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2)A cylindrical vessel of base radius 12cm contains water to a depth of 17cm.A steel ball of volume 432兀cm3 is dropped into the vessel.It is totally submerged and causes the water to rise the rim of the vessel without water overflow.Find the height of the vessel.3)The side of a metallic cube is 20cm in length.It... 顯示更多 2)A cylindrical vessel of base radius 12cm contains water to a depth of 17cm.A steel ball of volume 432兀cm3 is dropped into the vessel.It is totally submerged and causes the water to rise the rim of the vessel without water overflow.Find the height of the vessel .3)The side of a metallic cube is 20cm in length.It is melted and recast into a solid cylinder of base radius 15cm.If there is no loss of metal,find the height of the cylinder
最佳解答:
2) Let the rise of water level be hcm. 兀( 12^2 )( h ) = 432兀 144h = 432 h = 3 Therefore the height of the vessel = 17cm + 3cm = 20cm 3) Let the height of the cylinder be hcm. 20^3 = 15^2兀h h = 11.3 ( cor. to 3 s.f. ) Therefore the height is 11.3 cm. 2007-06-25 21:36:20 補充： 3) 11.32 cm for 4s.f.
其他解答:
2. The base area of the cylindrical vessel is πr2=122π =144πcm3 Since the volume of the steel ball is 432π cm3 So the height of water that was rised up by the steel ball is = 432π cm3 / 144π cm3 =3 cm Then the height of the vessel is = 17cm + 3 cm = 20 cm 3. The volume of the cube is =20cm x 20cm x 20cm =8000 cm3 The base area of the solid cylinder is = π r 2 = π 152 =225 π cm2 Let H be the heigth of the recast solid cylinder Then the volume of the solid cylinder is = base area of the solid cylinder x height of the solid cylinder = 225 π cm2 x H cm =225Hπ cm3 Since the volume of the cube is equal to the volume of the recast solid cylinder So 225Hπ cm3 = 8000 cm3 H=8000/225π =11.32 cm960E6207A465286D
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