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?

	0.379
	0.200
	zero
	0.621
	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? 36 N 43 N 18 N 24 N 30 N

Answer 1

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