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Pachacha [2.7K]
3 years ago
13

Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

d reflected beam. (Brewster angle).
Physics
1 answer:
nignag [31]3 years ago
5 0

Answer:

The refractive index of glass, \mu_{g} = 1.367

Solution:

Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.

To determine the refractive index of glass:

tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}                 (1)

where

\mu_{a} = refractive index of glass

\mu_{g} = refractive index of glass

Now, using eqn (1)

tan{53.8^{\circ}} = \frac{\mu_{g}}{1}

\mu_{g} = tan53.8^{\circ}

\mu_{g} = 1.367

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Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el
kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

E =\frac{Kq}{r^2}

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9} = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

8 0
3 years ago
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0
3 years ago
Gsjskebjwkksmndkkwksjdkdkskkskskkehdhjdj
DanielleElmas [232]

Answer:

I DON'T UNDERSTAND

Explanation:

GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.

3 0
3 years ago
Is the friction of the pendulum (catch mechanism, support axis, etc.) a random or systematic error? Will this source of error ca
Hatshy [7]

Answer:

l these errors believe that the speed of the system is less than that calculated

Explanation:

When we carry out any measurement in addition to the magnitude, the sources of uncertainty must also be analyzed.

We can have random uncertainties, correspondin

g to momentary errors, for example early warps during medicine, parallax errors, errors in the starting and ending points of the movement; I mean every possible random error. This error is the one that is analyzed and calculated in the statistical equations

There is another source of error, the systematic ones, these are much more complicated, they can be an error in the pendulum length, friction in the pendulum movement mechanism, deformities in the support systems, this errors are not analyzed by the statistic, in general They discover by looking at the results and comparing with the tabulated or real ones.

 

tith the explanation we see that the errors described are systematic.

In general these errors believe that the speed of the system is less than that calculated

5 0
3 years ago
2 masses of 4kg and 2.5kg have a separation of r =1.5m. Calculate the force of gravitational attraction between them (take g as
Evgesh-ka [11]

Answer:

F = 29.64 × 10-¹¹N

Explanation:

From newton's law of gravitation which states that every object in the universe will attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

That is,   F = (Gm1m2)/ r²

From the question m1=4kg, m2 = 2.5kg, r= 1.5m,G = 6.67×10-¹¹

F=( 6.67×10-¹¹ × 4 × 2.5) / 1.5²

F = (66.7×10-¹¹) / 2.25

F = 66.7/2.25 × 10-¹¹ N

F = 29.64 × 10-¹¹N

F = 29.64 × 10-¹¹N

I hope this was helpful, please mark as brainliest

5 0
3 years ago
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