X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
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
I believe one of these are called Elliptical, Spiral, and Irregular Galaxies.
