Asked by Ashleigh
a lead bullet is placed in a pool of mercury. what fractional part of the volume of the bullet is submerged?
Answers
Answered by
bobpursley
well, the lead displaces mercury such that the weight of the mercury is equal to the weight of the lead.
weight of mercury displaced= densitymercury*volumeleadsubmerged
weight of lead= densitylead*volumelead
so set them equal
Volumesubmerged/totalvolume= densitylead/densitymercury
weight of mercury displaced= densitymercury*volumeleadsubmerged
weight of lead= densitylead*volumelead
so set them equal
Volumesubmerged/totalvolume= densitylead/densitymercury
Answered by
Ashleigh
Perfect!
Since the density of lead is 11.3 x 10^3 kg/m^3 and the density of mercury is 13.6 x 10^3 kg/m^3, the answer is 0.831.
Thanks much!
Since the density of lead is 11.3 x 10^3 kg/m^3 and the density of mercury is 13.6 x 10^3 kg/m^3, the answer is 0.831.
Thanks much!
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