Question #256555

A stone is found to weigh 6.2 in air, 4.5 in water, and 5.3 N in an unknown liquid. Find a.) the density of the stone and the unknown liquid, b.) the buoyant force of the liquid on the stone, c.) the buoyant force of water on the stone, and d) the volume of the stone.


1
Expert's answer
2021-10-25T15:23:54-0400

Pair=6.2NP_{air} = 6.2N

Pwater=4.5NP_{water}= 4.5N

Punliquid=5.3NP_{unliquid}= 5.3N

g=9.8ms2g= 9.8\frac{m}{s^2}

in water:\text{in water:}

ρwater=1000kgm3\rho_{water} = 1000\frac{kg}{m^3}

BFwater=PairPwater=6.24.5=1.7NBF_{water} = P_{air}-P{water}= 6.2-4.5= 1.7N

BFwater=ρwatergVstoneBF_{water} = \rho_{water}gV_{stone}

Vstone=BFwaterρwaterg=1.710009.81.73104m3V_{stone} = \frac{BF_{water}}{ \rho_{water}g}=\frac{1.7}{1000*9.8}\approx 1.73*10^{-4}m^3

in air : \text{in air : }

Pair=mstonegP_{air} = m_{stone}g

ρstone=mstoneVstone=PairgVstone\rho_{stone}= \frac{m_{stone}}{V_{stone}}=\frac{P_{air}}{gV_{stone}}

ρstone=6.29.81.73104=3.66103kgm3\rho_{stone}= \frac{6.2}{9.8*1.73*10^{-4}}=3.66*10^{3}\frac{kg }{m^3}

in an unknown liquid:\text{in an unknown liquid:}

BFunliquid=PairPunliquid=6.25.3=0.9NBF_{unliquid}= P_{air} -P_{unliquid}= 6.2-5.3 = 0.9N

BFunliquid=ρunliquidgVstoneBF_{unliquid} = \rho_{unliquid}gV_{stone}

ρunliquid=BFunliquidgVstone=0.99.81.73104=0.53103kgm3\rho_{unliquid} =\frac{ BF_{unliquid}} {gV_{stone}}=\frac{0.9}{9.8*1.73*10^{-4}}=0.53*10^3\frac{kg}{m^3}


Answer:\text{Answer:}

a)ρstone=3.66103kgm3a)\rho_{stone}=3.66*10^{3}\frac{kg }{m^3}

ρunliquid=0.53103kgm3\rho_{unliquid} =0.53*10^3\frac{kg}{m^3}

b)BFunliquid=0.9Nb)BF_{unliquid}= 0.9N

c)BFwater=1.7Nc)BF_{water} = 1.7N

d)Vstone=1.73104m3d)V_{stone} = 1.73*10^{-4}m^3


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS