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Lina20 [59]
3 years ago
13

What phenomenon will take place as a light wave bends when it passes from one medium into another?

Physics
2 answers:
Licemer1 [7]3 years ago
3 0
Refraction
the fact or phenomenon of light, radio waves, etc. being deflected in passing obliquely through the interface between one medium and another or through a medium of varying density.
Ad libitum [116K]3 years ago
3 0

Answer:refraction

Explanation:

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Jessica is running a 10K. She alternates between running and walking each kilometer. She runs at a rate of 1 kilometer every 5 m
saveliy_v [14]

Answer:

c.100 minutes

Explanation:

Total distance = 10 km

Runs for 1 km every 5 minutes

walks 1 km every 15 min

She alternates between walking and running  so,  Jessica will walk 5 km and run 5 Km

Time taken by Jessica for walking : 5 km

Time taken to walk 1 km=5 minutes

Time taken to walk 5 km

=> 5 X 5

=>25 minutes

Time taken by Jessica for Running : 5km

Time taken to run 1 km = 15 minutes

 

=> 5 X 15

=>75 minutes

Total time taken   = Time taken by Jessica for walking + Time taken by Jessica for Running

=>25 minutes +75 minutes

=> 100 minutes

8 0
3 years ago
Read 2 more answers
Eat always flows from a place to a place.
Shkiper50 [21]

Explanation: Yes, If you digest something, it will flow from inside your neck into a tube that goes into stomach.

Hope this helps! :D

-TanqR

8 0
3 years ago
Read 2 more answers
Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it
Oksanka [162]
Coffee mug is cylindrical shape. Therefore, 
Volume of mug = volume of cylinder = πr²h 

r = inner radius of mug. h = height of coffee = 6 cm. 
Density of coffee = Density of water = 1 g/ml 

Therefore, volume of coffee = mass/density = 400/1 = 400 ml = 400 cm³
Volume of coffee = πr²h
400= πr²(6) 
r² = 21.23
r = inner radius of mug = 4.607 cm
6 0
3 years ago
Pay it forward! Answer a questio1) For a positive point charge, the lines radiate ………. . While, for a negative point charge, the
nikitadnepr [17]

Explanation:

1) For a positive point charge, the lines radiate outwards

for a negative point charge, the lines converge inwards

2) F = 2.3 X 10^-26 N

k = 9 X 10^9 N.m²/C

q1 = q2 = e = 1.6 X 10^-19 C

r = ?

F = kq1q2/r²

r² = kq1q2/F

r = √[kq1q2/F ]

r = √0.0100

r = 0.10m

The two protons are 0.10 m apart

3) The unit if electric field intensity is Newton-per-coulomb N/C

7 0
3 years ago
A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and  u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

8 0
3 years ago
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