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IgorC [24]
3 years ago
7

A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat, parallel plates of glass (n = 1.50). What mu

st be the thickness of the liquid layer if normally incident light with λ = 675 nm in air is to be strongly reflected?
Physics
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

t = 96.1 nm

Explanation:

For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength

now we know that the path difference of two reflected light from thin liquid layer is given as

2\mu t - \frac{\lambda}{2} = N\lambda

here we know that

\mu = 1.756

t = thickness of layer

N = 0 (for minimum thickness of layer)

\lambda = 675 nm

now we have

2(1.756) t = \frac{675 nm}{2}

t = 96.1 nm

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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
gizmo_the_mogwai [7]

Answer:

Explanation:

First of all we shall find the velocity at equilibrium point of mass 1.2 kg .

It will be ω A , where ω is angular frequency and A is amplitude .

ω = √ ( k / m )

= √ (170 / 1.2 )

= 11.90 rad /s

amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

= .5355 m /s

At middle point , no force acts so we can apply law of conservation of momentum

m₁ v₁ = ( m₁ + m₂ ) v

1.2 x .5355 = ( 1.2 + .48 ) x v

v = .3825 m /s

= 38.25 cm /s

Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

A₁ = .0379 m

New amplitude is .0379 m

7 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
2 years ago
Use the law of universal gravitation to describe the relationship between you and your desk
tangare [24]
The law of gravitation states that things closer to the core of gravitation, have a larger force pulling down on them. In relation of you to your desk, you and the desk are both drawn downwards towards the center of gravity.
6 0
3 years ago
Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca
Morgarella [4.7K]

Answer:

La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

6 0
3 years ago
A jet plane lands with a velocity of +107 m/s and can accelerate at a maximum rate of -5.18 m/s2 as it comes to rest. From the i
Marrrta [24]

Answer:

20.7 s

Explanation:

The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:

v = a*t + v₀

Solve for t:

t = (v - v₀)/a

3 0
3 years ago
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