Average speed of the gasoline as it leaves the nozzle?

AI Thread Summary
The discussion revolves around calculating the average speed of gasoline exiting a nozzle with a radius of 1 cm, given that it takes 2 minutes to fill a 40-liter tank. The flow rate equation Q = v * A is highlighted, where Q represents the flow rate and A is the cross-sectional area of the nozzle. Participants clarify that the average velocity can be derived from the volume of gasoline and the time taken to fill the tank. The final calculated average speed of the gasoline is confirmed to be approximately 1.1 meters per second. The conversation emphasizes understanding flow rate and velocity in fluid dynamics without delving into calculus.
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Homework Statement


It takes 2 minutes to fill a gas tank with 40 liters of gasoline. If the nozzle is 1cm in radius, what is the average speed of the gasoline as it leaves the nozzle?


Homework Equations


A1V1=A2v2 or 1/2pv^2 + pgh+p=constant


The Attempt at a Solution


I don't really know were to start I have the volume of the tank as being .04m^3 I am not sure what to do next.
 
Last edited:
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You want the flow rate equation.

Q=v*A

Where Q = flowrate in cm^3/s and A is the cross-sectional area.
 


Why is the flow rate in cm^3/s wouldn't it be squared because of area.
A would be = to 3.14159265*10^-4
 


Also the books says the anwser is 1.1 meters a second i just don't know how it got to that
 


Volume V = X *A
V/t = X/t*A = v*A, where v is the velocity of the liquid flow through the nozzle.
 


is that calculus because I haven't been taught how to use or apply it to these types of problems.
 


NO. Not at all.
In the problem what is flow rate V/t?
If you freeze the outflow of one second, whose length is L and area of cross section is A, what is the volume of the liquid?
(L/one second) is the average velocity of out flow.
 


the volume is .04m^3, cross sectional area is .01^2*pi
 


dQ/dt = Volume/time. What is the time taken to fill the tank?
Velocity = (dQ/dt)/Area.
 
  • #10


Thanks i got it for some reason i couldn't get to that equation before but now i get it sorry for taking up your time.
 
  • #11


It is OK.
 
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