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3 years ago

A ray of yellow light ( f8= 5.09 × 1014 Hz) travels at a speed of 2.04×10 meters per second in

Physics

2 answers:

Sveta_85 [38]3 years ago

7 0

I will assume that question in which medium the light of that frequency travels at that speed.
To find this answer we will have to calculate the refractive index of that medium.
Refractive index of a material is a number that tells us how light propagates through a medium. It tells us how much ray of light would bend entering the medium from a vacuum. It is defined as:
$n= \frac{c}{v}$
c is the speed of light in the vacuum and v is the speed of light in the medium. Now we can calculate refractive index: $n= \frac{3\cdot10^{8}}{2.04\cdot{10^8}}} =1.471$
We can search for a material that has this refractive index.
I found that Pyrex, glycerine and vegetable oil have a refractive index of 1.47.

Gala2k [10]3 years ago

5 0

The complete question in the attached figure
Let
c ------------------- > is the speed of light
v ------------------- > is the speed in medium
n ------------------- > is the refractive index of medium
we know that
c/v = n
n = (3 x 10^8)/(2.04 x 10^8) = 1.47

1.47 is the refractive index of glycerol.
therefore

the answer is (4) glycerol

You might be interested in

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

8 0

3 years ago

You hang a tv on your wall. What kind of energy does it have?

lara31 [8.8K]

Answer:

potential, not moving

Explanation:

5 0

3 years ago

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

Can someone help asap plz don’t understand how to do this

shtirl [24]

Homie I don’t know either‍♀️Drop out of schl ig ;-;

3 0

3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago