Answer:
Radius r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Explanation:
Area of a circle;
A = πr^2
A = area
r = radius
Making r the subject of formula;
r = √(A/π) ........1
Given;
A = 1300 cm^2
Substituting into the equation 1;
r = √(1300/π)
r = 20.34214472564 cm
r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
<span>The scientists use data
and measurement to obtain empirical evidence. Data can be collected through
direct observation or else experimentation. Empirical measurements and data can
be gathered by using qualitative and quantitative methods. Empirical evidence
contains the recording and analyzing the data which is a central part of scientific
method.</span>
Because there's no such thing as "really" moving.
ALL motion is always relative to something.
Here's an example:
You're sitting in a comfy cushy seat, reading a book and listening
to your .mp3 player, and you're getting drowsy. It's so warm and
comfortable, your eyes are getting so heavy, finally the book slips
out of your hand, falls into your lap, and you are fast asleep.
-- Relative to you, the book is not moving at all.
-- Relative to the seat, you are not moving at all.
-- Relative to the wall and the window, the seat is not moving at all.
-- But your seat is in a passenger airliner. Relative to people on the
ground, you are moving past them at almost 500 miles per hour !
-- Relative to the center of the Earth, the people on the ground are moving
in a circle at more than 700 miles per hour.
-- Relative to the center of the Sun, the Earth and everything on it are moving
in a circle at about 66,700 miles per hour !
How fast are they REALLY moving ?
There's no such thing.
It all depends on what reference you're using.
(a) 
The relationship beween centripetal acceleration and angular speed is
where
is the angular speed
r is the radius of the circular path
Here we gave
is the centripetal acceleration
r = 5.15 m is the radius
Solving for , we find:
(b) 21.3 m/s
The relationship between the linear speed and the angular speed is
where
v is the linear speed
is the angular speed
r is the radius of the circular path
In this problem we have
r = 5.15 m
Solving the equation for v, we find
To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.
Our values are given as
From the continuity equations in pipes we have to
Where,
= Cross sectional Area at each section
= Flow Velocity at each section
Then replacing we have,
From Bernoulli equation we have that the change in the pressure is
![7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)](https://tex.z-dn.net/?f=7.3%2A10%5E3%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%281000%29%28%5B%20%5Cfrac%7B%281.25%2A10%5E%7B-2%7D%29%5E2%20%7D%7B0.6%2A10%5E%7B-2%7D%29%5E2%7D%20v_1%20%5D%5E2-v_1%5E2%29)
Therefore the speed of flow in the first tube is 0.9m/s
