A person on a rocket traveling at 0.47 c (with respect to the Earth) observes a meteor come from behind and pass her at a speed
1 answer:
The concept to solve this problem is related to the relativistic physics for which the speed of the object in different frames of reference is related. This concept is called Velocity-addition formula
and can be written as,
Where,
u = Velocity of a body within a Lorentz Frame
v = Velocity of a second frame
u'= The transformed velocity of the body within the second frame
c = speed of light
Replacing we have to
Therefore the meteor moving with respect to the Earth to 230'700.000m/s
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Answer:
1.3 x 10⁻⁴ m
Explanation:
= wavelength of the light = 450 nm = 450 x 10⁻⁹ m
n = order of the bright fringe = 1
θ = angle = 0.2°
d = separation between the slits
For bright fringe, Using the equation
d Sinθ = n
Inserting the values
d Sin0.2° = (1) (450 x 10⁻⁹)
d (0.003491) = (450 x 10⁻⁹)
d = 1.3 x 10⁻⁴ m
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.