a collapsible plastic bag contains a gluccose solution. If the average gauge pressure in the atery is 1.33x10^4Pa, what must be the minimum height h, of the bag in order to infuse gluccose into the artery? Assume that the specific gravity of the solution is 1.02.
(1.02)(1.33x10^4)=(1x10^3kg/m^3)(9.8)(h)
h=1.38 (I don't know how to get the answer in the back of the book, 1.33m)
4 answers
the specific gravity should be on the mass side (right side) of the equation.
Gauge Pressure(P) = Density(p) x gravity(g) x height(h)
1.33x10^4Pa = 1.02* x 9.8m/s x h
*: Specific gravity = ratio of density of solution to water .:. Density = 1.02(no units)
.:. h = 1.33x10^4 / 1.02 x 9.8
= 1330.5m^3
= 1.33m
1.33x10^4Pa = 1.02* x 9.8m/s x h
*: Specific gravity = ratio of density of solution to water .:. Density = 1.02(no units)
.:. h = 1.33x10^4 / 1.02 x 9.8
= 1330.5m^3
= 1.33m
How did you find pressure...
If 1.33×10⁴Pa is for density...
And 1.02 is Specific Gravity
If 1.33×10⁴Pa is for density...
And 1.02 is Specific Gravity
1.33m
1 answer