Light is incident along the normal on face AB of a glass prismwith refractive index 1.52.
A. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in air.
B. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in water.
air refactive index = 1.00029
water refractive index = 1.33

Answers

Answer 1

Answer:

48.9°, 29°

Explanation:

We kno that snell's law States that

( sin i / sin r ) = ( n 2 / n 1 )

And here

i = 90 - α

r = 90

n 2 = n a = 1.00029

n 1 = n = 1.52

So

( sin ( 90 - α ) / sin 90 ) = (1.00029 / 1.52 )

= 0.658

sin ( 90 - α ) = 0.658

90 - α = sin -1 ( 0.658 )

= 41.15

α = 90 - 41.15

= 48.9°

( b ) to find angle when prism immerssed in water

Using snell's law again

( sin i / sin r ) = ( n 2 / n 1 )

here i = 90 - α

r = 90

n 2 = n w = 1.33

n 1 = n = 1.52

So

( sin ( 90 - α ) / sin 90 ) = (1.33 / 1.52 )

= 0.875

sin ( 90 - α ) = 0.875

90 - α = sin -1 ( 0.875 )

= 61

α = 90 - 61

= 29°


Related Questions

A machine carries a 100kg cargo to a boat at a rate of 10m/s2. The distance between the ground to the boat is 50ft. If the machine must transfer the cargo to the boat in 5 minutes, how much power should the machine exert?

Answers

Answer:

50.8 watt

Explanation:

we know that P=W÷t

W=F.S           S-->distance=50 ft= 15.24 m

F=ma

=100×10=1000 N

SO W= 1000×15.24

        =15240 J

NOW

P=W÷t          t=5 mints = 5×60=300 sec

P=15240÷300

P=50.8 watt

An ionized oxygen molecule (O2+) at point A has charge +e and moves at 1.24 ✕ 103 m/s in the positive x-direction. A constant electric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.766 mm past A on the x-axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of an oxygen molecule is 5.31 ✕ 10−26 kg and the fundamental charge is e = 1.60 ✕ 10−19 C.)

Answers

Answer:

[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]

[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]

Explanation:

Given that:

The charge on the ionized oxygen molecule = +e

The speed of the ionized oxygen molecule with this charge is 1.24 × 10³ m/s

distance travelled by the particle before rest is d = 0.766 m

According to the third equation of motion.

[tex]v^2 = u^2 +2as[/tex]

[tex]v^2 = u^2 +2(\dfrac{-eE}{m}) s[/tex]

[tex]0^2= u^2 +2(\dfrac{-eE}{m}) s[/tex]

[tex]E = \dfrac{mu^2}{2e* \ s}[/tex]

[tex]E = \dfrac{5.31 *10^{-26}* (1.24*10^3)^2}{2*1.6*10^{-19}*0.766*10^{-3}}[/tex]

[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]

Thus, the electric field shows to be in the negative x-direction.

The potential difference between point A and B now is:

[tex]\Delta V = E.d \\ \\ V_A - V_B = 3.33 \times 10^2 \times 0.766 \times 10^{-3}[/tex]

[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]

The volume of water and an egg in a graduated beaker is 200mL. After the egg is removed the volume of the water is found to be 125mL . What is the volume of the egg in cm ^3

Answers

Answer:

75 cm³

Explanation:

The volume of the egg is equal to the volume of the egg and water minus the volume of the water.

V = 200 mL − 125 mL

V = 75 mL

V = 75 cm³

Answer:

75 cm

Explanation:

The equation would be 200ml -125 ml thus finding the volume of the egg.

A woman on a snowmobile moving with a constant velocity east down the road fires a flare straight upward and the snowmobile continues to move with a constant
velocity as the flare is in the air. Assume no air resistance. Where will the flare land?

A) it will hit the person on the snowmobile who fired the flare
B) Somewhere behind the snowmobile, depends on velocity
C) in front of the snowmobile
D) Impossible to know
E) behind the snowmobile in exactly the same location (relative to the ground) from which it was fired


Answers

D) inmpossible to know

In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min , 43.6 s43.6 s . Suppose "Pre" was at the 7.85 km7.85 km mark at a time of 25.0 min25.0 min . If he accelerated for 60 s60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s60 s interval. Assume his instantaneous speed at the 7.85 km7.85 km mark was the same as his overall average speed up to that time.

