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vladimir2022 [97]
4 years ago
14

Match the description to the proper category to differentiate reflection and refraction

Physics
2 answers:
QveST [7]4 years ago
4 0

Answer:

The picture below for people on Edgenuity

Explanation:

Vladimir79 [104]4 years ago
3 0

Answer:

Reflection - 1) speed of ray does not change ,

2) ray bounces off the boundary , 3) Direction found by law of reflection

Refraction  - 4) direction found by Snell's law , 5) speed of ray changes .

Explanation:

In reflection , the reflected ray will bounce off the reflecting surface , in a direction opposite to the direction of the incident wave, in the same plane as the incident wave.

But in refraction, bending of the light ray occurs when there is a change in the medium in which the light ray travels.  The laws of refraction as given by Snell's law are followed for refraction. Speed of the light ray changes since there is an interaction of the ray with the particles of the medium separating the two media.

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A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from
ankoles [38]

Answer

current density is given by

I = \int J.da

where J = σ E

I = \int J.da\\ =\int \sigma E .da\\ = \sigma E \int da \\ =\sigma E \int_0^{2\pi} rL

now,

E=\dfrac{I}{2\pi r L \sigma}

a) Electric field strength at the inner surface of an iron cylinder

a = 0.5 cm = 0.005 m             b = 2.3 = 0.023

L = 10 cm = 0.1 m                     I = 27 A

E=\dfrac{I}{2\pi a L \sigma}

E=\dfrac{27}{2\pi\times 0.005\times 0.1\times 10^7}

      E = 8.59 x 10⁻⁴ V/m

b) Electric field strength at the outer surface of an iron cylinder

a = 0.5 cm = 0.005 m             b = 2.3 = 0.023

L = 10 cm = 0.1 m                     I = 27 A

E=\dfrac{I}{2\pi b L \sigma}

E=\dfrac{27}{2\pi\times 0.023\times 0.1\times 10^7}

      E =1.87 x 10⁻⁴ V/m

7 0
3 years ago
A roller coaster speeds up with constant acceleration for 2.3 s until it reaches a velocity of 35 m/s. During this time, the rol
Goshia [24]

Answer:

0.65 m/s

Explanation:

Applying the equation,

v = u + at

35 = u + a×2.3    -(1)

Again, applying the equation,

s = ut + \frac{1}{2}at^{2}

41 = u×2.3 +  \frac{1}{2} × (2.3)^{2}

35.65 = 2u + 2.3a -(2)

comparing first and second we get u= 0.65 m/s

3 0
3 years ago
A 1.5kw vacuum cleaner is used for half an hour. Calculate its energy usage.​
juin [17]

Answer:

its energy usage is 0.05

Explanation:

1.5kw/30=0.05

8 0
3 years ago
A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, cau
topjm [15]

Answer:

0.58 J

Explanation:

We know that Total energy is conserved.

Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy

Initial kinetic energy = 0 ( magnet is at rest initially)

Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J

Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J

Final potential energy = 0

∴ Dissipated heat energy = (0.69 -0.11) J = 0.58 J

4 0
3 years ago
A hummingbird flies 2.0 m along a straight path at a height of 5.3 m above the ground. Upon spotting a flower below, the humming
stepan [7]

To solve this problem we will apply the concepts of Pythagoras to find the net path. An easy way is to graph the path to trace the path, which resembles that of a right triangle. The information provided is equivalent to that of the two short sides of the triangle, so we will find the longest side by the Pythagorean theorem. In this way,

OC = \sqrt{2^2+2.4^2}

OC = 3.12m

Therefore the magnitude of the hummingbird’s total displacement is 3.12m

6 0
4 years ago
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