A piece of metal, mass 3.6kg is suspended from a spring balance. What is the reading of the balance (in Newton’s)
a) with the metal in air
b) with metal in water
c) with metal in brine and
d) what is the density of a liquid in which it gives a reading of 33N?
1
Expert's answer
2018-09-10T12:48:09-0400
a) W=mg=(3.6)(9.8)=35 N b) W=mg-ρ_w gV=mg-(ρ_w gm)/ρ_m =mg(1-ρ_w/ρ_m )=(3.6)(9.8)(1-1000/9000)=31 N c) W=mg-ρ_b gV=mg(1-ρ_b/ρ_m )=(3.6)(9.8)(1-1200/9000)=31 N d) (3.6)(9.8)(1-ρ/9000)=33 1-ρ/9000=33/(3.6)(9.8) ρ=9000(1-33/(3.6)(9.8) )=580 kg/m^3 .
Comments
Leave a comment