Weegh merc vay > 아 this U4 -5 13:25x10 9.81) W= 11:07N de ad blais Buoyant (ig) on Welght of the displaced mercury > the
Weegh merc vay > 아 this U4 -5 13:25x10 9.81) W= 11:07N de ad blais Buoyant (ig) on Welght of the displaced mercury > the
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i don't understand the solving can someone make this understandable and good writing
Transcribed Image Text:if cn obj ect qs submerged în
Jeiquid >
The buoyont Force
on the objiect>
Whore YE Volume
displa ced
P = Densty of
9= Gravitation al ace" - 981 m/s²
the body to be
of the
Eluid
Now For
în
m =Mess of the submer ged
object
Where
by >
mg
%3D
im=P,V (Where Po - Densily of object)
V6=Volume of dhject
-4. 3
10
V, - 100 cm= 10 x t0° m3
P, = Density of lead zh kg/ m3
3
0) x םן!
In question,
weighs
3
Iem of
morcuny
0:13 N;
=> VPg - 0 13
=7 IX10xPx 9.81 = 0·13
3
=) P= 13.25 × 10° K9/m3 → Vi
by >
mg
10
(byO and O
ILL
[3:25 x103
Densty of Jead -P= |1:29 ga/com?
3
(1290 kg/m
|1290x 10
13:25 x103
-5 3
8. 52 x10
V = 85:20 cm3
Transcribed Image Text:Weege OF
morc y -)
this od m
of
W- Vp P q (8-52 x1o 13:25 0% 981) N
5x10
W= 11:07N
Buoyant
tal on the
de ad bleais
equal = Wetght of the desplaced mesiciay >
So
F = W= I1:07 N
> = 11:07N
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