## 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
The passage describes the calculation of the force exerted on a pipe due to the difference in water speeds at the entry and exit points. It includes the derivation of the exit speed and the forces at both points, leading to the net force on the pipe.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## 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
user
The passage describes the calculation of the force exerted on a pipe due to the difference in water speeds at the entry and exit points. It includes the derivation of the exit speed and the forces at both points, leading to the net force on the pipe.

TYPE: SINGLE_QUESTION_WITH_ANSWER