Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
-- Accelerating at the rate of 8 m/s², Andy's speed
after 30 seconds is
(8 m/s²) x (30.0 s) = 240 m/s .
-- His average speed during that time is
(1/2) (0 + 240 m/s) = 120 m/s .
-- In 30 sec at an average speed of 120 m/s,
Andy will travel a distance of
(120 m/s) x (30 sec) = 3,600 m
= 3.6 km .
"But how ? ! ?", you ask.
How in the world can Andy leave a stop light and then
cover 3.6 km = 2.24 miles in the next 30 seconds ?
The answer is: His acceleration of 8 m/s², or about 0.82 G
is what does it for him.
At that rate of acceleration ...
-- Andy achieves "Zero to 60 mph" in 3.35 seconds,
and then he keeps accelerating.
-- He hits 100 mph in 5.59 seconds after jumping the light ...
and then he keeps accelerating.
-- He hits 200 mph in 11.2 seconds after jumping the light ...
and then he keeps accelerating.
-- After accelerating at 8 m/s² for 30 seconds, Andy and his
car are moving at 537 miles per hour !
We really don't know whether he keeps accelerating,
but we kind of doubt it.
A couple of observations in conclusion:
-- We can't actually calculate his displacement with the information given.
Displacement is the distance and direction between the starting- and
ending-points, and we're not told whether Andy maintains a straight line
during this tense period, or is all over the road, adding great distance
but not a lot of displacement.
-- It's also likely that sometime during this performance, he is pulled
over to the side by an alert cop in a traffic-control helicopter, and
never actually succeeds in accomplishing the given description.
Answer:
Explanation:
Time taken to complete one revolution is called time period.
So, Time period, T = 1 s
Diameter = 1.6 mm
radius, r = 0.8 mm
Let the angular speed is ω.
The relation between angular velocity and the time period is
ω = 2 x 3.14 = 6.28 rad/s
The relation between the linear velocity and the angular velocity is
v = r x ω
v = 0.8 x 10^-3 x 6.28
v = 0.005 m/s
Answer:
The velocity of the ball when its hit the ground will be 54.22 m/sec
Explanation:
We have given height from which ball is dropped h = 150 m
Acceleration due to gravity 
As the ball is dropped so initial velocity will be zero so u = 0 m/sec
According to third equation of motion we know that 
So the velocity of the ball when its hit the ground will be 54.22 m/sec
