It is diffraction
Explanation:
The opening is the aperture
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v
= ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v
= ( u + u ) / ( 1 + ( u×u / c²) )
v
= 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v
= 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v
= 1.904c / ( 1 + 0.906304 )
v
= 1.904c / 1.906304
v
= 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
It MUST be either glue or gravity.
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:
0.050 m
Explanation:
The strength of the magnetic field produced by a current-carrying wire is given by

where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.
In this problem, we have:
(current in the wire)
(strength of magnetic field)
Solving for r, we find the distance from the wire:
