Alarge rectangular raft has a density of 725.5 kg/m3) and floating on a lake. the surface area of the top of the raft is 39.6 m 2 and its volume is 5.2 m3. assume acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. how many meters "h" is the raft above the water?

h=0.036 m

Explanation:

First we calculate the volume of the raft :

Volume of the raft = raft surface x A

where A is the dimension of the raft that we don't know but we know that A= h+d where d is the submerged edge from the raft and h the distance asked

Volume of the raft = raft surface x A

5.2 m3=39.6 m2 x A

A = (13/99) m

Now let's calculate de weight of the raft :

ρ=m/v

ρ(raft)= m(raft)/v(raft)

725.5 kg/m3= m(raft)/5.2 m3

m(raft)=3772.6 kg

W=m x g

W (raft)=m(raft) x g

W = 3772.6 kg x 9.8 m/s2

W= 36971.48 N

The raft is floating so W(raft) =   water thrust

Water thrust = water volume x water density x gravity

36971.48 N = (39.6 m2 x d) x (1000 kg/m3) x 9.8 m/s2

d = 0.0953 m

Finally A=h+d

h= A - d

h = (13/99)m - 0.0953 m

h = 0.036 m

the awnser is d you right bro