Weight increases but mass stays the same
Answer:
θ = 4.716 10⁻⁶ rad
Explanation:
In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.
We use the diffraction equation for a slit
a sin θ = m λ
The minimum occurs at m = 1
sin θ = λ / a
Since the angles in these systems are very small, we can approximate the sine to its angle in radians
θ = λ / a
The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number
θ = 1.22 λ / a
Let's calculate
θ = 1.22 518 10⁻⁹ / 13.4 10⁻²
θ = 4.716 10⁻⁶ rad
It is indirectly proportional or inverse so it would be C
Answer:
total work is = 52450 J
Explanation:
given data
mass = 5000-lb
density = 10 lb/ft
height = 50 ft
solution
as we will treat here cable and ball are separate
and
here work need to lift cable is
w = (10Δy )(9.8 y ) j
and
now summing all segment of cable
so passing limit Δy to 0
so total work need
= 
= ![[49 y^2]^{50}_0](https://tex.z-dn.net/?f=%5B49%20y%5E2%5D%5E%7B50%7D_0)
= 2450J
so lifting 5000 lb wrcking 50 m required additional 5000 + 2450
so total work is = 52450 J
