How far apart are these graduations?

  • Thread starter Thread starter dami
  • Start date Start date
AI Thread Summary
The graduations on a graduated cylinder with a 4 cm diameter are spaced 0.0796 cm apart, calculated based on the formula for the height of a cylinder with a volume of 1.00 cm³. When a solid sphere with a radius of 1.2 cm is submerged, it displaces water, causing the water level to rise by 0.576 cm. This rise corresponds to the volume of the sphere, which is approximately 7.238 cm³. The calculations for both the graduation spacing and the water displacement are confirmed to be accurate. Overall, the discussion effectively addresses the relationships between volume, height, and displacement in cylindrical and spherical contexts.
dami
Messages
18
Reaction score
0
A certain graduated cylinder has an inside diameter of 4cm.
The graduations on its side are labeled "cc" (cubic centimeters).
1. How far apart are these graduations?
2. The cylinder is partially filled with water, and a solid sphere of radius 1.2 cm is then totally submerged therein. Through what distance does the water level rise when the sphere is so immersed?
 
Physics news on Phys.org


Is this correct?
1. Find the height of a cylinder whose radius is 2 cm and whose volume is 1.00 cm^3.

V = pi*r^2*h
h = V/(pi*r^2) = 1/
h = 0.0796 cm

The graduations would be 0.0796 cm apart

2. Find the height of a cylinder of radius 2 which has a volume equal to that of the sphere.

Using R = radius of sphere = 1.2 cm
V(sphere) = (4/3)pi*R^3 = 7.238229474 cm^3

V(sphere) = volume of cylinder = pi*r^2*h
h = V(sphere)/(pi*r^2) = 0.576 cm

So the level of water would rise by 0.576 cm
 


That all looks right to me...

12.566 == area of 4cm diameter circle
0.0795 == height to contain 1cc volume
7.238 == volume of 1.2cm radius sphere
0.576 == height to contain 7.238cc volume
 


Nice bit of homework?
 
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 '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