Answers

Answer:

a = 0.161 [tex]$m/s^2$[/tex]

Explanation:

Given :

[tex]$ d_{total}[/tex] = 10 km = 10000 m

[tex]$t_{total} $[/tex] = 27 min 43.6 s

        = 1663.6 s

[tex]$d_1$[/tex] = 7.85 km = 7850 m

[tex]$t_1$[/tex] = 25 min = 1500 s

[tex]$t_2$[/tex] = 60 s

Now the initial speed for the distance of 7.85 km is

[tex]$ v_1 = \frac{d_1}{t_1} = \frac{7850}{1500}$[/tex]  = 5.23 m/s

The velocity after 60 s after the distance of 7.85 kn is

[tex]$v_2 = v_1 + at_2$[/tex]

    = 5.23 + a(60)

The distance traveled for 60 s after the distance of 7.85 km is

[tex]$d_2 = v_1t_2+\frac{1}{2}at_2^2$[/tex]

[tex]$d_2 = (5.23)(60)+\frac{1}{2}a(60)^2$[/tex]

    = 313.8 + a(1800)

The time taken for the last journey where the speed is again uniform is

[tex]$t_3 = t_{total}-t_1-t_2 $[/tex]

   = 1663.6 - 1500 - 60

   = 103.6 s

Therefore, the distance traveled for the time [tex]$t_3$[/tex] is

[tex]$ d_3 = v_2 t_3$[/tex]

    = (5.23+60a)(103.6)

    = 541.8 + 6216 a

The total distance traveled,

[tex]$ d_{total}= d_1 + d_2 + d_3$[/tex]

Now substituting the values in the above equation for the acceleration a is

10000 = 7850 + (313.6 + 1800a) + (541.8 + 6216a)

10000 = 8706.5 + 8016a

1294.4 = 8016a

a = 0.161 [tex]$m/s^2$[/tex]

An open freight car rolls friction-free along a horizontal track in a pouring rain that falls vertically. As water accumalates in the car, its speed

Answers

Answer:

decrease

Explanation:

weight

As water accumulates in the car, its speed decrease according to Newton's second law of motion

What is newton's  second law of motion?

Newton's second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

according to newton's second law

force = mass * acceleration

as water accumulate in the car , mass will increase

since mass and acceleration are inversely proportional to each other

hence , acceleration will decrease and speed will decrease

learn more about  newton's second law:

https://brainly.com/question/13447525?referrer=searchResults

#SPJ3

A tennis ball is hit into the air and moves along an arc. (a) Neglecting air resistance, where along the arc is the speed of the ball a minimum? (Select all that apply.) at the initial position of motion at the highest point in the trajectory at the final position of motion

Answers

Answer:

at the highest point in the trajectory

Explanation:

When the tennis ball is hit, it moves in the air along a curve or an arc. This path is the parabola curve. Such a motion in the two dimension is known as projectile motion. It is constant accelerated motion in the downward direction.

The velocity of the ball is minimum at the highest point of the motion. When we hit the ball, the ball moves up to certain eight and then it gradually fall back to the earth surface along a curve.

The horizontal velocity of the ball is always the same along the curve. Only the vertical velocity varies. As the ball reaches the top of the curve or the maximum height, its vertical velocity becomes zero.

So, speed of the tennis ball is minimum at the highest point of the path.

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.A. Which laser has its first maximum closer to the central maximum?B. What is the distance delta ymax-min between the first maxima (on the same side of the central maximum) of the two patterns?C. What is the distance Delta ymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

Now

Using

ym = m x λ x L/d

therefore for each of first Max we have m =1

And laser 1 had

y = (d/20) x (6.0m)/d

y = 6.0m /20 = 0.3

Then laser 2 will be

y = 6.0m/15 = 0.4

y ( laser1 ) < y ( laser2 )

So the first maxima of laser 1 will be closer to the Central maxima

So

(0.4m -0 .3 m) = 0.1m

( C)

Now for laser 1 we say

y= 26.0 m / 20 = 0.6 m

Laser 2

We have

ym=(m+1/2) x λ x L/d

So

Because there is no central minimum the first minimum is at m = 0

We can way 3rd minimum is at m = 2

So

y = (2.5) x 6.0 / 15 = 1m

So

Δy= 1m - 0.6m = 0.4 m

What is the force applied by the ground called? A. applied force B. friction C. gravity

Answers

Answer:

gravity i think hope this helps

Explanation:

You are walking at 3.75 km per hour across a frozen lake in the snow. You do not realize that with each step you turn 0.350 degrees to your right. If your step length is 74.0 cm what is the diameter, in meters, of the circle that you are inadvertently tracing out?

