Arectangular wooden block of weight w floats with exactly one-half of its volume below the waterline.?
part
a. what is the buoyant force acting on the block?
a. 2w
b. w
c. 1/2 *(w)
d. the buoyant force cannot be determined. part
b. the density of water is 1.00 {g/cm^3}. what is the density of the block?
a. 2.00 {g/cm^3}
b. between 1.00 and 2.00 {g/cm^3}
c. 1.00 {g/cm^3}
d. between 0.50 and 1.00 {g/cm^3} e. 0.50 {g/cm^3} f. the density cannot be determined.

For part a:
Archimedes' principle states that the buoyancy force acting on an immersed object is equal to the weight of the fluid displaced. Because half of W displaces the water, the force must also equal half of W.
So the answer is c] (1/2) ×W.

For part b:
For an object to float entirely above water, the density of that object has to be greater or equal to the density of the water (1.00 g/cm³). So for half of it to lie below the waterline, it must be half as dense (0.5 g/cm³).
So the answer is e] 0.5 g/cm³.

Hope this helps!