Question

A boat floating in fresh water displaces water weighing 35.0 kN.
(a) What is the weight of the water that this boat displaces when floating in salt water of density 1.10 103 kg/m3?
kN

(b) What is the difference between the volume of fresh water displaced and the volume of salt water displaced (Vfresh - Vsalt)? (Take the density of fresh water to be998 kg/m3.)
m3

Answer 1

Weight of boat = 35.6 kN
mass of boat = 35.6 kN / 9.807 m/s^2 = 3.63 kg

using density = mass / volume

you find the volume of the water displaced:

volume = mass / density
V = 3.63 kg / (1.1(10^3)) kg/m^3 = .0033 m^3

to find new weight of water displaced:

W = density * volume * gravity = 1.1(10^3) * .0033 * 9.807
weight of salt water displaced = 35 kN

B) change of volume:
need to find volume displaced in fresh water
35.6 kN= 1 kg/m^3 * V * 9.897
Vfresh = .00363 m^3

Vf - Vi = .0033 m^3 - 00363 = -.00033 m^3

the volume decreased by 10 percent

Answer 2

a) = wt of boat = 35 kN
b) = 350(1/998 - 1/1100) = 376.7 x10^3 m^3

Answer 3

1min

Answer 4

Giventhat the weight of the boat is W = 36.9 kN =36.9*103N
The density of the water is ρsalt = 1.10*103kg/m3
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(a) ArchimedesPrinciple makes it clear that a body in order to float displaces anammount of the liquid which
corresponds to weight of the body . Hence the in the salt water of density 1.10*103 kg/m3 weight of the water that this boatdisplaces is equal to the

weight W = 36.9 kN

(b) The displaced volume of the salt water
V' = W / ρsalt*g
= (36.9*103N ) / (1.10 x103kg/m3)(9.8m/s2)

= 3.42 m3

In water the displaced volume is V =W/ρg
= (36.9*103N) / (1.0 x 103kg/m3)(9.8m/s2)
= 3.76 m3

Then we get Vfresh - Vsalt = V -V' = 0.34 m3