submerged in water. What is the density of the object?
Density of water = 1000 kg/m^3
2007-11-20 15:23:39 · 4 answers · asked by Pascal 4 in Science & Mathematics ➔ Physics
weight of water displaced = 98-75 = 23N.
mass of 23N. of water = 23/9.8 = 2.35 kg.
volume of 2.35 kg. of water = 2.35 l.
mass of object = 98/9.8 = 10 kg.
density of object = 10 /2.35 = 4.26 kg./l.
2007-11-20 15:38:57 · answer #1 · answered by J C 5 · 0⤊ 0⤋
I may be wrong, but there are a couple of factors here that haven't been revealed. It's been 30 years since I've dealt with this subject but something in my memory says it does matter what the object is made of and the temperature of the water would then effect the density of the object.
Perhaps I'm suffering from dementia, so someone smarter than me should feel free to jump on in!
2007-11-20 23:33:40 · answer #2 · answered by odechiro 3 · 0⤊ 1⤋
Did the question gave you the depth or the area of the object
2007-11-20 23:33:11 · answer #3 · answered by Murtaza 6 · 0⤊ 0⤋
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density of the object /density of water =relative density(RD)
relative density(RD)=wt of object in air/loss in wt in water
wt of object in air = 98 N
wt of object in water = 75 N
RD = wt in air / [wt in air - wt in water ]
R D = 98 / [ 98 - 75 ]
R D = 98 / 23=4.2608
density of the object =density of water *relative density(RD)
density of the object =1000*4.2608 =4206.8 kg / m^3
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2007-11-20 23:40:32 · answer #4 · answered by ukmudgal 6 · 1⤊ 0⤋