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Alem2000
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Can someone tell me what I am doing wrong? "A glass is a cylinder, 15cm tall and with a radius of 3cm. It is filled 12.5cm high with water d=1000kg/m^3.
A straw is placed in the glass, with the top of the straw 5cm above the top of the glass. How much work does it take to drink through the straw? Note..for the purpose of computing work, it is safe to assume that you are always drinking from the top most "layer" of water, no matter where the straw is placed."
I started using a system that the bottom of the cup is 0 height..and the top of the cup 15...I think some people do it the other way around
By the way I converted from cm to m randomly in the problem whenever the units came up in the problem.
This is my work...
[tex]\vec{W}=\vec{F}d[/tex]
[tex]\vec{F}=m\vec{a}[/tex]
[tex]d=m/V[/tex]
[tex]d=1,000kg/m^3[/tex]
[tex]V=9\pi,cm^2[/tex]
using the equations form above [tex]m=.9\pi,kg/m[/tex]
[tex]\vec{F}=\vec{w}=8.82N\pi[/tex]...weight
[tex]\vec{W}=(8.82)(h+.05)J\pi[/tex]..work
[tex]\vec{W}=8.82h=.441)J\pi[/tex]
[tex]\pi\int_{0}^{.125}8.82h+.441dh[/tex]
so far so good?
after integrating...i said that it takes [tex]2182.1J [/tex] to do the work I am assuming that's wrong...thats a lot of Joules just to drink some water ay?
A straw is placed in the glass, with the top of the straw 5cm above the top of the glass. How much work does it take to drink through the straw? Note..for the purpose of computing work, it is safe to assume that you are always drinking from the top most "layer" of water, no matter where the straw is placed."
I started using a system that the bottom of the cup is 0 height..and the top of the cup 15...I think some people do it the other way around
By the way I converted from cm to m randomly in the problem whenever the units came up in the problem.
This is my work...
[tex]\vec{W}=\vec{F}d[/tex]
[tex]\vec{F}=m\vec{a}[/tex]
[tex]d=m/V[/tex]
[tex]d=1,000kg/m^3[/tex]
[tex]V=9\pi,cm^2[/tex]
using the equations form above [tex]m=.9\pi,kg/m[/tex]
[tex]\vec{F}=\vec{w}=8.82N\pi[/tex]...weight
[tex]\vec{W}=(8.82)(h+.05)J\pi[/tex]..work
[tex]\vec{W}=8.82h=.441)J\pi[/tex]
[tex]\pi\int_{0}^{.125}8.82h+.441dh[/tex]
so far so good?
after integrating...i said that it takes [tex]2182.1J [/tex] to do the work I am assuming that's wrong...thats a lot of Joules just to drink some water ay?
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