Answers

Answer:

242.27929622673 meters

The morning after a massive snowstorm, Michaela gets into her car to drive to work. The storm caused her windows to freeze, so she first needs to defrost the car. While the engine is running, she checks the thermometer. It shows the air inside of her car has a temperature of 0 °C. Does this mean the air inside of her car has no kinetic energy? Explain your answer.

Answers

Answer:

Hope it helps

A Brainliest please

PLEASEEEEEE HELP
A jet makes a landing traveling due east with a speed of 120 m/s. If the jet comes to rest in 13.5 s , what’s the magnitude of its average acceleration?
Part B: What is the direction of its average acceleration? (North, south, east, west??)

Answers

Answer:

8.89 m/s² west

Explanation:

Assume east is +x.  Given:

v₀ = 120 m/s

v = 0 m/s

t = 13.5 s

Find: a

v = at + v₀

0 m/s = a (13.5 s) + 120 m/s

a = -8.89 m/s²

a = 8.89 m/s² west

A ball is thrown straight upward with a speed of 36 m/s. How long does it take to return to its starting point, assuming negligible air resistance?

Answers

Answer:

The time taken for the ball to return to the starting point is is 7.4 s

Explanation:

Given;

initial velocity of the ball, u = 36 m/s

the final vellocity at maximum height, v = 0

let time taken for the ball to reach maxmimum height = t

Time taken for the ball to return to the starting point is known as time of flight, calculated as;

[tex]t = \frac{v-u}{-g} \\\\T = 2t\\\\T = \frac{2(0-u)}{-g}\\\\T = \frac{-2u}{-g}\\\\ T = \frac{2u}{g}[/tex]

T = (2 x 36) / 9.8

T = 7.4 s

Therefore, the time taken for the ball to return to the starting point is is 7.4 s

What are the benefits of living in a country with a growing population rate?

Answers

Answer:

The benefits are the place you live in will have more oppurtunities for new jobs as well as growth in residential areas.

Explanation:

when the temperature of matter decrease , the particles do what

Answers

Answer:

When the temperature decreases the particals start to slow down.

A falling object satisfies the initial value problem dv dt = 9.8 − v 5 , v(0) = 0 where v is the velocity in meters per second. (a) Find the time that must elapse for the object to reach 95% of its limiting velocity. (Round your answer to two decimal places.) s (b) How far does the object fall in the time found in part (a)? (Round your answer to two decimal places.) m Additional Materials

Answers

Answer:

a.  t [tex]\simeq[/tex] 14.98 sec

b.   x = 501.27 m

Explanation:

From the given information:

[tex]\dfrac{dv}{dt}=9.8-(\dfrac{v}{5 })[/tex]  and   [tex]v(0)=0[/tex]

[tex]\dfrac{dv}{dt}=\dfrac{49-v}{5 }[/tex]

[tex]\dfrac{dv}{49-v}=\dfrac{dt}{5 }[/tex]

Taking  Integral of  both sides

[tex]- ln(49-v) = \dfrac{t}{5} + C[/tex]  

at t=0 we have v=0

This implies that

[tex]- ln(49-0) = \dfrac{0}{5} + C[/tex]

[tex]C= - ln(49)[/tex]

Thus:

[tex]\dfrac{t}{5} - In (49) = - In (49 -v) \\ \\ In(49) - \dfrac{t}{5} = In (49-v)[/tex]

[tex]49-v = e^{(-\frac{t}{5} +ln(49))}\\ \\ v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]

The limiting velocity when the time is infinite is :

95% of 49 = 46.55

[tex]0.05= e^{(-\dfrac{t}{5})}[/tex]

[tex]\dfrac{t}{5}= In(\dfrac{1}{0.05})[/tex]

[tex]\dfrac{t}{5}=2.9957[/tex]

t = 5 × 2.9957

t [tex]\simeq[/tex] 14.98 sec

b.) [tex]v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]

[tex]v = \dfrac{dx}{dt}=49 - 49e^{(-\dfrac{t}{5})}[/tex]

[tex]dx=(49 - 49e^{(-\frac{t}{5})}) \ dt[/tex]

Taking integral of both sides.

