To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:

Where,
V = the linear speed
r = Radius
Angular speed
The angular speed is given by


Replacing at our first equation we have that the centripetal acceleration would be



To transform it into multiples of the earth's gravity which is given as
the equivalent of 1g.


PART B) Now the linear speed would be subject to:



Therefore the linear speed of a point on its edge is 51.05m/s
Judging on the fact that this is middle school physics, I would say that it is a beneficial relationship.
Left. Opposite of the direction the box is pushed.
Answer:
The speed of the block when it is 5.00 m from the top of the incline is 3.04 m/s
Explanation:
given information:
s = 7.80 m
v = 3.8 m/s
if s = 5 m, v?
first we have to find the acceleration of the block using the following equation:
v² = v₀² + 2as, v₀² = 0 thus
3.8² = 2 a 7.8
a = 0.93
so, if s = 5m the final speed is
v² = 2 (0.93) (5)
= 9.26
v = √9.26
= 3.04 m/s
Answer:
20 degrees.
Explanation:
From Snell’s law of refraction:
sinθ1•n1 = sinθ2•n2
where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.
The red light is in air as medium1, so n1 (air) = 1.00029
So, to find θ2, the refracted angle:
sinθ1•1.00029 = sinθ2•1.464
sin(30)•1.00029 / 1.464 = sinθ2
0.5•1.00029 / 1.464 = sinθ2
sinθ2 = 0.3416291
θ2 = arcsin(0.3416291)
θ2 = 19.976 degrees
To the nearest degree,
θ2 = 20 degrees.