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The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees

Answers

Answer 1

Answer:

Answer:

Angle of transmitted ray is $29.14^(o)$

Explanation:

According to snell's law we have

$n_(1)sin(\theta _(i))=n_(2)sin(\theta r)$

Since the incident medium is air thus we have $n_(1)=1$

By definition of refractive index we have

$n=(c)/(v)$

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

$n_(2)=(3* 10^(8))/(2.5* 10^(8))=1.2$

Thus using the calculated values in Snell's law we obtain

$sin(\theta _(r))=(sin(31.3)/(1.2)\n\n\therefore sin(\theta _(r))=0.4329\n\n\theta_(r)=sin^(-1)(0.4329)\n\n\theta _(r)=29.14$

Answer 2

Answer:

Answer:

Angle made by the transmitted ray = 25.65°

Explanation:

Speed of light in plastic = v = 2.5 × 10⁸ m/s

refractive index of plastic (n₂) / refractive index of air (n₁)

= speed of light in air c / speed of light in plastic v.

⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2

Angle of incidence = 31.3° = i

n₁ sin i = n₂ sin r

⇒ sin r = (1)(0.5195) / 1.2 = 0.4329

⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°

Related Questions

If you drew magnetic field lines for this bar magnet, which statement would be true
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A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.b. What should be its sign, so that all three charges will be in equilibrium?c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.
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A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) $h*(1 - 1/2 g * h/v_0^2)$

b) $h = v_0^2/ g$

c) $h = v_0^2/ g$

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

$y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t$

where:

y = height at time t

y0 = initial height

v0 = initial velocity

a = acceleration

t = time

v = velocity

a) When the balls collide, h1 = h2. Then,

$h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0$

Replacing in the equation of the height of the first ball:

$h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)$

b) that the balls collide at t = h/v0. Then:

$h/ v_0 = -v_0/-g\nh = v_0^2/ g$

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

$h = v_0^2/ g$

See more about velocity at brainly.com/question/862972

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be

Answers

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_(atm)$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_(1) + \rho\cdot (v_(1)^(2))/(2) + \rho\cdot g \cdot z_(1) = P_(2) + \rho\cdot (v_(2)^(2))/(2) + \rho\cdot g \cdot z_(2)$

Where:

$P_(1)$ , $P_(2)$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(1)$ , $v_(2)$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_(1)$ , $z_(2)$ - Heights of the tank and ground level, measured in meters.

Given that $P_(1) = P_(2) = P_(atm)$ , $\rho = 1000\,(kg)/(m^(3))$ , $g = 9.807\,(m)/(s^(2))$ , $v_(1) = 0\,(m)/(s)$ , $z_(1) = 6.9\,m$ and $z_(2) = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (6.9\,m) = (v_(2)^(2))/(2) + \left(9.807\,(m)/(s^(2)) \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_(2) = \sqrt{2\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot (6.9\,m-4.9\,m)}$

$v_(2) \approx 6.263\,(m)/(s)$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by:

Vx,f = sqrt (2*g*x*sin(Q))

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Answers

A. The rms value of electric field be "1.05 × 10⁶ N/C".

B. The rms value of magnetic field will be "3.5 × 10⁻³ T".

Magnetic and Electric field

According to the question,

Intensity of the wave, S = 2.93 × 10⁹ W/m²

Free space permittivity, $\epsilon_0$ = 8.86 × 10⁻¹²

Speed of light, c = 3 × 10⁸

A. We know that,

The rms value of electric field,

→ $E_(rms)$ = $\sqrt{(S)/(\epsilon_0 c) }$

By substituting the values,

= $\sqrt{(2.93* 10^9)/((8.85* 10^(-12))(3* 10^8)) }$

= 1.05 × 10⁶ N/C

and,

B. We know that,

The rms value of magnetic field,

→ $B_(rms)$ = $(E_(rms))/(c)$

By substituting the values,

= $(1.05* 10^6)/(3* 10^8)$

= 3.5 × 10⁻³ T

Thus the above response is appropriate.

Find out more information about magnetic field here:

brainly.com/question/25801845

To solve this problem, it is necessary to apply the concepts related to the electric field according to the intensity of the wave, the permittivity constant in free space and the speed of light.

As well as the expression of the rms of the magnetic field as a function of the electric field and the speed of light.

PART A) The expression for the rms of electric field is

$E_(rms) = \sqrt{(S)/(\epsilon_0 c)}$

Where,

S= Intensity of the wave

$\epsilon_0$ = Permitivitty at free space

c = Light speed

Replacing we have that,

$E_(rms) = \sqrt{((2.93*10^9))/((8.85*10^(-12))(3*10^8))}$

$E_(rms) = 1.05*10^6N/C$

The RMS value of electric field is $1.05*10^6N/C$

PART B) The expression for the RMS of magnetic field is,

$B_(rms) = (E_(rms))/(c)\nB_(rms) = (1.05*10^6)/(3*10^8)\nB_(rms) =3.5*10^(-3)T$

The RMS of the magnetic field is $3.5*10^(-3)T$

Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?

Answers

The particle reach its minimum velocity at time 1.06 sec.

The function is given as

x=5t^3-8t^2+12

Differentiating the above equation with respect to time, to obtain the velocity

dx/dt=v=15t^2-16t

For maximum and minimum values, put dx/dt=0

15t^2-16t=0

On solving the equation, t=0, 1.06

Therefore at time t=1.06 sec, the particle has the minimum value of velocity.

The particle reaches its minimum velocity at t = 0 s or t = 16/15 s

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = (v - u)/(t) } }$

$\large {\boxed {d = (v + u)/(2)~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

$x = ( 5t^3 - 8t^2 + 12) ~ m$

To find the velocity function, we will derive the position function above.

$v = (dx)/(dt)$

$v = 5(3)t^(3-1) - 8(2)t^(2-1)$

$v = ( 15t^2 - 16t ) ~ m/s$

Next to calculate the time to reach its minimum speed, then v = 0 m/s

$0 = ( 15t^2 - 16t )$

$0 = t( 15t - 16)$

$\large {\boxed {t = 0 ~s ~ or ~ t = 16/15 ~ s} }$

Learn more

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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