[tex]x = 49t + 245 e^{(\frac{-t}{5})} +C[/tex]

 at time t = 0 , distance x traveled = 0

C= - 245

Therefore

[tex]x = 49t + 245 e^{(\frac{-t}{5})} -245[/tex]

replacing the value of t = 14.98

[tex]x = 49(14.98) + 245 e^{(\frac{-14.98}{5})} -245[/tex]

x = 501.27 m

which of the following has the greatest inertia ping pong ball, golf ball, softball, and a bowling ball

Answers

Answer:

bowling ball

Explanation:

A bowling ball has more mass than the others, thus having more inertia.

Starting from rest, a car accelerates at a rate of 7.8m/s^2 for 4.9 seconds. What is it’s velocity at the end of this time?

Answers

Answer:

7.8 m/'s = Change of accelerates / time taken =7.8/4.9=

Explanation:

please solve that answer

An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of the transverse and longitudinal waves in rock are 8.8 km/s and 5.9 km/s respectively. A seismograph records the arrival of the transverse waves 69 s before that of the longitudinal waves. Assuming the waves travel in straight lines, how far away is the center of the earthquake

Answers

Answer:

1239.216 km

Explanation:

The speed of the transverse = 8.8km/s

The speed of the longitudinal = 5.9km/s

distance = speed x time,

8.8km/s x trans_time = 5.9km/s x long_time

8.8 / 5.9 = long_time / trans_time

1.49 = long_time / trans_time

long_time = 1.49 trans_time

the transverse wave was 69s faster than longitudinal,

trans_time - long_time = 69s

trans_time - 1.49trans_time  = 69s

0.49 trans_time = 69

trans_time = 69 / 0.49 = 140.82s

long_time = 140.82 - 69 = 71.82s

the distance of the earthquake;

distance = 8.8 x 140.82 = 1239.216 km

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −0.0200 m in the same direction as the electric field. The electron's speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.04 ✕ 10−17 J. The electron continues to x = −0.210 m within the constant electric field. If there's a change in potential energy of −9.60 ✕ 10−17 J as it goes from x = 0.190 m to x = −0.210 m, find the electron's speed (in m/s) at x = −0.210 m.

Answers

Answer:

The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]

Explanation:

Given that,

Electric field [tex]E=1.50\times10^{3}\ N/C[/tex]

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy [tex]P.E=5.04\times10^{-17}\ J[/tex]

Change in potential energy [tex]\Delta P.E=-9.60\times10^{-17}\ J[/tex] as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

[tex]W=-eE\Delta x[/tex]

Put the value into the formula

[tex]W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))[/tex]

[tex]W=-5.04\times10^{-17}\ J[/tex]

We need to calculate the initial velocity

Using change in kinetic energy,

[tex]\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2[/tex]

[tex]\Delta K.E=\dfrac{-3mv^2}{8}[/tex]

Now, using work energy theorem

[tex]\Delta K.E=W[/tex]

[tex]\Delta K.E=\Delta U[/tex]

So, [tex]\Delta U=W[/tex]

Put the value in the equation

[tex]\dfrac{-3mv^2}{8}=-5.04\times10^{-17}[/tex]

[tex]v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}[/tex]

Put the value of m

[tex]v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}[/tex]

[tex]v=1.21\times10^{7}\ m/s[/tex]

We need to calculate the change in potential energy

Using given potential energy

[tex]\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})[/tex]

[tex]\Delta U=-4.56\times10^{-17}\ J[/tex]

We need to calculate the speed of electron

Using change in energy

[tex]\Delta U=-W=-\Delta K.E[/tex]

[tex]\Delta K.E=\Delta U[/tex]

[tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}[/tex]

Put the value into the formula

[tex]v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}[/tex]

[tex]v_{f}=1.5\times10^{7}\ m/s[/tex]

Hence, The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]

A student is sitting at rest in a chair. How does the force that the student exerts
on the chair compare to the force the chair exerts on the student? *
•a larger magnitude but the opposite direction
•a smaller magnitude but the same direction
•the same magnitude but the opposite direction
•the same magnitude and the same direction

Answers

Answer:

the same magnitude but the opposite direction

Explanation:

Newton's third law of motion states that there is always an equal and opposite reaction to every action. This means that the amount of force exerted upon an object is equal to the amount of force the object exerts but in an opposite direction.

