How Does Sucking on a Straw Affect Pressure in a Child's Mouth?

[gills]

Homework Statement

A drinking straw 20cm long and 3.0mm in diameter stands vertically in a cup of juice 8.0cm in diameter. A section of straw 6.5cm long extends above the juice. A child sucks on the straw, and the level of juice in the glass begins dropping at 20cm/s.

(a) by how much does the pressure in the child's mouth differ from atmospheric pressure?

(b) What is the greatest height from which the child could drink, assuming the same mouth pressure?

Homework Equations

P = P_o + $$\rho$$gh

p + $$\rho$$gh + $$\frac{1}{2}$$$$\rho$$v$$^{2}$$

The Attempt at a Solution

I'm just reviewing for my final exam. Am i in the right direction, and if yes, guide me a little. Thanks

 

[Shooting Star]

Equate the quantities in your 2nd expression for the values at the surface of the liquid and at the point where the liquid exits the straw, viz., the mouth. Take h as the height from some reference level, e.g., the bottom of the straw.

 

[ogb p]

!

I can provide a response to this content by analyzing the situation and applying the relevant equations. First, we can use the equation P = P_o + \rhogh to calculate the pressure inside the straw and inside the child's mouth. The atmospheric pressure (P_o) is constant and can be ignored in this case. The density of the juice (ρ) can be assumed to be the same as water, which is 1000 kg/m^3. The acceleration due to gravity (g) is 9.8 m/s^2. Using these values and the given dimensions, we can calculate the pressure inside the straw and inside the child's mouth.

(a) The pressure inside the straw can be calculated as follows:
P = \rhogh = (1000 kg/m^3)(9.8 m/s^2)(0.065 m) = 637 Pa
The pressure inside the child's mouth will be slightly lower due to the suction created by sucking on the straw. This pressure difference can be calculated using the equation p + \rhogh + \frac{1}{2}\rhov^2, where p is the atmospheric pressure and v is the velocity of the juice being sucked up the straw. Since the velocity is given as 20 cm/s, we can convert it to m/s and calculate the pressure difference as follows:
p + \rhogh + \frac{1}{2}\rhov^2 = (637 Pa) + (1000 kg/m^3)(9.8 m/s^2)(0.08 m) + (0.5)(1000 kg/m^3)(0.2 m/s)^2 = 794 Pa
Therefore, the pressure in the child's mouth is 794 Pa lower than atmospheric pressure.

(b) To determine the greatest height from which the child could drink, we need to consider the pressure difference and the maximum height the child can create using suction. The maximum suction pressure a person can create is around 100 Pa, so we can use this value to calculate the maximum height as follows:
P = \rhogh = (1000 kg/m^3)(9.8 m/s^2)(h) = 100 Pa
h = 10.2 m
Therefore, the greatest height from which the child could drink is 10.2 m, assuming the same mouth pressure. However, this is not a realistic scenario as most people cannot create