A girl is floating in a freshwater lake with her head just above water.if she weighs 610N,what is the volume of the submerged part of her body (density of warer is 1000kg/m³).
Answers
Answer:
The volume of the submerged part of her body is 0.0622m^{3}0.0622m
3
Explanation:
Let's define the buoyant force acting on a submerged object.
In a submerged object acts a buoyant force which can be calculated as :
B=B= ρ.V.g
Where ''B'' is the buoyant force
Where ''ρ'' is the density of the fluid
Where ''V'' is the submerged volume of the object
Where ''g'' is the acceleration due to gravity
Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.
We can write :
We can write :
W_{girl}=BW
girl
=B (I)
Where ''W'' is weight
⇒ If we consider ρ = 1000\frac{kg}{m^{3}}1000
m
3
kg
(water density) and g=9.81\frac{m}{s^{2}}g=9.81
s
2
m
and replacing this values in the equation (I) ⇒
B=W_{girl}B=W
girl
B=610NB=610N
ρ.V.g = 610N
1000\frac{kg}{m^{3}}.V.(9.81\frac{m}{s^{2}})=610N1000
m
3
kg
.V.(9.81
s
2
m
)=610N (II)
The force unit ''N'' (Newton) is defined as
N=kg.\frac{m}{s^{2}}N=kg.
s
2
m
Using this in the equation (II) :
(9810\frac{N}{m^{3}}).V =610N(9810
m
3
N
).V=610N
V=\frac{610N}{9810\frac{N}{m^{3}}}V=
9810
m
3
N
610N
V=0.0622m^{3}V=0.0622m
3
We find that the volume of the submerged part of her body is 0.0622m^{3}0.0622m