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lana66690 [7]
3 years ago
12

the refractive index is the blank of the speed of light in a medium to the speed of light in a vacuum

Physics
2 answers:
andrew11 [14]3 years ago
8 0

Answer:ratio

Explanation:

creativ13 [48]3 years ago
3 0

Answer:

ratio

Explanation:

Apex Verified

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"An object at rest starts accelerating. If it is going 120 m/s after traveling 202 meters, how quickly did it speed up?"
Marizza181 [45]

Answer:

this is just a guess bc i only looked at it for 5 seconds but i think 150 m/s

4 0
2 years ago
The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this met
Svetach [21]

Answer:

The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

Explanation:

The relationship between energy and wavelength is expressed below:

E = hc/λ

λ = hc/EK - EL

Considering the condition of Bragg's law:

2dsinθ = mλ

For the first order Bragg's law of reflection:

2dsinθ = (1)λ

2dsinθ = hc/EK - EL

d = hc/2sinθ(EK - EL)

Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.

Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

5 0
2 years ago
Sarah launches her purse straight up in air with a velocity of 30.2m/s<br> How high will it go?
yaroslaw [1]

Answer: 46.53

Explanation:

6 0
3 years ago
9. According to statistical theory, 68% of your measurements of time should fall within the range of and of therefore about 3 of
Nady [450]

Answer:

Yes

Explanation:

Yes, this is a random error generating because of statistical constraint. We only have finite number of data points. As per this, if we plot our observation we will get a gaussian (inverse bell ) shaped curve with mean equal to central value.

5 0
2 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
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