Asked by Ademide
A solid weigh 45N and 15N respectively in air and water. Calculate the relative density
Answers
Answered by
Damon
mass = 45/g
change in weight = 45 -15 = 30 N
so
mass of water displaced = 30/g
volume of water displaced = 30/g / water density
which is volume of the solid since it is totally submerged
so
mass of solid / volume of solid
= (45/g) /[(30/g)/water density]
= (45/30)*water density
so
density relative to water = 45/30 = 1.5
change in weight = 45 -15 = 30 N
so
mass of water displaced = 30/g
volume of water displaced = 30/g / water density
which is volume of the solid since it is totally submerged
so
mass of solid / volume of solid
= (45/g) /[(30/g)/water density]
= (45/30)*water density
so
density relative to water = 45/30 = 1.5
Answered by
Scott
water is 1
the solid displaced 30N of water
45N / 30N = ?
the solid displaced 30N of water
45N / 30N = ?
Answered by
Damon
No fair saying water is 1.
1000 kg/m^3
but not needed because relative requested.
1000 kg/m^3
but not needed because relative requested.
Answered by
Damon
They should not have said water, just some fluid :)
Answered by
Mohini
It is the very good explanation of my question... Thnx...
Answered by
Lukman
Thanks
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