Answer:
2.63 Hz
Explanation:
m = mass of the object = 8.0 kg
x = stretch in the spring = 3.6 cm = 0.036 m
k = spring constant of the spring
using equilibrium of force
Spring force = weight of object
k x = m g
k (0.036) = (8) (9.8)
k = 2177.78 N/m
frequency of oscillation is given as
= 2.63 Hz
Ok so we are given the radius of 7cm and time of 5 seconds.
From the data we got we can calculate speed, frequency, perimeter and area of the semicircle.
Let's start with perimeter.
We know that perimeter of circle is so the perimeter of semicircle is
or simply
So the perimeter is equal to:
So this is the length of a curve or let's say the distance.
Now let's look at the linear speed where d is distance and t time.
We know the distance and we know the time.
So let's calculate it.
Hope this helps.
r3t40
Answer:
= 22.41rad/s
Explanation:
First, we know that:
a = 4 rad/s^2
S = 10 rev = 62.83 rad
Now we know that:
where is the final angular velocity,
the initial angular velocity, a is the angular aceleration and S the radians.
Replacing, we get:
Finally, solving for :
= 22.41rad/s
Answer:
Explanation:
Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .
Then length of the platform = length of the train rocket .
A )
Time to cross a particular point on the platform
= length of rocket train / .96 x 3 x 10⁸
= 90 / .96 x 3 x 10⁸
= 31.25 x 10⁻⁸ s
B) Rest length of the rocket = length of platform = 90 m
C ) length of platform as viewed by moving observer =
=
= 321 m
D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s
for the observer in the rocket , time will be dilated so time recorded by observer in motion ,
8.75 x 10⁻⁸ s .