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lora16 [44]
3 years ago
11

Different _______ of light through two separate mediums causes the bending of wave fronts associated with light rays.

Physics
2 answers:
AveGali [126]3 years ago
7 0
<h3><u>Answer;</u></h3>

Reflection

Different <em><u>reflections</u></em> of light through two separate mediums causes the bending of wave fronts associated with light rays.

<h3><u>Explanation;</u></h3>
  • <em><u>Reflection is the bouncing back of light upon hitting a boundary or a surface. It involves abrupt change in direction of propagation of a wave that strikes the boundary between two different media. </u></em>
  • When a wave front undergoes reflection at an interface between two different media it changes direction such that it returns into the medium it originated from.
sweet-ann [11.9K]3 years ago
6 0
<span>Different _______ of light through two separate mediums causes the bending of wave fronts associated with light rays.

velocity

</span>
You might be interested in
When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde
Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

F=\dfrac{kq_1q_2}{r^2}

r is the separation between charges  

F\propto \dfrac{1}{r^2}

r=\sqrt{\dfrac{1}{F}}

If F'= 2F

r'=\dfrac{1}{\sqrt{2F} }

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r'=\dfrac{1}{\sqrt{2F} }. Hence, this is the required solution.                                                                                    

6 0
2 years ago
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released qu
Kruka [31]

Answer: a) 1766 sec. b) 55.5 MJ c) 13.9 MW d) -12,944 Nm

Explanation:

a) The torque and  the angular acceleration, are related by the following expression, which resembles very much to the Newton's 2nd Law for point masses:

ζ = I . γ, where ζ=external torque, I = rotational inertia and γ = angular acceleration.

We also know that a flywheel is a solid disk, so the rotational inertia for this type of body is equal to MR² / 2.

By definition, angular acceleration is the rate of change of angular velocity with time, so we can write the following:

γ = ωf -ω₀ / t

Assuming that the flywhel starts from rest, we know that ω₀ = 0, and ωf = 12,000 rpm.

As all the units are given in SI units, it is advisable to convert the rpm to rad/sec, as follows:

12,000 rpm = 12,000 rev. (2π/rev) . (1min/60 sec) = 400 π rad/sec

Returning to the original equation, we have:

ζ = MR² / 2 . (ωf/ t)

Replacing by the values, and solving for t, we have:

t = 250 Kg. (0.75)² m² . 400 π / 2. 50 Nm = 1,766 sec.

b) Due to the flywheel is just rotating, all the stored energy is rotational kinetic energy, which can be written as follows:

K = 1/2 I ωf² = 1/2 (MR²/2) ωf² = 1/4. 250 Kg. (0.75)² m². (400π)²

K= 55.5 MJ

c) Power is defined as energy delivered in a given time.

The energy delivered, is just the half of the originally stored value, i.e. , 55.5 MJ /2, equal to 27.75 MJ.

Dividing this value by 2.0 sec, we have the average power delivered to the machine, that we found to be equal to 27.75 MJ / 2s =  13. 9 MW

d) Using the same relationship than in a), we can write the following:

ζ = I. γ

I remains the same (as the flywheel is the same), so the only unknown is the angular acceleration.

Angular acceleration, by definition, is as follows:

γ = ωf - ω₀ / t

We know the value of ω₀, as it is the top speed value that we have already got,i.e., 400 π rad/sec.

We don't know the value for ωf, but we know the value of the rotational kinetic energy after 2.0 secs, which is equal to the half of the one we obtained in step b).

So, we can write the following:

Kf = 1/2 I ωf² = 1/2 (1/2 I ω₀²) ⇒ 1/ 2 ωf² = 1/4 ω₀² ⇒ωf = ω₀/√2

Replacing in the expression for angular acceleration:

γ = (ω₀/√2 - ω₀) / t = -0.29. 400. π/ 2 rad/sec²= -184.1 rad/sec²

Finally, we can get the torque as follows:

ζ = (250 kg. (0.75)² m² /2) . 184.1 rad/sec² = -12,944 Nm

6 0
3 years ago
Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two
slava [35]

Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?

Answer:

The force = 38.4 N

Explanation:

From coulombs law,

F = kq₁q₂/r² ............................ Equation 1

Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.

When, F = 6.4 N, r = d m.

6.4 = kq₁q₂/d²......................... Equation 1

When q₁' = 2q₁, q₂' = 3q₂, r = d cm

F = k(2q₁)(3q₂)/d²

F = 6kq₁q₂/d².......................... Equation 2

Dividing Equation 1 by equation 2

6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)

6.4/F = 1/6

F = 6.4×6

F = 38.4 N.

Thus the force = 38.4 N

6 0
3 years ago
Using our understanding of the law of gravity. What happens to the gravity as we triple the distance between two objects?
MA_775_DIABLO [31]

Answer:

1/9

Explanation:

<em>Newton’s Law of Universal Gravitation </em>

Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.

F = GMm/r²

where  

F - the gravitatioal force in Newtons,  

M   and m  -two masses in kilograms  

r  - the separation in meters.  

G  - the gravitational constant (6.674*10 ⁻¹¹ N (m/kg) ² )

Because of the magnitude of  G , gravitational force is very small unless large masses are involved.

So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance

F1 ∝ 1/r² ---------------(1)

F2 ∝ 1(3r)²

F2 ∝ 1/9r²--------------(2)

(2)/(1)

\frac{F_2}{F_1} =\frac{1}{9}\\ F_2 =\frac{F_1}{9}

From their you get as the distance tripled, Force reduce by a factor of 9(3³)

for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)

3 0
3 years ago
La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0
2 years ago
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