Determine the medicine's flow speed through the needle

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Homework Help Overview

The discussion revolves around fluid dynamics, specifically focusing on the flow speed of a medicine through a hypodermic needle and the buoyancy of a submerged block of metal. Participants are exploring the application of Bernoulli's principle and the continuity equation in these contexts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between pressure, area, and flow speed, referencing Bernoulli's equation and the continuity equation. There are attempts to apply these principles to determine the flow speed and clarify the setup of the problems.

Discussion Status

Several participants are actively engaging with the equations and concepts, questioning each other's reasoning and calculations. There is a mix of attempts to solve the problems and requests for clarification on specific steps, indicating a collaborative effort to understand the underlying physics.

Contextual Notes

Some participants express uncertainty about the assumptions made, such as the pressure conditions and the dimensions provided in the problems. There are also references to atmospheric pressure and the implications of the syringe's orientation.

LesVampires
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A hypodermic syringe contains a medicine with the density of water (see figure below). The barrel of the syringe has a cross-sectional area of 2.20 10-5 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force of magnitude 1.80 N is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the needle. Assume the pressure in the needle remains equal to 1.00 atm and that the syringe is horizontal.

second question

An 11.3 kg block of metal is suspended from a scale and immersed in water, as in the figure below. The dimensions of the block are 12.0 cm 8.7 cm 8.7 cm. The 12.0 cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the wateri know that P=Po + pgh p=m/v and g i can't figure h out
can u direct me a bit with this ty verymuch

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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LesVampires said:
A hypodermic syringe contains a medicine with the density of water (see figure below). The barrel of the syringe has a cross-sectional area of 2.20 10-5 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force of magnitude 1.80 N is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the needle. Assume the pressure in the needle remains equal to 1.00 atm and that the syringe is horizontal.

http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html

height stays same so it cancels

solve for v
 
hmm i know that Av= Av and the equations but i keep gettin a wrong answer... A= 2.2x10^-5 m^2
 
LesVampires said:
hmm i know that Av= Av and the equations but i keep gettin a wrong answer... A= 2.2x10^-5 m^2

P_1 + 1/2 \rho v^2 + \rho gy_1 = P_2 + 1/2\rho v^2 + \rho g y_2

don't forget atmospheric pressure is at the end of the needle solve for v2

P=F/A to find the pressure exerted over the area of the plunger (force exerted on plunger )

After finding the equivalent pressure that is due to the plunger, add that to atm pressure.

show me what you did and I'll tell you what your doing incorrect.
 
Last edited:
well the pgys are 0 u have v= 2change in P/pv pv is A right? P=1.01 x10^5 and
 
LesVampires said:
well the pgys are 0 u have v= 2change in P/pv pv is A right? P=1.01 x10^5 and


yes pgy is 0 so what is the equation then?

not sure what your saying when you say
v=2change in P/pv pv is A
:confused:


and the area is small on the end of the syringe but did they give it to you?


Look at the post above (last post by me) I added stuff.

I have to go but I'll see what you did tommorow (post your work)

A2/A1<<1 -> note
 
Last edited:
isnt that what's left of the equation?
 
LesVampires said:
isnt that what's left of the equation?

what are you talking about.

No the A1V1=A2V2 is the continuity equation and that is substituted into the bernoulli equation. And then you can solve for v2 which you want.
 

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