A board that is 20.0 cm wide, 5.00 cm thick, and 3.00 m long has a density 379 k
ID: 1468783 • Letter: A
Question
A board that is 20.0 cm wide, 5.00 cm thick, and 3.00 m long has a density 379 kg/m3. The board is floating partially submerged in water of density 1000 kg/m3. What fraction of the volume of the board is above the surface of the water?
The answer depends on which edge of the board is vertical.
A 3.1-kg solid sphere, made of metal whose density is 2500 kg/m3, is suspended by a cord. When the sphere is immersed in water (of density 1000 kg/m3), what is the tension in the cord?
0.379 0.200 zero 0.621The answer depends on which edge of the board is vertical.
A 3.1-kg solid sphere, made of metal whose density is 2500 kg/m3, is suspended by a cord. When the sphere is immersed in water (of density 1000 kg/m3), what is the tension in the cord?
36 N 43 N 18 N 24 N 30 NExplanation / Answer
Given that
A board that of width (w) =20.0 cm wide
The thickness of the board (t) =, 5.00 cm thick,
The length of the board (L) = 3.00 m long
Density of the board (pb) = 379 kg/m3.
The board is floating partially submerged in water of density(pw)= 1000 kg/m3.
Now we consider here that
Buoyant force =Weight
(20*10-2*3*t)*1000*9.81 =(20*10-2*5*10-2*3)*379*9.81
Then t =0.01895m
The fraction is given by 1-t/0.05
1 -0.01895/0.05 =0.621
2)
Given that
A sphere of mass (m) = 3.1-kg
A solid sphere, made of metal whose density is(ps)= 2500 kg/m3
When the sphere is immersed in water (pw) =1000 kg/m3
We know that density =mass/volume
Now the volume of the sphere is given by
V =mass/density =3.1kg/2500 kg/m3 =1.24*10-3m3
The mass of water dispalced is given by
m =density *volume =1000kg/m3*1.24*10-3m3=1.24kg
Now the weight of the sphere w =mg =3.1kg*9.81 =30.411N
and weight of sphere in water W =1.24*9.81 =12.164N
Therefore the net tension is given by 30.411-12.164 =18.247N nearly 18N
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