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
The rate in witch ditermans the speed or vibration of the movment under the waves witch couses vibrational freequencys to be disrupted.
Answer:
0.33m/
East.
Hope this helps you. Do mark me as brainliest.
Answer:
a) —0.5 j m/s
b) 4.5 i + 2.25 j m
Explanation:
<u>Givens:</u>
v_0 =3.00 i m/s
a= (-3 i — 1.400 j ) m/s^2
The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :
<em>v_x = v_0 + at = 0 </em>
(3.00 i m/s) + (-3 i m/s^2)t=0
t = (3 m/s)/-3 i m/s^2
t = -1 s
Therefore the particle reaches the maximum x-coordinate at time t = 1 s.
Part a The velocity-of course- is all in the y-direction,therefore:
v_y =v_0+ at
We have that v_0 = 0 in the y-direction.
v_y = (-0.5 j m/s^2)(1 s)
= —0.5 j m/s
Part b: While the position of the particle at t = 1 s is given by:
r=r_0+v_0*t+1/2*a*t^2
Where r_0 = 0 since the particle started from the origin.
Its position at t = 1 s is then given by :
r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2
=4.5 i + 2.25 j m
conclusion as last step. then see what happens next
