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King
Group Title
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.
 2 years ago
 2 years ago
King Group Title
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.
 2 years ago
 2 years ago

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King Group TitleBest ResponseYou've already chosen the best response.0
do u know how to do this?
 2 years ago

Gab02282 Group TitleBest ResponseYou've already chosen the best response.0
\[d=\frac{m}{V}\] Therefore, \[m=dV\] Hope this helps, somehow. I'm unsure.
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
how?can u solve it?
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
help tomas......
 2 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.1
i think you need to use Archimedus law or something
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
dw:1326975917374:dw
 2 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.1
dw:1326976017906:dw
 2 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.1
also \[F_a>mg\] if that matters lol
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
i know but then i am stuck at hw to find the missin volume..............
 2 years ago

sachin_vishaul Group TitleBest ResponseYou've already chosen the best response.0
u hv the same problem?
 2 years ago

sachin_vishaul Group TitleBest ResponseYou've already chosen the best response.0
nanbenda
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
jacob help man!
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
I think you have to use Archimedes Principle and Density Mass relationship. Still I am not getting the equation.
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
i know but how????
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Sorry, i was saying whether we can use density = mass/volume?
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
yeah we can use anythin
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Then it is simple. Mass= volume X density, so Mass= 10 * 0.15= 1.5 gm (Still u check the answer)
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
AADARSH!did u see my diagram and read the question again ....
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
answer is wrong
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Saw it, but confused about the equation
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
volume is 10+some x
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
I will consult and say u tomorrow
 2 years ago

nikvist Group TitleBest ResponseYou've already chosen the best response.3
\[\rho=0.15\,g/cm^3\quad,\quad \rho_w=1\,g/cm^3\quad,\quad V\uparrow=10\,cm^3\]\[\rho Vg=\rho_w V\downarrow g\quad\Rightarrow\quad \frac{V}{V\downarrow}=\frac{V\uparrow+V\downarrow}{V\downarrow}=\frac{V\uparrow}{V\downarrow}+1=\frac{\rho_w}{\rho}\]\[V\downarrow=\frac{\rho}{\rho_w\rho}V\uparrow\]\[m=\rho(V\uparrow+V\downarrow)=\rho\left(1+\frac{\rho}{\rho_w\rho}\right)V\uparrow=\frac{\rho\rho_w}{\rho_w\rho}V\uparrow\]\[m=\frac{0.15}{0.85}\cdot 10\,g=1.765\,g\]
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.2
dw:1326987424332:dwFor 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)
 2 years ago

King Group TitleBest ResponseYou've already chosen the best response.0
nikvist could u explain ure answer in words or JamesJ could u do it fr him?
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.2
hmmm..... that is wat i did.............
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Please explain in simple words. Are we to use that Density formula or not?
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.2
knowingV it is easy to calculate the mass of that part of cork+mass of upper part of cork...........got it adarsh?
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.0
Yeah, that's right.
 2 years ago

salini Group TitleBest ResponseYou've already chosen the best response.2
adarsh see this link for better understabding http://ocw.mit.edu/courses/physics/801physicsiclassicalmechanicsfall1999/videolectures/lecture28/
 2 years ago

unicorn Group TitleBest ResponseYou've already chosen the best response.0
here is yours ans attachment below hope it will help you
 2 years ago
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