Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
Answer: MR²
is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center
Explanation:
Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.
I = ∫ r² dm
= R²∫ dm
MR²
where M is total mass and R is radius of the hoop .
Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone, 
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
Able to be hammered or pressed permanently out of shape without breaking or cracking.
-- Speed = (distance) / (time to cover the distance) = 840/2.5 = 336 m/s
-- Frequency = (speed) / (wavelength) = 336/0.70 = 480 Hz.
