Glycerin flows through a tube that expands from a 1.00 cmcross-section area at point 1 to a 4.00 cm² cross-section area farther downstream at point 2. The pressure difference between points 1 and 2 is 9.45 kPa. 2 Part A What is the speed of the glycerin at point 1? Assume that the glycerin flows as an ideal fluid. Express your answer with the appropriate units. uue μΑ ? V1 = Value Units Part B What is the speed of the glycerin at point 2?Assume that the glycerin flows as an ideal fluid. Express your answer with the appropriate units. μΑ ? V2 = Value Units

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20-12-2022
Physics

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Glycerin flows through a tube that expands from a 1.00 cmcross-section area at point 1 to a 4.00 cm² cross-section area farther downstream at point 2. The pressure difference between points 1 and 2 is 9.45 kPa. 2 Part A What is the speed of the glycerin at point 1? Assume that the glycerin flows as an ideal fluid. Express your answer with the appropriate units. uue μΑ ? V1 = Value Units Part B What is the speed of the glycerin at point 2?Assume that the glycerin flows as an ideal fluid. Express your answer with the appropriate units. μΑ ? V2 = Value Units

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contexto1220 contexto1220 · Answer 1 · 2022-12-23T04:30:15-05:00

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.

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 * (cross-sectional area * 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:

V1 = constant / (density * A1) = constant /

( densit y^ * 1.00 cm^ ^ 2 At point 2, the velocity can be calculated as follows:

V2= constant /(density^ * A2)=constant/

( densit y^ * 4.00 cm^ ^ 2)We can find the value of the constant by using the pressure difference between the two points and the ideal gas law:

P 1-P2=(density^ * velocity^ ^ 2)/2

Substituting the known values, we have:

9.45 kPa=( densit y^ * Vl^ ^ 2)/2

Solving for V1, we find that the velocity at

point 1 is:

V1 = sqrt(2* 9.45 kPa / density)

Similarly, we can solve for V2:

V1=sqrt(2^ * 9.45kPa / density )

Similarly, we can solve for V2:

V * 2 = sqrt(2 * 9.45kPa / d * ensity) / 2 = V * 1/2.

Note that the density of glycerin is not giver so we cannot calculate the exact values of V1 and V2. However, it is possible to determine the relationship between velocities at two points. In particular, we can see that the velocity at point 2 is half the velocity at point 1.