King
  • King
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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King
  • King
do u know how to do this?
anonymous
  • anonymous
\[d=\frac{m}{V}\] Therefore, \[m=dV\] Hope this helps, somehow. I'm unsure.
King
  • King
how?can u solve it?

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More answers

King
  • King
help tomas......
anonymous
  • anonymous
i think you need to use Archimedus law or something
King
  • King
|dw:1326975917374:dw|
anonymous
  • anonymous
|dw:1326976017906:dw|
King
  • King
ya so now?
anonymous
  • anonymous
also \[F_a>mg\] if that matters lol
King
  • King
i know but then i am stuck at hw to find the missin volume..............
anonymous
  • anonymous
hi
anonymous
  • anonymous
u hv the same problem?
King
  • King
i posted fr u
anonymous
  • anonymous
nanbenda
King
  • King
jacob help man!
King
  • King
wait sachin
anonymous
  • anonymous
I think you have to use Archimedes Principle and Density- Mass relationship. Still I am not getting the equation.
King
  • King
i know but how????
King
  • King
how is d=m/g
anonymous
  • anonymous
Sorry, i was saying whether we can use density = mass/volume?
King
  • King
yeah we can use anythin
anonymous
  • anonymous
Then it is simple. Mass= volume X density, so Mass= 10 * 0.15= 1.5 gm (Still u check the answer)
King
  • King
AADARSH!did u see my diagram and read the question again ....
King
  • King
answer is wrong
anonymous
  • anonymous
Saw it, but confused about the equation
King
  • King
volume is 10+some x
anonymous
  • anonymous
I will consult and say u tomorrow
nikvist
  • 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
  • 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
  • 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
  • King
nikvist could u explain ure answer in words or JamesJ could u do it fr him?
anonymous
  • anonymous
hmmm..... that is wat i did.............
anonymous
  • anonymous
Please explain in simple words. Are we to use that Density formula or not?
anonymous
  • 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
  • anonymous
knowingV it is easy to calculate the mass of that part of cork+mass of upper part of cork...........got it adarsh?
anonymous
  • anonymous
Yeah
anonymous
  • 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
  • anonymous
Yeah, that's right.
anonymous
  • 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
  • anonymous
Seeing.
unicorn
  • unicorn
here is yours ans attachment below hope it will help you
1 Attachment

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