Ask question

Ask question

All categories

3 years ago

10

Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted

ray?
A.
The refracted ray is unpolarized whereas the reflected ray is horizontally polarized.
B.
The refracted ray is unpolarized whereas the reflected ray is vertically polarized.
C.
The refracted ray is horizontally polarized whereas the reflected ray is unpolarized.
D.
The refracted ray is vertically polarized whereas the reflected ray is unpolarized.
E.
The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

1 answer:

kkurt [141]3 years ago

7 0

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

#PLATOLIVESMATTER

You might be interested in

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

$E=U=\frac{1}{2}kx^2$

where

k = 7.91 N/m is the spring constant

$x=5.08 cm = 0.0508 m$

Substituting,

$E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J$

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

$E+W_f=K$

where

$K_f=\frac{1}{2}mv^2$ is the kinetic energy of the ball, with

$m=5.38 g = 5.38\cdot 10^{-3} kg$ being the mass of the ball

v being the final speed

$W_f = -F_f d$ is the work done by friction (which is negative since the force of friction is opposite to the motion), where

$F_f = 0.0323 N$ is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

$W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J$

So, the kinetic energy of the ball as it leaves the cannon is

$K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s$

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

$U=\frac{1}{2}kx^2$

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

$W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J$

So, the kinetic energy of the ball at the point of maximum speed is

$K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s$

7 0

3 years ago

There are many well-documented cases of people surviving falls from heights greater than 20.0 m. In one such case, a 55.0 kg wom

bixtya [17]

1a) -192.7g

1b) 0.0126 s

2) 1309 kg m/s

3) $1.04\cdot 10^5 N$

Explanation:

1a)

First of all, we have to find the velocity of the womena just before hitting the ground.

Since the total mechanical energy is conserved during the fall, the initial gravitational potential energy of the woman when she is at the top is entirely converted into kinetic energy.

So we can write:

$mgh=\frac{1}{2}mv^2$

where

m = 55.0 kg is the mass of the woman

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 29.0 m is the initial height of the woman

v is her final speed

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(29.0)}=23.8 m/s$

Then, when the woman hits the soil, she is decelerated until a final velocity

$v'=0$

So we can find the deceleration using the suvat equation:

$v'^2-v^2=2as$

where

s = 15.0 cm = 0.15 m is the displacement during the deceleration

Solving for a,

$a=\frac{v'^2-v^2}{2s}=\frac{0-23.8^2}{2(0.15)}=-1888.3 m/s^2$

In terms of g,

$a=\frac{-1888.3}{9.8}=-192.7g$

1b)

Here we want to find the time it takes for the woman to stop.

Since her motion is a uniformly accelerated motion, we can do it by using the following suvat equation:

$v'=v+at$

where here we have:

v' = 0 is the final velocity of the woman

v = 23.8 m/s is her initial velocity before the impact

$a=-1888.3 m/s^2$ is the acceleration of the woman

t is the time of the impact

Solving for t, we find:

$t=\frac{v'-v}{a}=\frac{0-23.8}{-1888.3}=0.0126 s$

So, the woman took 0.0126 s to stop.

2)

The impulse exerted on an object is equal to the change in momentum experienced by the object.

Therefore, it is given by:

$I=\Delta p =m(v'-v)$

where

$\Delta p$ is the change in momentum

m is the mass of the object

v is the initial velocity

v' is the final velocity

Here we have:

m = 55.0 kg is the mass of the woman

v = 23.8 m/s is her initial velocity before the impact

v' = 0 is her final velocity

So, the impulse is:

$I=(55.0)(0-23.8)=-1309 kg m/s$

where the negative sign indicates the direction opposite to the motion; so the magnitude is 1309 kg m/s.

3)

The impulse exerted on an object is related to the force applied on the object by the equation

$I=F\Delta t$

where

I is the impulse

F is the average force on the object

$\Delta t$ is the time of the collision

Here we have:

$I=1309 kg m/s$ is the magnitude of the impulse

$\Delta t = 0.0126 s$ is the duration of the collision

Solving for F, we find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{1309}{0.0126}=1.04\cdot 10^5 N$

7 0

4 years ago

What's the difference mass and weight

yanalaym [24]

Answer:

Mass doesn't change.

Weight is measured based on gravitational pull.

Explanation:

4 0

3 years ago

A circuit has an AC voltage source in series with a 50 ohm resistor and a 113 mH inductor. The frequency is 100 cycles/sec, and

user100 [1]

Answer:

Explanation:

The rms voltage = 140/√2 = 140/1.414 = 99 V.

Reactance of inductor = wL = 2 X 3.14 X 100 X 113 X 10⁻³ =70.96 ohm.

Total resistance in terms of vector = 50+70.96j

j is imaginary unit number

Magnitude of this resistance = √ 50² + 70.96² = 86.80 ohm

current in resistance (rms) ( I ) = 99/86.80 = 1.14 A.

Power dissipated in resistor = I² R = 1.14 X 1.14 X 50 = 65 W( approx)

7 0

4 years ago

The time that is required for a vibrating object to complete one full cycle is called : Group of answer choices

disa [49]

Answer:

Correct option : (c) period.

Explanation:

The time that is required for a vibrating object to complete one full cycle is called the time period. If f is the frequency of a wave, then the relation between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

These are the characteristics of a wave. Some other characteristics are wavelength, amplitude, intensity etc. So, the correct option is (c) "period".

4 0

4 years ago