/*! 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 1 (1) Calculate the force exerted ... [FREE SOLUTION] | 91Ó°ÊÓ

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(1) Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 \(\mathrm{kg} / \mathrm{s}\) with a speed of \(4.5 \times 10^{4} \mathrm{m} / \mathrm{s}\) .

Short Answer

Expert verified
The force exerted is \(5.85 \times 10^7 \, \text{N}\).

Step by step solution

01

Understand the problem

We are given a rate at which gases are expelled from a rocket and the speed of these gases. Our task is to find the force exerted on the rocket caused by this expulsion.
02

Identify the formula

The force exerted by expelled gases can be calculated using the formula for thrust force, which is given by Newton's third law of motion: \( F = \dot{m} \times v \), where \( \dot{m} \) is the mass flow rate (in kg/s) and \( v \) is the velocity of the expelled gases (in m/s).
03

Plug in the values

Substitute the given values into the formula: \( \dot{m} = 1300 \, \text{kg/s} \) and \( v = 4.5 \times 10^4 \, \text{m/s} \). Thus, the force \( F \) is calculated as follows: \( F = 1300 \, \text{kg/s} \times 4.5 \times 10^4 \, \text{m/s} \).
04

Perform the calculation

Compute the force using the values substituted in the previous step: \( F = 1300 \times 45000 = 58500000 \, \text{N} \).
05

Conclusion

After calculating, we find that the force exerted on the rocket by the expulsion of gases is \( 58,500,000 \, \text{N} \), or \( 5.85 \times 10^7 \, \text{N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thrust Force
Thrust force is a critical concept in understanding how rockets and jet engines propel themselves. In simple terms, thrust is the force that pushes an object in a specific direction. For rockets, it's the force generated by expelling gases at high speed. This force occurs because when the rocket's engines burn fuel, they shoot out gases. These gases move backward at a high velocity, consequently pushing the rocket forward due to the force created.

Thrust force is calculated using the formula:
\[ F = \dot{m} \times v \]
  • \( \dot{m} \) is the mass flow rate, the mass of gas expelled per second in kilograms per second (kg/s).
  • \( v \) is the velocity at which the gas is expelled in meters per second (m/s).

When combined, these components quantify the force that propels the rocket. Understanding thrust force helps engineers design engines that can propel rockets effectively into space or across the atmosphere.
Mass Flow Rate
The mass flow rate is a crucial measurement in determining the thrust force a rocket can achieve. It tells us how much mass (or weight) of gas is being expelled by the rocket per second. In the context of rocketry, having a high mass flow rate can significantly impact the thrust force produced. The more gas expelled at a higher rate, the greater the thrust.

In our example, the mass flow rate is given as 1300 kg/s. This means that every second, the rocket expels 1300 kilograms of propellant gases. Knowing the mass flow rate is a vital part of understanding overall rocket dynamics, as it is directly proportional to the thrust. The relationship between the expulsion rate and the required thrust determines how much fuel the rocket needs and how long it can sustain propulsion.

Understanding this concept ensures that rockets have the right balance of speed and efficiency to meet their mission objectives.
Newton's Third Law
Newton's third law of motion is fundamental to the science of rocket propulsion. The law states: "For every action, there is an equal and opposite reaction." This principle is what makes rocket propulsion possible.

When a rocket engine expels gas, it exerts a force backward on the gas (action). According to Newton's third law, the gas exerts an equal and opposite force propelling the rocket forward (reaction). This is why the expulsion of gases results in thrust.
  • The gases expelled backward push the rocket in the opposite direction, creating motion.
  • This reaction causes the rocket to accelerate forward, demonstrating an application of Newton's laws.

Understanding this law not only helps explain why rockets travel into space but also underscores the importance of meticulously calculating the expelled mass and velocity to achieve the desired thrust and trajectory. Newton's third law is thus the cornerstone of all propulsion systems, from rockets to jet planes.

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Most popular questions from this chapter

The space shuttle launches an \(850-\) kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.0 s to give it a velocity of 0.30 \(\mathrm{m} / \mathrm{s}\) in the \(z\) -direction relative to the shuttle. The mass of the shuttle is \(92,000 \mathrm{kg}\) . (a) Determine the component of velocity \(v_{\mathrm{f}}\) of the shuttle in the minus z-direction resulting from the ejection. \((b)\) Find the average force that the shuttle exerts on the satellite during the ejection.

You have been hired as an expert witness in a court case involving an automobile accident. The accident involved car A of mass \(1500 \mathrm{~kg}\) which crashed into stationary car \(\mathrm{B}\) of mass \(1100 \mathrm{~kg} .\) The driver of car \(\mathrm{A}\) applied his brakes \(15 \mathrm{~m}\) before he skidded and crashed into car \(\mathrm{B}\). After the collision, car A slid \(18 \mathrm{~m}\) while car \(\mathrm{B}\) slid \(30 \mathrm{~m}\). The coefficient of kinetic friction between the locked wheels and the road was measured to be \(0.60 .\) Show that the driver of car A was exceeding the \(55-\mathrm{mi} / \mathrm{h}(90 \mathrm{~km} / \mathrm{h})\) speed limit before applying the brakes.

(II) Billiard ball \(\mathrm{A}\) of mass \(m_{\mathrm{A}}=0.120 \mathrm{~kg}\) moving with speed \(v_{\mathrm{A}}=2.80 \mathrm{~m} / \mathrm{s}\) strikes ball \(\mathrm{B}\), initially at rest, of mass \(m_{\mathrm{B}}=0.140 \mathrm{~kg} .\) As a result of the collision, ball \(\mathrm{A}\) is deflected off at an angle of \(30.0^{\circ}\) with a speed \(v_{\mathrm{A}}^{\prime}=2.10 \mathrm{~m} / \mathrm{s}\) (a) Taking the \(x\) axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the \(x\) and \(y\) directions separately. ( \(b\) ) Solve these equations for the speed, \(v_{\mathrm{B}}^{\prime},\) and angle, \(\theta_{\mathrm{B}}^{\prime},\) of ball B. Do not assume the collision is elastic.

(I) The \(\mathrm{CM}\) of an empty \(1250-\mathrm{kg}\) car is \(2.50 \mathrm{~m}\) behind the front of the car. How far from the front of the car will the \(\mathrm{CM}\) be when two people sit in the front seat \(2.80 \mathrm{~m}\) from the front of the car, and three people sit in the back seat \(3.90 \mathrm{~m}\) from the front? Assume that each person has a mass of \(70.0 \mathrm{~kg}\).

(II) A square uniform raft, \(18 \mathrm{~m}\) by \(18 \mathrm{~m}\), of mass \(6200 \mathrm{~kg}\), is used as a ferryboat. If three cars, each of mass \(1350 \mathrm{~kg}\), occupy the NE, SE, and SW corners, determine the CM of the loaded ferryboat relative to the center of the raft.

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