Density: Cube of Aluminum to Copper Sphere

AI Thread Summary
To determine the diameter of a copper sphere with the same mass as a 9.00 cm aluminum cube, the mass of the aluminum cube is calculated to be 1968.3 grams using its density. The volume of the copper sphere is then found to be 219.676 cm3 based on copper's density. The radius is incorrectly calculated as 52.4438 cm due to a misstep in taking the cubic root. The correct approach requires revisiting the volume equation for a sphere to find the accurate radius and subsequently the diameter. The poster acknowledges the error and seeks clarification on the calculations.
JustinDaniels
Messages
8
Reaction score
0

Homework Statement


What is the diameter of a copper sphere that has the same mass as a 9.00 cm× 9.00 cm× 9.00 cm cube of aluminum?

Density of Aluminum = P(al) = 2.70g/cm3
Density of Copper = P(cu) = 8.96 g/cm3

Volume of Aluminum Cube = Vcube = 729 cm3

Homework Equations


Volume of a Sphere = 4/3pi(r3) - Note: We are solving for 2r (i.e. the diameter).
Density = Mass/Volume

The Attempt at a Solution


1) Find the mass of the aluminum cube.
Density = Mass/Volume - substitute in known values
2.70g/cm3 = mass/729cm3
Mass = 1968.3 grams

2) Find the volume of the sphere.
Density = Mass/Volume
8.96 g/cm3 = 1968.3g/Volume
Volume = 219.676 cm3

3) Find the radius of the sphere.
Volume of a Sphere = 4/3pi(r3) - substitute in known values
219.676 = 4/3pi(r3)
Radius = 52.4438 cm

Diameter = Radius * 2 = 104.8877 cm

Everything looks pretty spot on to me; however, my online homework says this is incorrect. Any help would be greatly appreciated.

Thanks guys,
Justin Daniels.

P.S. This is my first post. Please let me know if I've failed to follow the format for posting questions, so I can correct this moving forward.
 
Physics news on Phys.org
Edit: I forgot to take the cubic root of the radius when solving in step 3. Sorry for the pointless thread, and thank you guys anyway!
 
JustinDaniels said:
219.676 = 4/3pi(r3)
Radius = 52.4438 cm
Your equation cannot possibly produce a radius that large. Check the calculation.
 
JustinDaniels said:
Edit: I forgot to take the cubic root of the radius when solving in step 3. Sorry for the pointless thread, and thank you guys anyway!
We crossed in the post.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
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...
Back
Top