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A wholesaler has three bags of sugar weighing 54, 45 and 63 kg. He wants to make as many bags of equal weight as large as possible without mixing the contents of the bags. > How many bags can he fill?I want minimum comen multiple |
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Answer» Radius of the steel CABLE, r=1.5 cm = 0.015 m Radius of the steel cable, r=1.5 cm = 0.015 mMaximum ALLOWABLE stress = 10 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−section Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 = 7.065×10 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 = 7.065×10 4 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 = 7.065×10 4 N Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 = 7.065×10 4 NHence, the cable can support the maximum load of 7.065×10 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 = 7.065×10 4 NHence, the cable can support the maximum load of 7.065×10 4 Radius of the steel cable, r=1.5 cm = 0.015 mMaximum allowable stress = 10 8 N/m 2 Maximum stress = Areaofcross−sectionMaximumforce Maximum force = Maximum stress × Area of cross-section = 10 8 ×π×(0.015) 2 = 7.065×10 4 NHence, the cable can support the maximum load of 7.065×10 4 N. |
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