Ablock of wood floats in fresh water with 0.663 of its volume v submerged and in oil with 0.913 v submerged. find the density of (a) the wood and (b) the oil.

(a). 663kg/m³.

(b). 726.18kg/m³

Explanation:

(a).  Let V be the volume of the block of wood,

Then the submerged volume in fresh water is

V₍s₎=0.663V

Where V₍s₎= submerged volume in water.

Since the block is floating, according to Archimedes principle,

The weight of the displaced fresh water is equal to the weight of block.

∴D₍w₎V₍s₎ = DV (1)

Where D₍w₎ = Density of fresh water, D = Density of Block of wood

Since V₍s₎=0.663V, and D₍w₎=1000kg/m³.

We substitute the value of V₍s₎ and D₍w₎ in equation(1) above.

⇒ 1000×0.663V = D × V

∴ D × V = 1000 × 0.663V

Dividing both side of the equation by V

∴ D = 663kg/m³.

(b). Let V be the volume of the wood block. and V₍o₎ is the submerged volume in oil.

V₍o₎ = 0.913V

Since the wood block is floating, according to Archimedes principle,

The weight of the displaced oil is equal to the weight of the wood block.

IF D₍o₎ = Density of  oil.

D₍o₎V₍o₎  = DV(2)

D=663kg/m³, and V₍o₎ =0.913V,

Substituting the value of D and V₍o₎  in equation(2)

D₍o₎ × 0.913V = 663 × V

Dividing both side of the equation by the coefficient of D₍o₎

∴D₍o₎ × 0.913V/0.913V = 663V/0.913V

∴ D₍o₎ = 663/0.913

   D₍o₎ = 726.177

  D₍o₎ ≈ 726.18kg/m³

insects!

explanation:

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)