A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
A piece of metal is released under water. The volume of the metal is 50.0 cm^3 and its specific gravity is 5.0. What is its initial acceleration?
I mostly just need some validation here because the answer isn't in the back of the book. I took the volume, converted to m^3. Using definition of s.g. solved for mass of object. Found the weight of object, which is equal to buoyant force. F=ma and solved for acceleration. Is that right?
anonymous
 5 years ago
A piece of metal is released under water. The volume of the metal is 50.0 cm^3 and its specific gravity is 5.0. What is its initial acceleration? I mostly just need some validation here because the answer isn't in the back of the book. I took the volume, converted to m^3. Using definition of s.g. solved for mass of object. Found the weight of object, which is equal to buoyant force. F=ma and solved for acceleration. Is that right?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmmm, maybe not. I just end up with 9.8...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The buoyant force a fluid applies to an object is equal to the weight of the volume of fluid displaced, not the weight of the object. I've just done this very quickly so it might not be entirely correct, but... Based on what we know of specific gravity and density and their respective equations, we can work out the mass of the object: \[\Large \begin{array}{l} SG = \frac{{{\rho _o}}}{{{\rho _f}}}\\ {\rho _f} = 1000{\rm{ kg/}}{{\rm{m}}^3}\\ {\rho _o} = SG \cdot {\rho _f} = 5.0 \cdot 1000 = 5000{\rm{ kg/}}{{\rm{m}}^3}\\ m = {\rho _o}V\\ = 5000 \cdot (50 \times {10^{  6}}) = 0.25{\rm{ kg}} \end{array}\] Note the conversion from g/cm^3 to kg/m^3 (10^6), and the assumption that the density of the water is 1000kg/m^3. We can work out the buoyancy force based on the density of water, gravity and the volume of displaced water: \[\Large \begin{array}{l} B = {\rho _f}gV\\ = 1000 \cdot 9.81 \cdot (50 \times {10^{  6}})\\ = 0.4905{\rm{ N}} \end{array}\] And the rest is simple forces: \[\Large \begin{array}{l} F = mg\\ = 0.25 \cdot 9.81 = {\rm{2}}{\rm{.4525 N}}\\ \Delta F = B  F\\ = 0.4905  2.4525\\ =  1.962{\rm{ N}}\\ F = ma\\ a = \frac{F}{m} = \frac{{  1.962}}{{0.25}} =  7.848{\rm{ m}}{{\rm{s}}^{2}} \end{array}\] Noting that the 'negative force' is simply relative to its direction, where negative would be down.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Aha! The weight of the displaced water is where it got me! Thanks
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.