/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 A jet plane is flying at 130 m/s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 86

A jet plane is flying at 130 m/s relative to the ground. There is no wind. The engines take in \(81 \mathrm{kg}\) of air per second. Hot gas (burned fuel and air) is expelled from the engines at high speed. The engines provide a forward force on the plane of magnitude \(6.0 \times 10^{4} \mathrm{N} .\) At what speed relative to the ground is the gas being expelled? [Hint: Look at the momentum change of the air taken in by the engines during a time interval \(\Delta t .\) ] This calculation is approximate since we are ignoring the \(3.0 \mathrm{kg}\) of fuel consumed and expelled with the air each second.

Short Answer

Expert verified
Answer: The speed of the hot gas being expelled from the engines relative to the ground is approximately 742 m/s.

Step by step solution

01

Write the momentum conservation equation.

We start by writing the conservation of momentum equation: \(\Delta momentum = m_{air}(v_{final} - v_{initial})\) where \(\Delta momentum\) is the change in momentum, \(m_{air}\) is the mass of the air taken in by the engines per second, \(v_{final}\) is the speed of the expelled gas relative to the ground, and \(v_{initial}\) is the initial speed of the jet plane.
02

Calculate the change in momentum.

Since we know that the engines provide a forward force of \(6.0 \times 10^{4} \mathrm{N}\), we can use the equation: \(\Delta momentum = F_{engine}\Delta t\) Where \(F_{engine}\) is the forward force of the engines, and \(\Delta t\) is the time interval. Therefore, \(\Delta momentum = m_{air}(v_{final} - v_{initial}) = 60,000 \mathrm{N}\ \Delta t\)
03

Solve for the speed of expelled gas relative to the ground.

Next, we'll solve for \(v_{final}\) in the conservation of momentum equation: \(v_{final} = \frac{6.0 \times 10^{4} \mathrm{N}\Delta t}{81\ \mathrm{kg}} +130\ \mathrm{m/s}\) We know that \(81\ \mathrm{kg}\) of air is taken in by the engines per second, so the mass flow rate of the gas remains constant. Thus, we can divide both sides of the equation by \(\Delta t\) and solve for \(v_{final}\): \(v_{final} = \frac {6.0 \times 10^{4}\ \mathrm{N}}{81\ \mathrm{kg/s}} + 130\ \mathrm{m/s}\)
04

Calculate the speed of expelled gas relative to the ground.

Now, we just plug in the numbers and calculate \(v_{final}\): \(v_{final} = \frac {6.0 \times 10^{4}\ \mathrm{N}}{81\ \mathrm{kg/s}} + 130\ \mathrm{m/s}= 741.98\ \mathrm{m/s}\) Therefore, the speed of the hot gas being expelled from the engines relative to the ground is approximately \(742\ \mathrm{m/s}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Block \(A,\) with a mass of \(220 \mathrm{g},\) is traveling north on a frictionless surface with a speed of \(5.0 \mathrm{m} / \mathrm{s} .\) Block \(\mathrm{B}\) with a mass of \(300 \mathrm{g}\) travels west on the same surface until it collides with A. After the collision, the blocks move off together with a velocity of \(3.13 \mathrm{m} / \mathrm{s}\) at an angle of \(42.5^{\circ}\) to the north of west. What was \(\mathrm{B}\) 's speed just before the collision?
A ball of mass \(5.0 \mathrm{kg}\) moving with a speed of $2.0 \mathrm{m} / \mathrm{s}\( in the \)+x$ -direction hits a wall and bounces back with the same speed in the \(-x\) -direction. What is the change of momentum of the ball?
Puck 1 sliding along the \(x\) -axis strikes stationary puck 2 of the same mass. After the elastic collision, puck 1 moves off at speed \(v_{1 f}\) in the direction \(60.0^{\circ}\) above the \(x\)-axis; puck 2 moves off at speed $v_{2 f}\( in the direction \)30.0^{\circ}\( below the \)x\(-axis. Find \)v_{2 \mathrm{f}}\( in terms of \)v_{1 f}$
A boy of mass \(60.0 \mathrm{kg}\) is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was \(8.0 \mathrm{m}\) above the net. The firefighters lower their arms as he lands in the net so that he is brought to a complete stop in a time of \(0.40 \mathrm{s}\). (a) What is his change in momentum during the 0.40 -s interval? (b) What is the impulse on the net due to the boy during the interval? [Hint: Do not ignore gravity.] (c) What is the average force on the net due to the boy during the interval?
the ball of mass \(145 \mathrm{g},\) and hits it so that the ball leaves the bat with a speed of \(37 \mathrm{m} / \mathrm{s}\). Assume that the ball is moving horizontally just before and just after the collision with the bat. (a) What is the magnitude of the change in momentum of the ball? (b) What is the impulse delivered to the ball by the bat? (c) If the bat and ball are in contact for \(3.0 \mathrm{ms}\), what is the magnitude of the average force exerted on the ball by the bat?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Energy Physics

Read Explanation

Thermodynamics

Read Explanation

Magnetism

Read Explanation

Dynamics

Read Explanation

Astrophysics

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.