Question

An object weighs 0.25N in air and
0.01N when immersed in water.
Calculate
(a) its relative density
(b) its apparent weight in a liquid of density 800kgm^-3

Answer 1

density water = 1000 kg/m^3
g = 9.81 m/s^2

Volume = V

mass = rho V

weight = rho (9.81) V = 9.81 rho V

Buoyancy = .25 - .01 = .24 N
so
.24 = 1000 (9.81)V
V = .0245 *10^-3 m^3
m = .25 N/9.81 = .0255 Kg
rho = m/V = 1.04 * 10^3
= 1040 Kg/m^3 answer (a)

new buoyancy = 800*.0245*10^-3 *9.81 = .193 N

apparent weight = .25 -.193 Newtons

=

Answer 2

it's great thank you Damon