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)

the specific gravity should be on the mass side (right side) of the equation.

1.33m

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

How did you find pressure...

If 1.33×10⁴Pa is for density...
And 1.02 is Specific Gravity

Well, it seems like you have a "pressuring" question there! But don't worry, I'm here to help... with humor!

So, you have a collapsible plastic bag filled with glucose solution, and you want to know the minimum height of the bag to infuse the glucose into the artery. Now, let's break this down.

First, we have the average gauge pressure in the artery, which is 1.33x10^4 Pa. Now, let's convert that into a language we all understand - humor. That's enough pressure to make a clown cry!

Now, we need to consider the specific gravity of the glucose solution, which is given as 1.02. This means the glucose solution is a bit denser than water. Maybe it's been on a high-calorie diet?

Next, we have to account for gravity itself, which is always pulling things down. Just like a clown and his pants that are too big! So we have 9.8 m/s^2.

Now, let's put all of this into a comedic equation:

(1.02)(1.33x10^4) = (1x10^3)(9.8)(h)

Now, by solving this equation, you should get a minimum height h of about 1.38 meters. So it seems like you're really close to the answer in the back of the book, which is 1.33 meters.

But hey, remember that humor and calculations are like clowns and funny hats – sometimes things don't fit perfectly. So let's give you a round of applause for getting pretty close! Just remember, if all else fails, make sure you have a clown on hand to lighten the mood.

To find the minimum height "h" of the bag, we'll use the given information and the concept of gauge pressure.

First, let's understand the given equation: (specific gravity) * (gauge pressure) = (density of the liquid) * (acceleration due to gravity) * (h)

In this case, the specific gravity of the solution is given as 1.02, the gauge pressure is given as 1.33x10^4 Pa, the density of water is 1x10^3 kg/m^3, and the acceleration due to gravity is 9.8 m/s^2. We need to solve for "h".

Starting with the equation:
(1.02)(1.33x10^4) = (1x10^3)(9.8)(h)

Now, let's solve for "h":
h = (1.02)(1.33x10^4) / ((1x10^3)(9.8))

Using a calculator, h ≈ 1.3825 ≈ 1.38 m (rounded to two decimal places).

The answer you obtained, 1.38m, seems quite close to the answer provided in the back of the book, 1.33m. It is possible that the discrepancy may be due to rounding errors or slight variations in the given values.