help math about spheres an cylinders?

Question

Calculate, in terms of pi, the volume of a sphere with radius of length 3cm. ( I got 36picm^3)

A cylinder of height9cm has volume to 4 times the volume of the sphere. Calcuate the legth of the radius of the cylinder.

the cylinder is partly filled with water. Nine identical soild cubes with edge of length 2cm are then submerage in the water. The water rises in the cylinder but does not over flow. how far does the water rise?
Give your answer correct to one decimals.

Show how you got it please I think the first bit is done right but not sure

Answer 1

Your first answer is correct. V = 4/3 pi r^3 = 4/3 pi (3)^3 = 36 pi cm^3.



The cylinder: V = pi r^2 h. V = pi r^2 (9). The volume is 4 times that of the sphere. So: 4(36) pi = 9 pi r^2. Solve for r. Divide both sides by 9 pi and you get (4)(4) = r^2. r = 4



Drop 9 solid cubes each with a 2 cm length. The volume of each cube is (2)^3 = 8 cm^3. Nine would have a total volume of 72 cm^3.



If the volume of the nine cubes, 72 cm^3 displaces water in the cylinder, the shape of the cylinder remains which means you use the equation for the volume of the cylinder again. V = pi r^2 h. You have the value of the displaced volume = 72 cm^2; you have the value of r = 4, so solve for h which is the height of water displaced by the cubes.



So V = pi r^2 h = 72 cm^3 = pi (16 cm^2) h. h = (72 cm^3) / (16 cm^2)(pi). h = 1.433 or 1.4 to one decimal point.

Answer 2

The volume of a sphere is given by V = 4/3*pi*r^3. A radius of 3 yields V = 36*pi, as you said.



The volume of a cylinder is given by the area of the base (pi*r^2) times the height. If it has the volume of four of our spheres, this is 144*pi. We can set 9*pi*r^2 = 144*pi to see that r = 4.



A cube has volume V = r^3. So, each cube has a volume of 8 cm^3. Thus, the 9*8 = 72 cm^3 of water are displaced. Since our cylinder has a base area of pi*4^2 = 50.265 (approx.), we can set 50.265*h = 72 to get h = 1.432 (approx.). This means that the water rose 1.432 centimeters.

Answer 3

Assuming the pyramid could be a wide-unfold sq. pyramid, the biggest sq. pyramid interior a cube with edge length d, could have height f = d and base edge e = d (such that it shares a cube's edge as its base), and its apex on the middle of the different cube edge, such that its volume V[pyramid] = (a million/3) e^2 f = d^3 / 3 = one-third the quantity V[cube] of the cube. to that end, V[pyramid] is maximized as V[cube] is maximized. A cube interior a cylinder with radius b and height c, could have volume V[cube] such that: V[cube] = min(2b, c)^3 *** Eq. a million when you consider that c is inversely proportional Assuming the two caps of the cylinder are such that their around perimeters are parallel "small circles" of the sector, (it truly is, for the cylinders optimal radius b[max] = a million, c = 0; and additionally, for the cylinder's optimal height c[max] = 2, b = 0). enable element o be the middle of the sector, and enable element p be on the fringe of a around cap. enable oq be a radius of the cylinder such that pq is perpendicular to the around cap. Then, op = the around radius a = a million, pq = 0.5 the cylinder's height = c/2, and oq = the cylinder's radius b opq is a maximum suitable triangle with hypotenuse op. by applying the Pythagorean Theorem, c^2 / 4 + b^2 = a million => c = 2 sqrt(a million - b^2) *** Eq. 2 when you consider that 0 < b < a million and nil < c < 2, we prefer purely evaluate the effective branches of the sqrts. Now, b is inversely proportional to c, and the two are monotonic, meaning that (by applying Eq. a million), V[cube] is maximized the place b = c. So our Pythagorean equation could be simplified to: b^2 / 4 + b^2 = a million => b = 2 sqrt(5) / 5 And, so V[pyramid, max] = V[cube, max] / 3 = b^3 / 3 = (2 sqrt(5) / 5)^3 / 3 = 8 sqrt(5) / seventy 5 *** answer ? 5.7% of the quantity of the sector --- by using fact the inradius of a unit cube is a million/2 and a circumradius is sqrt(3)/2, i could think of packing the sector interior the cube could maximize area utilization. the main awkward volume to fill seems to be the pyramid. So i could guess a maximal order could bypass like this (from innermost to outermost): pyramid, cylinder, cube, sphere i could guess that a minimum order could be cylinder, cube, sphere, pyramid