Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 560

An engine pump is used to pump a liquid of density \(\rho\) continuously through a pipe of cross-sectional area \(\mathrm{A}\). If the speed of flow of the liquid in the pipe is \(\mathrm{v}\), then the rate at which kinetic energy is being imparted to the liquid is (A) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{3}\) (B) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{2}\) (C) \((1 / 2) \mathrm{A} \rho \mathrm{V}\) (B) \(\mathrm{ApV}\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{1}{2} \rho A v^3\).

Step by step solution

01

Determining the mass flow rate

In this case, the mass flow rate can be found by multiplying the density (ρ) of the liquid by the cross-sectional area (A) of the pipe and the flow velocity (v). The formula for mass flow rate is given by: Mass flow rate = ρAv
02

Applying the kinetic energy formula

Now, let's substitute the mass flow rate into the kinetic energy formula to determine the rate of imparted kinetic energy. Rate of imparted kinetic energy = (1/2)(Mass flow rate)v^2 Substitute the mass flow rate (ρAv) into the formula: Rate of imparted kinetic energy = (1/2)(ρAv)v^2
03

Simplifying the expression

Now, we'll simplify the expression: Rate of imparted kinetic energy = (1/2)ρAv^3
04

Comparing the expression with given options

After simplifying the expression, we have found that the rate of imparted kinetic energy to the liquid is (1/2)ρAv^3. Comparing this expression with the given options, we can see that the correct answer is: (A) (1 / 2) ρAV^3

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electric motor develops \(5 \mathrm{KW}\) of power. How much time will it take to lift a water of mass \(100 \mathrm{~kg}\) to a height of $20 \mathrm{~m}\( ? \)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4 \mathrm{sec}\) (B) \(5 \mathrm{sec}\) (C) \(8 \mathrm{sec}\) (D) \(10 \mathrm{sec}\)

A gun fires a bullet of mass \(40 \mathrm{~g}\) with a velocity of $50 \mathrm{~m} / \mathrm{s}$. Because of this the gun is pushed back with a velocity of \(1 \mathrm{~m} / \mathrm{s}\). The mass of the gun is (A) \(1.5 \mathrm{~kg}\) (B) \(3 \mathrm{~kg}\) (C) \(2 \mathrm{~kg}\) (D) \(2.5 \mathrm{~kg}\)

A sphere of mass \(\mathrm{m}\) moving the velocity \(\mathrm{v}\) enters a hanging bag of sand and stops. If the mass of the bag is \(\mathrm{M}\) and it is raised by height \(\mathrm{h}\), then the velocity of the sphere was (A) \([\\{\mathrm{m}+\mathrm{M}\\} / \mathrm{m}] \sqrt{(2 \mathrm{gh})}\) (B) \((\mathrm{M} / \mathrm{m}) \sqrt{(} 2 \mathrm{gh})\) (C) \([\mathrm{m} /\\{\mathrm{M}+\mathrm{m}\\}] \sqrt{(2 \mathrm{gh})}\) (D) \((\mathrm{m} / \mathrm{M}) \sqrt{(2 \mathrm{gh})}\)

A particle of mass \(0.5 \mathrm{~kg}\) travels in a straight line with velocity \(\mathrm{v}=\mathrm{ax}^{3 / 2}\), Where $\mathrm{a}=5 \mathrm{~m}^{[(-1) / 2]} \mathrm{s}^{-1}$. The work done by the net force during its displacement from \(\mathrm{x}=0\) to $\mathrm{x}=2 \mathrm{~m}$ is (A) \(50 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(25 \mathrm{~J}\) (D) None of these

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)
See all solutions

Recommended explanations on English Textbooks

Lexis and Semantics

Read Explanation

Textual Analysis

Read Explanation

Phonology

Read Explanation

Discourse

Read Explanation

Essay Plan

Read Explanation

English Grammar Summary

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free