text: ## Force Exerted on a Pipe Carrying Water
  
  It is possible for a water pipe to exerts a force on the user apart from the force of gravity - if the water entry points have different areas. All the water entering the pipe must exit it, but if the outlet point has a smaller area, the water must emerge faster. This increase in speed as the water exits the pipe over the speed at which the water enters the pipe produces a force pushing the pipe back. Suppose a pipe has an entry point of cross sectional area 10 cm2 = 10-3 m2 and an exit point of point of cross sectional area 1 cm2 = 10-4 m2.
  Water entering the pipe at 1 m/s.Water must entr and leave the pipe at the same rate of mass or volume per second, since water is virtually in compressible. The volume of water that enters the pipe each second is the cross sectional area of the entry point times the speed at which water enters the pipe and the rate at which water leaves at the exit point is equal to the cross sectional area at the exit point times the speed with which water exits the pipe. These entry and exit rates are equal.
  
  $A_{Entry}v_{Entry}= A_{Exit}v_{Exit} \rightarrow v_{Exit}= \frac{A_{Entry}}{A_{Exit}} v_{Entry}=\frac{10^{-3}}{10^{-4}} \times 1=10 m/s$
  
  The force exerted on the pipe at the entry point is equal to the rate of change of momentum at that point:
  \begin{aligned}F_{Entry} &= m_{Entry}v_{Entry} \\ &= ( \rho A_{Entry} v_{Entry})v_{Entry} \\ &= (1000 \times 10^{-3} \times 1) \times 1 \\ &= 10^{-2} N \end{aligned}
  .
  The force exerted on the pipe at the exit point is equal to the rate of change of momentum at that point:
  \begin{aligned}F_{Exit} &= m_{Exit}v_{Exit} \\ &= ( \rho A_{Exit} v_{Exit})v_{Exit} \\ &= (1000 \times 10^{-4} \times 10) \times 10 \\ &= 10^{-1} N \end{aligned}
  
  The net force on the pipe is the difference between these two and is equal to
  $9 \times 10^{-2} N$
  .
  
  Refresh
explanation: The document describes the calculation of the force exerted on a pipe carrying water due to the difference in cross-sectional areas at the entry and exit points. It includes detailed calculations for the speed of water at the exit point, the force exerted at the entry and exit points, and the net force on the pipe. The document does not contain a question but rather provides a complete explanation and solution to a problem.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## Force Exerted on a Pipe Carrying Water
  
  It is possible for a water pipe to exerts a force on the user apart from the force of gravity - if the water entry points have different areas. All the water entering the pipe must exit it, but if the outlet point has a smaller area, the water must emerge faster. This increase in speed as the water exits the pipe over the speed at which the water enters the pipe produces a force pushing the pipe back. Suppose a pipe has an entry point of cross sectional area 10 cm2 = 10-3 m2 and an exit point of point of cross sectional area 1 cm2 = 10-4 m2.
  Water entering the pipe at 1 m/s.Water must entr and leave the pipe at the same rate of mass or volume per second, since water is virtually in compressible. The volume of water that enters the pipe each second is the cross sectional area of the entry point times the speed at which water enters the pipe and the rate at which water leaves at the exit point is equal to the cross sectional area at the exit point times the speed with which water exits the pipe. These entry and exit rates are equal.
  
  $A_{Entry}v_{Entry}= A_{Exit}v_{Exit} \rightarrow v_{Exit}= \frac{A_{Entry}}{A_{Exit}} v_{Entry}=\frac{10^{-3}}{10^{-4}} \times 1=10 m/s$
  
  The force exerted on the pipe at the entry point is equal to the rate of change of momentum at that point:
  \begin{aligned}F_{Entry} &= m_{Entry}v_{Entry} \\ &= ( \rho A_{Entry} v_{Entry})v_{Entry} \\ &= (1000 \times 10^{-3} \times 1) \times 1 \\ &= 10^{-2} N \end{aligned}
  .
  The force exerted on the pipe at the exit point is equal to the rate of change of momentum at that point:
  \begin{aligned}F_{Exit} &= m_{Exit}v_{Exit} \\ &= ( \rho A_{Exit} v_{Exit})v_{Exit} \\ &= (1000 \times 10^{-4} \times 10) \times 10 \\ &= 10^{-1} N \end{aligned}
  
  The net force on the pipe is the difference between these two and is equal to
  $9 \times 10^{-2} N$
  .
  
  Refresh
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document describes the calculation of the force exerted on a pipe carrying water due to the difference in cross-sectional areas at the entry and exit points. It includes detailed calculations for the speed of water at the exit point, the force exerted at the entry and exit points, and the net force on the pipe. The document does not contain a question but rather provides a complete explanation and solution to a problem.
  
  Label: no-question