A cork of density 0.15 g/cm^3 floats in a bucket of water with 10 cm^3 of its volume above the surface of water.Find the mass of the cork.

- King

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- King

do u know how to do this?

- anonymous

\[d=\frac{m}{V}\]
Therefore, \[m=dV\]
Hope this helps, somehow. I'm unsure.

- King

how?can u solve it?

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## More answers

- King

help tomas......

- anonymous

i think you need to use Archimedus law or something

- King

|dw:1326975917374:dw|

- anonymous

|dw:1326976017906:dw|

- King

ya so now?

- anonymous

also
\[F_a>mg\]
if that matters lol

- King

i know but then i am stuck at hw to find the missin volume..............

- anonymous

hi

- anonymous

u hv the same problem?

- King

i posted fr u

- anonymous

nanbenda

- King

jacob help man!

- King

wait sachin

- anonymous

I think you have to use Archimedes Principle and Density- Mass relationship. Still I am not getting the equation.

- King

i know but how????

- King

how is d=m/g

- anonymous

Sorry, i was saying whether we can use density = mass/volume?

- King

yeah we can use anythin

- anonymous

Then it is simple. Mass= volume X density, so Mass= 10 * 0.15= 1.5 gm
(Still u check the answer)

- King

AADARSH!did u see my diagram and read the question again ....

- King

answer is wrong

- anonymous

Saw it, but confused about the equation

- King

volume is 10+some x

- anonymous

I will consult and say u tomorrow

- nikvist

\[\rho=0.15\,g/cm^3\quad,\quad \rho_w=1\,g/cm^3\quad,\quad V\uparrow=10\,cm^3\]\[\rho Vg=\rho_w V\downarrow g\quad\Rightarrow\quad \frac{V}{V\downarrow}=\frac{V\uparrow+V\downarrow}{V\downarrow}=\frac{V\uparrow}{V\downarrow}+1=\frac{\rho_w}{\rho}\]\[V\downarrow=\frac{\rho}{\rho_w-\rho}V\uparrow\]\[m=\rho(V\uparrow+V\downarrow)=\rho\left(1+\frac{\rho}{\rho_w-\rho}\right)V\uparrow=\frac{\rho\rho_w}{\rho_w-\rho}V\uparrow\]\[m=\frac{0.15}{0.85}\cdot 10\,g=1.765\,g\]

- JamesJ

nikvist, this a good answer. I understand your notation. But perhaps for the others could you explain in words what you've done here? Thanks.

- anonymous

|dw:1326987424332:dw|For practical purposes water is incompressible so the submerged part would displace an amount of water equal to its own volume.
so same mass that is what nikivist has done ...equate the mass of cork as a whole to mass of water displaced
then manipulations gives us everything written in terms of density which we know(data in the question)

- King

nikvist could u explain ure answer in words or JamesJ could u do it fr him?

- anonymous

hmmm..... that is wat i did.............

- anonymous

Please explain in simple words. Are we to use that Density formula or not?

- anonymous

I WILL TELL EVERYTHING FROM FIRST
by archimedes principle,
the wt of cork=wt of water displaced(buoyant force)
this is because the body is floating upward forces=downward
then wt of cork=volume*density*g
wt of water displaced=volume(V)*density oif water*g
write volume of cork as volume of upper immersed portion+vol of downward immersed portion(V)
REMEMBER IT IS V that we need to calculate
do it in simple terms after this
|dw:1327073074048:dw|

- anonymous

knowingV it is easy to calculate the mass of that part of cork+mass of upper part of cork...........got it adarsh?

- anonymous

Yeah

- anonymous

u need to know here that by archimedes principle,buoyant force=wt of water displaced in this case as the cork is floating the wt mg equals the buoyant force so to to keep the cork at rest

- anonymous

Yeah, that's right.

- anonymous

adarsh see this link for better understabding http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-28/

- anonymous

Seeing.

- unicorn

here is yours ans attachment below hope it will help you

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