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bearhunter [10]

2 years ago

11

If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac

tor of 4? (enter an integer, as in it will be rotating ________ times faster). (3 pts)

1 answer:

maks197457 [2]2 years ago

5 0

In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁

Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁

Hence, the protostar will be rotating 16 </span><span>times faster.</span>

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hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0

2 years ago

Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

dimulka [17.4K]

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is $m= 10 g = 0.01 \ kg$

The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

So

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

5 0

2 years ago

An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir

Natalija [7]

A photoelectric cell is an electronic device which is used to convert light energy into electric energy.The operation of this device is based on photoelectric effect.

Light of suitable frequency i.e greater or equal to threshold frequency will fall on the cathode maintained at negative potential.The electron emission will take place and these electrons are drifted towards the anode which is at positive potential.

Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.

We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.

$Mathematically\ energy\ E=\frac{hc}{\lambda}$

Here h is the Planck's constant,c is the velocity of light.

Here we have been given red light and blue light.

In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.

Hence, the blue light will activate the most and red the least.

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A swimmer does 3,560 J of work in 55 s. What is the swimmer’s power output? Round your answer to two significant figures. The po

Natasha2012 [34]

The value of the swimmer's power output is calculated by dividing the work done by the time it took for the work to be completed. From the given in this item,
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The image on the left shows the charges on a balloon after it’s been rubbed with a wool cloth. The image on the right is a piece

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