/*! 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 66 A jet aircraft is traveling at \... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 66

A jet aircraft is traveling at \(223 \mathrm{~m} / \mathrm{s}\) in horizontal flight. The engine takes in air at a rate of \(80.0 \mathrm{~kg} / \mathrm{s}\) and burns fuel at a rate of \(3.00 \mathrm{~kg} / \mathrm{s}\). The exhaust gases are ejected at \(600 . \mathrm{m} / \mathrm{s}\) relative to the speed of the aircraft. Find the thrust of the jet engine.

Short Answer

Expert verified
Answer: The thrust of the jet engine is 50,369 N.

Step by step solution

01

Identify the mass flow rates

We are given the mass flow rates for the intake air \(m_{air}\) and the burning fuel \(m_{fuel}\): \(m_{air} = 80.0\,\mathrm{kg/s}\) \(m_{fuel} = 3.00\,\mathrm{kg/s}\) Now, we determine the mass flow rate of the exhaust gases, \(m_{exhaust}\), which is just the sum of the intake air and burning fuel mass flow rates: \(m_{exhaust} = m_{air} + m_{fuel}\)
02

Calculate the mass flow rate of the exhaust gases

Using the given mass flow rates, we calculate the mass flow rate of the exhaust gases: \(m_{exhaust} = 80.0\,\mathrm{kg/s} + 3.00\,\mathrm{kg/s} = 83.0\,\mathrm{kg/s}\)
03

Determine the relative exhaust gas velocity

We are given the exhaust gas velocity relative to the aircraft's velocity \(u_{relative}\): \(u_{relative} = 600\,\mathrm{m/s}\) Keep in mind that the aircraft is traveling horizontally at a velocity \(u_{aircraft}\): \(u_{aircraft} = 223\,\mathrm{m/s}\) The actual exhaust gas velocity, \(u_{exhaust}\), will be: \(u_{exhaust} = u_{aircraft} + u_{relative}\)
04

Calculate the exhaust gas velocity

Using the given aircraft and relative exhaust gas velocities, we calculate the actual exhaust gas velocity: \(u_{exhaust} = 223\,\mathrm{m/s} + 600\,\mathrm{m/s} = 823\,\mathrm{m/s}\)
05

Determine the thrust of the jet engine

The thrust of the jet engine, \(T\), is determined by applying the conservation of linear momentum principle to the engine system: \(T = m_{exhaust}\times u_{exhaust} - m_{air}\times u_{aircraft}\) Plugging in the values, we get: \(T = (83.0\,\mathrm{kg/s})(823\,\mathrm{m/s}) - (80.0\,\mathrm{kg/s})(223\,\mathrm{m/s})\)
06

Calculate and report the thrust

Performing the calculations, we find the thrust of the jet engine: \(T = 68209\,\mathrm{N} - 17840\,\mathrm{N} = 50369\,\mathrm{N}\) Thus, the thrust of the jet engine is \(50369\,\mathrm{N}\).

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Most popular questions from this chapter

A system consists of two particles. Particle 1 with mass \(2.0 \mathrm{~kg}\) is located at \((2.0 \mathrm{~m}, 6.0 \mathrm{~m})\) and has a velocity of \((4.0 \mathrm{~m} / \mathrm{s}, 2.0 \mathrm{~m} / \mathrm{s}) .\) Particle 2 with mass \(3.0 \mathrm{~kg}\) is located at \((4.0 \mathrm{~m}, 1.0 \mathrm{~m})\) and has a velocity of \((0,4.0 \mathrm{~m} / \mathrm{s})\) a) Determine the position and the velocity of the center of mass of the system. b) Sketch the position and velocity vectors for the individual particles and for the center of mass.

The center of mass of an irregular rigid object is always located a) at the geometrical center of c) both of the above the object. d) none of the above b) somewhere within the object.

A man standing on frictionless ice throws a boomerang, which returns to him. Choose the correct statement::: a) Since the momentum of the man-boomerang system is conserved, the man will come to rest holding the boomerang at the same location from which he threw it. b) It is impossible for the man to throw a boomerang in this situation. c) It is possible for the man to throw a boomerang, but because he is standing on frictionless ice when he throws it, the boomerang cannot return. d) The total momentum of the man-boomerang system is not conserved, so the man will be sliding backward holding the boomerang after he catches it.

An artillery shell is moving on a parabolic trajectory when it explodes in midair. The shell shatters into a very large number of fragments. Which of the following statements is true (select all that apply)? a) The force of the explosion will increase the momentum of the system of fragments, and so the momentum of the shell is not conserved during the explosion. b) The force of the explosion is an internal force and thus cannot alter the total momentum of the system. c) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the last fragment touches the ground. d) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the first fragment touches the ground. e) The center of mass of the system of fragments will have a trajectory that depends on the number of fragments and their velocities right after the explosion.

Two objects with masses \(m_{1}\) and \(m_{2}\) are moving along the \(x\) -axis in the positive direction with speeds \(v_{1}\) and \(v_{2}\), respectively, where \(v_{1}\) is less than \(v_{2}\). The speed of the center of mass of this system of two bodies is a) less than \(v_{1}\). b) equal to \(v_{1}\). c) equal to the average of \(v_{1}\) and \(v_{2}\). d) greater than \(v_{1}\) and less than \(v_{2}\). e) greater than \(v_{2}\).
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