Answer:
option C
Explanation:
given,
diameter of circular room = 8 m
rotational velocity of the rider = 45 rev/min
=
=4.712 rad/s
here in this case normal force is equal to centripetal force
N = m r ω²
N = m x 4 x 4.712²
N = 88.83m
frictional force = μ N
= 88.83m x μ
now, for the body to not to slide
gravity force is equal to frictional force
m g = 88.83 m x μ
g = 88.83 x μ
9.8 = 88.83 x μ
μ = 0.11
hence, the correct answer is option C
To solve this problem it is necessary to apply the concepts related to optical magnification (is the process of enlarging the apparent size, not physical size, of something.). Specifically the angular magnification of an optical telescope is given by
Where,
Focal length of the objective lens in a refractor
Focal length of the eyepiece
Our values are given as
71cm
2.1cm
Replacing we have
Therefore the magnification of this astronomical telescope is -33.81
C. The downward component of the projectile's velocity continually increases
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion (with constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction
Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:
where
u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)
downward is the acceleration of gravity
t is the time
So the equation becomes
This means that
C. The downward component of the projectile's velocity continually increases
Because every second, it increases by in the downward direction.
Learn more about projectile motion:
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