### Educational Content#### Problem DescriptionDuring the processing of a polymer, it is pumped through the plant. The pump must produce enough power to move the material to a height of 10 meters at a volumetric flow rate of 0.2 cubic meters per second. The specific gravity of the polymer is 0.6262. If the pump is 75% efficient, determine the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower (hp).1. Click the icon to view the table of common derived units in the SI system.2. Click the icon to view the conversion table.3. Click the icon to view the density of water.The power rating is ___ hp. (Round your answer to one decimal place.)#### Density of Water - \( = 1 \, \text{lb/ft}^3 \)- \( = 1 \, \text{g/cm}^3 \)- \( = 1000 \, \text{kg/m}^3 \)- \( = 62.4 \, \text{lb}/\text{ft}^3 \)- \( = 1.94 \, \text{slug/ft}^3 \)#### Unit Conversion Table- **Length** - \( 1 \, \text{m} = 3.28 \, \text{ft} \) - \( 1 \, \text{km} = 0.621 \, \text{mi} \) - \( 1 \, \text{in} = 2.54 \, \text{cm} \) - \( 1 \, \text{mi} = 5,280 \, \text{ft} \) - \( 1 \, \text{yd} = 3 \, \text{ft} \)- **Force** - \( 1 \, \text{N} = 0.225 \, \text{lb}_f \)- **Power** - \( 1 \, \text{W} = 3.412 \, \text{BTU/h} \) - \( 1 \, \text{W} = 0.00134 \, \text{hp} \) - \( 1 \, \text{hp} =

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Question list ✔Question 6 K During processing, a polymer is pumped through the plant. The pump must produce enough power to move the material a height of 10 meters [m] at a volumetric flowrate of 0.2 cubic meters per second [m³/s]. Assume the specific gravity of the polymer is 0.6262. If the pump is 75% efficient, what is the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower [hp]? Click the icon to view the table of common derived units in the Sl system. Click the icon to view the conversion table. Click the icon to view density of water.

Question list ✔Question 6 K During processing, a polymer is pumped through the plant. The pump must produce enough power to move the material a height of 10 meters [m] at a volumetric flowrate of 0.2 cubic meters per second [m³/s]. Assume the specific gravity of the polymer is 0.6262. If the pump is 75% efficient, what is the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower [hp]? Click the icon to view the table of common derived units in the Sl system. Click the icon to view the conversion table. Click the icon to view density of water.

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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Question

### Educational Content

#### Problem Description
During the processing of a polymer, it is pumped through the plant. The pump must produce enough power to move the material to a height of 10 meters at a volumetric flow rate of 0.2 cubic meters per second. The specific gravity of the polymer is 0.6262. If the pump is 75% efficient, determine the power rating on the pump (input power) to supply the necessary potential energy in units of horsepower (hp).

1. Click the icon to view the table of common derived units in the SI system.
2. Click the icon to view the conversion table.
3. Click the icon to view the density of water.

The power rating is ___ hp. (Round your answer to one decimal place.)

#### Density of Water
- \( = 1 \, \text{lb/ft}^3 \)
- \( = 1 \, \text{g/cm}^3 \)
- \( = 1000 \, \text{kg/m}^3 \)
- \( = 62.4 \, \text{lb}/\text{ft}^3 \)
- \( = 1.94 \, \text{slug/ft}^3 \)

#### Unit Conversion Table
- **Length**
- \( 1 \, \text{m} = 3.28 \, \text{ft} \)
- \( 1 \, \text{km} = 0.621 \, \text{mi} \)
- \( 1 \, \text{in} = 2.54 \, \text{cm} \)
- \( 1 \, \text{mi} = 5,280 \, \text{ft} \)
- \( 1 \, \text{yd} = 3 \, \text{ft} \)

- **Force**
- \( 1 \, \text{N} = 0.225 \, \text{lb}_f \)

- **Power**
- \( 1 \, \text{W} = 3.412 \, \text{BTU/h} \)
- \( 1 \, \text{W} = 0.00134 \, \text{hp} \)
- \( 1 \, \text{hp} =

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