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dalvyx [7]
3 years ago
15

Pre-laboratory Assignment: Experiment 20 Reflection and Refraction of Light 1. When light is incident on a reflective surface, w

hat can be said about the angle and speed at which the light is reflected? (Information is in your ‘General Physics Laboratory Manual’ Chapt. 20) 2. At what angle is the normal drawn to the reflective surface or air-medium interface? 3. How are angles of incidence, angles of reflection and of refraction measured? 4. Describe what happens to a light ray as it enters from a medium of greater refractive index to a medium of lesser refractive index
Physics
1 answer:
shusha [124]3 years ago
6 0

Answer:

1) ngle of incidence and reflection are equal,  light carries does not change

2) the angle of this line with respect to the surface is 90º

3) protractor

4)    n₂  sin θ₂ = n_1 sin θ₁,  light ray must have a greater angle than the incident ray ,

Explanation:

1) When light falls on a reflective surface, the angle of incidence and reflection are equal and as it travels in the same medium, the speed that the light carries does not change

2) The normal is a line perpendicular to the point of incidence of light, so the angle of this line with respect to the surface is 90º

3) Angles are measured with a protractor

4) When light passes from one medium to another, the speed of the ray changes due to the difference in the refractive index in each medium, due to this change in speed the transmitted light ray must have a greater angle than the incident ray , since the speed increases as the density of the medium decreases

           \frac{sin \theta _2}{ sin \theta_1} = \frac{v_2}{v_1}

          \frac{c}{v_2} \  sin \theta_2 = \frac{c}{v_1}  \ sin \theta_1

          n₂  sin θ₂ = n_1 sin θ₁

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Light from a sodium lamp 1l = 589 nm2 illuminates two narrow slits. the fringe spacing on a screen 150 cm behind the slits is 4.
BartSMP [9]

d = 0.221 mm is the spacing (in mm) between the two slits..

here,

wavelength  = 589 nm

D = 150 cm  =  1.5 m

y  = 4 mm

then ,

y = mλD/d

for m  = `1

0.004  = 589*10^-9 * 1.5/d

d = 2.21*10^-4 m  

d = 0.221 mm

slit spacing is 0.221 mm

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7 0
2 years ago
Parts of a transverse wave include:
iris [78.8K]

Answer:

Parts of a transverse wave include:

Crest - the highest point on the wave

Trough - the lowest point on the wave

Wavelength - the distance from one crest to the next crest, or one trough to the next trough

Amplitude - the displacement of the wave from the midpoint to the highest point (crest) or the lowest point (trough)

Explanation:

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8 0
3 years ago
Read 2 more answers
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a
ryzh [129]

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = 0

m₁u₁ = - m₂u₂

u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

3 0
3 years ago
Which only list metalloids
HACTEHA [7]

Answer: I think the answer <em><u>MIGHT</u></em> be: Boron, germanium, and tellurium....

Explanation:I know this cuz I had this question on my unit test and I had to look it up online....Hope this helps

I really hope you guys find this helpful and thx so much for rating :D

       pls say something in chat if you found this helpful

                                              UwU           ;-;

                                                         

7 0
3 years ago
Mario golpea el balón con el pie para lanzárselo a Tamara que está situada a 18 m de distancia. El ángulo de salida del balón es
Brilliant_brown [7]

Answer:

y = 0.99 m

Explanation:

This is a projectile launching exercise, let's start by finding the components of the initial velocity, using trigonometry

         cos θ = v₀ₓ / v₀

         sin θ = v_{oy} / v₀

         v₀ₓ = vo cos θ

         v_{oy} = I go sin θ

         v₀ₓ = 15 cos 30 = 12.99 m / s

         v_{oy} = 15 sin 30 = 7.5 m / s

Let's find the time it takes to travel x = 18 m

         x = v₀ₓ t

         t = x / v₀ₓ

         t = 18 / 12.99

         t = 1,385 s

at this point it is at a height of

         y = v_{oy} - ½ g t²

         y = 7.5 1.385 - ½ 9.8 1.385²

         y = 0.99 m

therefore the camera must place the foot 99 cm from the ground

7 0
3 years ago
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