00:01
Hi there, so for this problem we have a rectangular block that is gradually push face down into a liquid, as is shown in the part a of this figure.
00:13
Now the block has a hide d on the bottom and top the face area.
00:22
That area is given and that is 5 .67.
00:32
To the squared.
00:35
And the figure b of this problem gives us the apparent weight of the block as a function of the depth age of its lower face.
00:49
Now the scale on this, on this, the vertical axis for this graph, we have that this is equal, this weight weight adds is equal to 0 .2 newtons.
01:10
So what we need to determine is the density of the liquid.
01:24
Now, from the king in the graph, it is clear that the distance d is 1 .5, this point right here, because each of these squares measure 0 .5.
01:46
0 .5 centimeters.
01:48
So we can conclude from third that the distance is equal to 1 .5 centimeters.
02:01
And also the height h point, the height at, the point at h equals to zero, makes it clear that the true weight is going to be this value right here.
02:24
Value right here if this measures 2 0 .2 then each one of this measures 0 .05 neutrons so this position in here measures 0 .25 so we know that the weight is 0 .25 5 newtons...