1. A pipe under your sink is filled with water. The water flows through the pipe's cross-sectional area 7.50 x 10-² m² at a velocity of 2.50 m/s. Find the water speed at point in the pipe where the cross-sectional area is 0.107 m² Express your answer in meters per second.
1. A pipe under your sink is filled with water. The water flows through the pipe's cross-sectional area 7.50 x 10-² m² at a velocity of 2.50 m/s. Find the water speed at point in the pipe where the cross-sectional area is 0.107 m² Express your answer in meters per second.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement:
A pipe under your sink is filled with water. The water flows through the pipe's cross-sectional area of \(7.50 \times 10^{-2} \, m^2\) at a velocity of \(2.50 \, m/s\).
Find the water speed at a point in the pipe where the cross-sectional area is \(0.107 \, m^2\).
Express your answer in meters per second.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa49c042c-e715-46d7-b1af-568fb3890c2a%2F4926b389-7341-472f-bc4e-e792ac4aa23c%2Fb66xhjg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A pipe under your sink is filled with water. The water flows through the pipe's cross-sectional area of \(7.50 \times 10^{-2} \, m^2\) at a velocity of \(2.50 \, m/s\).
Find the water speed at a point in the pipe where the cross-sectional area is \(0.107 \, m^2\).
Express your answer in meters per second.
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Follow-up Question
![### Problem Statement:
2) A 2600 lb truck is moving on a road at 60 mph. Find the magnitude of its momentum.
### Explanation:
To solve this problem, we need to use the formula for momentum \( p \), which is given by:
\[ p = m \cdot v \]
where
- \( p \) is the momentum,
- \( m \) is the mass of the object, and
- \( v \) is the velocity of the object.
First, convert the weight of the truck from pounds (lbs) to a mass in slugs (since 1 slug = 32.174 lb):
\[ \text{Mass of truck } (m) = \frac{2600 \text{ lbs}}{32.174 \text{ lb/slug}} \approx 80.8 \text{ slugs} \]
Next, use the given speed \( v = 60 \text{ mph} \) in the momentum formula:
\[ p = 80.8 \text{ slugs} \times 60 \text{ mph} = 4848 \text{ slug}\cdot\text{mph} \]
Therefore, the magnitude of the truck's momentum is 4848 slug·mph.](https://content.bartleby.com/qna-images/question/a49c042c-e715-46d7-b1af-568fb3890c2a/a31a461b-c784-4740-82b3-b3628ef6bd45/jy00xik_thumbnail.jpeg)
Transcribed Image Text:### Problem Statement:
2) A 2600 lb truck is moving on a road at 60 mph. Find the magnitude of its momentum.
### Explanation:
To solve this problem, we need to use the formula for momentum \( p \), which is given by:
\[ p = m \cdot v \]
where
- \( p \) is the momentum,
- \( m \) is the mass of the object, and
- \( v \) is the velocity of the object.
First, convert the weight of the truck from pounds (lbs) to a mass in slugs (since 1 slug = 32.174 lb):
\[ \text{Mass of truck } (m) = \frac{2600 \text{ lbs}}{32.174 \text{ lb/slug}} \approx 80.8 \text{ slugs} \]
Next, use the given speed \( v = 60 \text{ mph} \) in the momentum formula:
\[ p = 80.8 \text{ slugs} \times 60 \text{ mph} = 4848 \text{ slug}\cdot\text{mph} \]
Therefore, the magnitude of the truck's momentum is 4848 slug·mph.
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