Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.

4 0
3 years ago
A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m

3 0
3 years ago
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Sindrei [870]
7.5m/s

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3 years ago
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