To solve this problem, we can use the continuity equation, which states that the mass flow rate (the rate at which mass flows through a point in a system) must be constant throughout the system. In other words, the mass flow rate at point 1 must equal the mass flow rate at point 2.
The mass flow rate is equal to the density of the fluid times the flow rate (also known as the volume flow rate). The flow rate is equal to the cross-sectional area of the tube times the velocity of the fluid. Therefore, we can write the continuity equation as:
density [tex]${ }^*\left(\right.$[/tex] cross-sectional area [tex]${ }^*$[/tex] velocity )=constant
We can rearrange this equation to solve for the velocity at each point:
velocity = constant (density* cross-sectional area)
At point 1 , the velocity can be calculated as follows:
[tex]$\mathrm{V} 1=$[/tex]constant[tex]$/\left(\right.$[/tex] density[tex]$\left.{ }^* \mathrm{~A} 1\right[/tex])= constant [tex]$/\left(\right.$[/tex] density [tex]$\left.* 1.00 \mathrm{~cm}^{\wedge} 2\right)$[/tex]
At point 2, the velocity can be calculated as follows:
V 2 = constant[tex]$/($ density $* A 2)[/tex]= constant [tex]$/\left(\right.$[/tex] density [tex]$\left.* 4.00 \mathrm{~cm}^{\wedge} 2\right)$[/tex]
We can find the value of the constant by using the pressure difference between the two points and the ideal gas law:
[tex]$\mathrm{P} 1-\mathrm{P} 2=\left(\text { density }^* \text { velocity }^{\wedge} 2\right) / 2$[/tex]
Substituting the known values, we have:
[tex]$9.45 \mathrm{kPa}=\left(\text { density }{ }^* \mathrm{~V} 1^{\wedge} 2\right) / 2$[/tex]
Note that the density of glycerin is not given, so we cannot calculate the exact values of V1 and V2. However, we can still determine the relationships between the velocities at the two points. Specifically, we can see that the velocity at point 2 is half the velocity at point 1 .
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