Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
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
That depends on which angle you picked first, because that determines
which angles "the other two" are.
If you picked the right angle (90°) first, before you asked the question,
then the other two are acute angles, and they're also complementary
Answer:
Explanation:
= Refractive index of bubble = 1.33
f = Frequency of light = 
c = Speed of light = 
The wavelength of light is given by
Wavelength is also given by
m = 1 for minimum thickness
The minimum thickness is
