A low pressure system has moved away from the equator and is moving directly over water. When the system reaches 600 N latitude,
1 answer:
You might be interested in
The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.
<h3>Their speed at the bottom</h3>Apply the principle of conservation of energy,
E(up) - E(friction) = E(bottom)
mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²
where;
- u is the speed of the after 28 m
u = √2gh
u = √(2gL sin15)
u = √(2 x 9.8 x 28 x sin 15)
u = 11.92 m/s
Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.
Learn more about speed here: brainly.com/question/6504879
#SPJ1
The acceleration experienced by the particle is given by
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed
of the particle and its distance r from the axis by the relationship
In our problem,
, so we can solve for :
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to
revolutions, while 
. So we the conversion is
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed
In our problem,
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to