This is the case in this scenario where a student sits at rest in a chair. The student is supplying the action force being exerted on the chair. According to the third law of Newton, the chair will exert the same size of force back in the student but in an opposite direction.

Hence, the force the chair exerts on the students compare with that of the student in the sense that they are the same magnitude (size) but the opposite directions.

A mass on a string of unknown length oscillates as a pendulum with a period of 1.8 s. What is the period in the following situations? (Parts (a) to (d) are independent questions, each referring to the initial situation.)(a) The mass is doubled?(b) The string length is doubled?(c) The string length is halved?(d) The amplitude is doubled?

Answers

Answer:

Using

Period ( P) is given as

P~√(L/g).

a) since mass has no effect on the period of a pendulum. So, the period will remain 1.8seconds

b) using the formula above ,period varies with the square root of the length. Thus , when the length doubles, the period is multiplied by √2. So, the period is 1.8s*√2 = 2.54s

c) in this case, the period is multiplied by √(1/2).

1.8√(1/2)=1.27s.

d) amplitude of the pendulum doesn't affect the period (unless itsvery high, so, the period is still 1.8s

Which is not standing in the way of astronomers getting a good view of distant stars? A. some stars are too far away for our telescope to see B. lights get distorted by gas and dust as it goes through the atmosphere C. Redshift makes stars difficult to see D. there is alot of light pollution on earth

Answers

Answer: D. There is a lot of light pollution on earth

Explanation: The light pollution on Earth has nothing to do with the stars in the sky

Answer: C. Redshift makes stars difficult to see.

Explanation:

I did the test

Is it possible to accelerate and not speed up or slow down?

Answers

Answer: No,

explanation: When the object is neither speeding up or slowing down, we can say that its speed is constant.

Hope this helps

Plz mark brainlesit

The slope of the x-t curve at any point represents:_______a. Displacementb. Velocityc. Accerlationd. None of these

Answers

Answer:

Velocity

Explanation:

x-t curve means position vs time graph in which position (x) is in x-axis and time is in y-axis.

Slope of a graph = [tex]\dfrac{\Delta y}{\Delta t}[/tex].

We know that, velocity = displacement/time

or we can say that slope = displacement/time = velocity

Hence, the correct option is (b) "velocity".

HELLO CAN SOMEONE HELP ME PLS

A car is moving at 35 mph and comes to a stop in 5 seconds.

Find the acceleration of the car.

Answers

Answer:

I do believe it's 7

1.
Which of the follow
following
cannot be zero?
A. Distance B. Displacement
" Velocity D. Acceleration
C.​

Answers

A.Distance , because distance cannot be zero

If a moving clock is ticking half as fast as normal what speed is the clock traveling?

Answers

Answer:

Speed of moving clock is

 [tex]V2[/tex] =πr/(86400) m/s

Here,

m and s are SI units of distance and time respectively.

Explanation:

If

radius of circular clock= r

than,

total distance covered on clock=S=2πr m

here, m=SI unit of distance

and required time for covering the total distance=t= 86400s

speed of normal clock=[tex]V1[/tex]=S/t

                                       [tex]V1[/tex]=2πr/86400 m/s

As, moving moving clock is ticking half as fast as normal clock so,

speed of moving clock=[tex]V2[/tex]=[tex]V1[/tex]/2

                                           [tex]V2[/tex]=2πr/(86400)*2 m/s

                                            [tex]V2[/tex] =πr/(86400) m/s

what would the answer be ?

Answers

The second one since you’re changing the soil up by adding different fertilisers. This will be you’re independent variable. And you’re dependent variable is your result = the plant height .
Hope this helps :)

How does gamma radiation differ from alpha or beta particle radiation?
1) it does not consist of matter
2) it only consists of space
3) it does not consist of energy
4) it only consists of matter

Answers

Answer:

1.

Explanation:gamma rays are the most powerful in the electromagnetic spectrum and they are a result of a radioactive atom.they aren't made of matter but just energy as a wave.

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