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4 years ago

Why would you expect the speed of light to be slightly less in the atmosphere then in a vacuum?

1 answer:

azamat4 years ago

7 0

The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material

Explanation:

When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.

But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.

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HELP ASAPPP which statement explains the first law of thermodynamics a)energy is created and destroyed or b) energy is transform

Aliun [14]

B I believe... :) good luck!

7 0

4 years ago

Read 2 more answers

You are working at a company that manufactures electrical wire. Gold is the most ductile of all metals: it can be stretched into

devlian [24]

Answer: $R=1.424 M\Omega$

Explanation:

Given

mass of gold $m=1.6 gm$

Length of wire $L=2.2 km$

Resistivity of gold $\rho =2.44\times 10^{-8}$

density of gold $=19.3\times 10^3 kg/m^3$

and $mass=volume\times density$

$1.6\times 10^{-3}=volume\times 19.3\times 10^3$

$volume=8.29\times 10^{-8} m^3$

And $Resistance R=\frac{\rho L}{A}$

also be written as

$R=\frac{\rho L^2}{V}$

where $L=length$

$V=volume$

$\rho =resistivity\ of\ gold$

$R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}$

$R=1.42\times 10^{6} \Omega$

$R=1.424 M\Omega$

5 0

3 years ago

A total electric charge of 6.75 nC is distributed uniformly over the surface of a metal sphere with radius 20.0 cm. If the poten

djverab [1.8K]

Answer:

a) 60 V

b) 125 V

c) 125 V

Explanation:

Given

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.

Required

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

Solution

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4 $\pi$ ∈_o (1)

r is the distance where the potential is measured and the term 1/4 $\pi$ ∈_o equals 9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m

V= 1*q/4 $\pi$ ∈_o*r

=60 V

(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by

V = 1*q/4 $\pi$ ∈_o*R (2)

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m

V = 1*q/4 $\pi$ ∈_o*R

= 125 V

(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)

V= 125 V

7 0

3 years ago

You are looking down on a 20 kg beam resting on a horizontal, frictionless surface. The beam is 2 m long and can pivot about one

Korolek [52]

The angular speed of the rod after the impact is 1.49 rad/s

What is angular speed?

The rate of change of angular displacements is known as angular speed.

Angular speed is a scalar measure of the rotating object.

What is Angular momentum?

It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object.

Angular momentum is expressed as follows:

L=m*v*r

Here,

mass of beam, M =20 kg

mass of rock, m =0.1 kg

length of the beam, L =2 m

length where rock slides, l = (L / 2), l = 1 m

velocity of rock, v =400 m/s

As here the Torque on which the system is zero implies that the angular momentum is conserved.

Initial angular momentum for rock: I(ri) = m*v*r

Final angular momentum for rock: I(rf) = m*w*r^2

Final angular momentum for beam: I(bf) = 1/3 (M*L^2w)

Now, According to the conservation of momentum:

m*v*r = m*w*r^2 + 1/3 (ML^2w)

w = m*v*r / ( mr^2 + 1/3 ML^2 )

w = 0.1 *400*1 / ( (0.1 * 1) + 1/3 20* 2^2 )

w = 1.49 rad / s

The angular speed of the rod after the impact is 1.49 rad/s

Learn more about Angular speed here:

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2 years ago

Electric field lines moves away from positive to wards negative?

Archy [21]

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

6 0

2 years ago