The Professor's centripetal acceleration is 0.044 m/s²
Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.
It is given by:
a = v²/r
where v is the velocity and r is the radius.
Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:
a = v²/r = 0.419²/4 = 0.044 m/s²
The Professor's centripetal acceleration is 0.044 m/s²
Find out more at: brainly.com/question/6082363
Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration 
, so:
where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:
W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2
)= 1.976 rpm
Answer:
The order of increasing energy is as follows
"microwave < infrared < visible < ultraviolet"
Option (A) is correct.
Explanation:
Given:
Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.
From the formula of energy in terms of frequency.
Where 
planck constant, 
frequency of light.
From above formula we can conclude that higher frequency means higher energy.
In our case ultraviolet has higher frequency and microwave has lower frequency.
So ultraviolet has higher energy and microwave has lower energy.
microwave < infrared < visible < ultraviolet
Therefore, the order of increasing energy is as follows
"microwave < infrared < visible < ultraviolet"
A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.
So, the hang up time, or the time the object stayed on air is calculated using this equation:
a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
