A pendulum is swinging back and forth with a period of 2.0 seconds here on Earth. This pendulum is then brought to the Moon, whe
1 answer:
You might be interested in
a) Distance moved by the top lid during a blink: 1 cm (estimate)
b) The acceleration is
c) The final speed is 0.83 m/s
Explanation:
a)
For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.
b)
Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
For the eyelid, we have:
u = 0 (it starts from rest)
s = 1 cm = 0.01 m is the distance covered
t = 0.024 s is the time taken
Solving for a, we find the acceleration:
c)
The final speed of the upper eyelid can be found by using another suvat equation:
where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
For the eyelid in this problem, we have
u = 0
t = 0.024 s
Substituting, we find the final speed:
Learn more about accelerated motion:
#LearnwithBrainly
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
where
is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
(c) 5550 m
The maximum playing time of the CD is
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
(d)
The angular acceleration of the CD is given by
where
is the final angular speed (when the CD is scanned at the outermost part)
is the initial angular speed (when the CD is scanned at the innermost part)
is the time elapsed
Substituting into the equation, we find