It is determined by immersing a crown in water that it\'svolume is 39in 3 =0.002

Question

It is determined by immersing a crown in water that it'svolume is 39in3=0.00225ft3. (a) What would its weight be if it were made of puregold? (b) What would its weight be if half of its volume was goldand half lead? Worth 14 Thank you. It is determined by immersing a crown in water that it'svolume is 39in3=0.00225ft3. (a) What would its weight be if it were made of puregold? (b) What would its weight be if half of its volume was goldand half lead? Worth 14 Thank you. (a) What would its weight be if it were made of puregold? (b) What would its weight be if half of its volume was goldand half lead? Worth 14 Thank you.

Explanation / Answer

It is determined by immersing a crown in water that it'svolume is 39in3=0.00225ft3.=

0.02831 cubic meter

density of gold = 19320kg/m3
density of lead = 11340 kg/m3





(a) What would its weight be if it were made of pure gold?
weight=volume*density*g=0.02831*19320*9.8=5360.10N
(b) What would its weight be if half of its volume was goldand half lead?
effective density=[d1+d2]/2=[19320+11340]/2 = 15330kg/m3
weight=volume*density*g=0.02831*15330*9.8=42531.24N
(a) What would its weight be if it were made of pure gold?
weight=volume*density*g=0.02831*19320*9.8=5360.10N
(b) What would its weight be if half of its volume was goldand half lead?
effective density=[d1+d2]/2=[19320+11340]/2 = 15330kg/m3
weight=volume*density*g=0.02831*15330*9.8=42531.24N

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