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vodka [1.7K]
2 years ago
12

1. Light traveling in air (n1=1) hits a piece of glass at an angle of 30 degrees. The light refracts in the glass at an angle of

15 degrees. What is the index of refraction (RI) of the glass?
2. Light traveling in air (n1=1) hits a piece of glass at an angle of 45 degrees. If the glass has a refractive index of 1.5, what angle should the light bend at?

3. Light traveling in air (n1=1) hits a piece of glass at an angle of 37 degrees. The light refracts in the glass at an angle of 30 degrees. What is the index of refraction (RI) of the glass?

4. Light traveling in air (n1=1) hits a piece of glass at an angle of 60 degrees. The light refracts in the glass at an angle of 15 degrees. What is the index of refraction (RI) of the glass?

5. What is the angle of incidence for a light ray traveling from water (n2 = 1.33) into flint glass, if the angle of refraction is 30 degrees?
Physics
1 answer:
ollegr [7]2 years ago
5 0

Explanation:

hope it helps

mark brainliest please

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15. A volleyball player who weighs 650 Newtons jumps 0.500 meters vertically off the floor. What is her kinetic energy just befo
allochka39001 [22]
<h2>Kinetic energy just before hitting the floor is 324.57 J</h2>

Explanation:

Weight of volleyball player = 650 N

That is

            Mass x Acceleration due to gravity = 650

            Mass x 9.81 = 650

            Mass = 66.26 kg

We also have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0.5 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 9.81 x 0.5

v = 3.13 m/s

Velocity with which he lands on ground is 3.13 m/s

We have kinetic energy = 0.5 x Mass x Velocity²

Substituting

          Kinetic energy = 0.5 x 66.26 x 3.13²

           Kinetic energy = 324.57 J

Kinetic energy just before hitting the floor is 324.57 J

3 0
2 years ago
Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth.
bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}

3 0
3 years ago
When is the universal theme of a story often revealed?
elena-14-01-66 [18.8K]
It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>
7 0
3 years ago
Read 2 more answers
the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

3 0
3 years ago
Calculate the heat flux (in W/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°C
ExtremeBDS [4]

Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

5 0
3 years ago
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