Measuring Spheres: How to Use the Formula (4/3 π r3)

AI Thread Summary
The discussion centers on understanding the formula for the volume of a sphere, (4/3 π r3), where "r" is the radius. Participants clarify that the formula calculates the volume and suggest that the original poster should be able to derive the problem from this equation. There is some debate about whether the poster needs a derivation or proof of the formula. The conversation also touches on the relevance of referencing older threads. Overall, the focus remains on how to apply the formula for homework related to measuring spheres.
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I have homework, the teacher gave us measuring spheres. He gave us the formula (4/3 π r3) But he didn't show us how to figure out the problem.
I don't know how your suppose to figure it out.
 
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What's the problem? Do you want to know the derivation of the formula or a proof?
 
The equation you stated it the equation for the volume of a sphere where "r" represents the radius of the sphere (the distance from the centre of the sphere to edge) and Pi is the constant 3.14...

You should be able to work things from there.

Leila
 
I would really doubt if the OP thought the same way as you are explaining :confused: :cool:
 
Don't get dexter started on spheres..

Edit: was there any need to bring up threads that are nearly a month old?
 

Thread 'Errors and significant figures'

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Thread 'A cylinder connected to a hanging mass'

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Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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