Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
Hi there!
This is an example of a recoil collision.
Using the conservation of momentum:

The initial momentum is 0 kgm/s (objects start from rest), so:

We are given that the 15 kg block has a velocity of 12 m/s to the left, so:

Solve for v2':

The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.
Considering also the formula for Gravitational Potential Energy GPE = mgh
For the bird on the ground, h =0, therefore GPE = m*9.8*0 = 0
For that on the tree = mgh = m*9.8*h
Of course the one on the tree has a value greater than zero.
K.E = 1/2mv^2
The kinetic energy is 50,000 J and the mass of the car is 10,000 kg.
50,000 J = 1/2(10,000kg)v^2
Solve for v (velocity)
50,000 J = 5,000 kg x v^2
10 = v^2

about 3 m/s (in terms of significant figures)