I am assuming that the context is that of a thrown projectile (a ball, a bullet, etc.) in the gravitational field of the Earth.
Answer:
Angular velocity of an ultra centrifuge is 17146.42 rad/s.
Explanation:
Given that,
Acceleration of an ultra centrifuge, 
Distance from axis, r = 2 cm = 0.02 m
We need to find the angular speed. We know that the relation between angular speed and angular acceleration is given by :
So, the angular velocity of an ultra centrifuge is 17146.42 rad/s.
75km=75000m
35min=2100 seconds
Velocity=Displacement/Time
=75000/2100
=750/21
=35.71 m/s
Answer:
4.4 m
Explanation:
We are told the light from his flashlight, 1.3 m above the water level. Thus; h1 = 1.3m
Also,we are told that the light shone 2.5 m from his foot at the edge of the pool. Thus, L1 = 2.5 m
Angle of incidence θ1 is given by;
tan θ1 = L1/h1
tan θ1 = 2.5/1.3
tan θ1 = 1.9231
θ1 = tan^(-1) 1.9231
θ1 = 62.53°
Using Snell's law, we can find the angle of refraction from;
Sin θ2 = (η_air/η_water) Sin θ1
Where;
η_air is Refractive index of air = 1
η_water is Refractive index of water = 1.33
Thus;
Sin θ2 = (1/1.33) × sin 62.53°
Sin θ2 = 0.6671
θ2 = sin^(-1) 0.6671
θ2 = 41.84°
We want to find where the spot of light hit the bottom of the pool if the pool is 2.1 m deep. Thus, h2 = 2.1 m
Now, the spot can be found from;
L = L1 + L2
Where L2 = (h2) tan θ2
L = 2.5 + 2.1 tan 48.84
L = 2.5 + (2.1 × 0.8954)
L ≈ 4.4 m
