/*! 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 11 Meteor Crater. About \(50,000\) ... [FREE SOLUTION] | 91影视

91影视

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Recent \((2005)\) measurements estimate that this meteor had a mass of about \(1.4 \times 10^{3} \mathrm{kg}\) (around \(150,000\) tons) and hit the ground at 12 \(\mathrm{km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 -megaton nuclear bomb? (A megaton bomb releases the same energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{J}\) of energy.)

Short Answer

Expert verified
The meteor delivered about \( 1.008 \times 10^{11} \) J; this is \( 2.41 \times 10^{-5} \) times the energy of a 1.0-megaton bomb.

Step by step solution

01

- Understanding the Problem

We are tasked with finding the kinetic energy of a meteor that crashed into the Earth. We must also compare this energy with that of a 1.0-megaton nuclear bomb.
02

- Formula for Kinetic Energy

Kinetic energy \( KE \) can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
03

- Calculate the Meteor's Kinetic Energy

Substitute the mass \( m = 1.4 \times 10^{3} \) kg and the velocity \( v = 12 \) km/s (or \( 12,000 \) m/s) into the formula: \[ KE = \frac{1}{2} \times 1.4 \times 10^{3} \text{ kg} \times (12,000 \text{ m/s})^2 \] \[ KE = 0.7 \times 10^{3} \times 144 \times 10^{6} = 100.8 \times 10^{9} \text{ J} \] Thus, the meteor delivered approximately \( 1.008 \times 10^{11} \) Joules of kinetic energy to the ground.
04

- Calculate Energy of a 1.0-Megaton Nuclear Bomb

A 1.0-megaton bomb releases the energy equivalent to one million tons of TNT. Since 1 ton of TNT releases \( 4.184 \times 10^{9} \) J, a megaton can be calculated as:\[ Energy = 1,000,000 \times 4.184 \times 10^{9} \text{ J} = 4.184 \times 10^{15} \text{ J} \]
05

- Compare the Energies

Compare the kinetic energy of the meteor with the energy of a 1.0-megaton bomb:\[ \frac{1.008 \times 10^{11} \text{ J}}{4.184 \times 10^{15} \text{ J}} \approx 2.41 \times 10^{-5} \] This means the meteor's kinetic energy is about \( 2.41 \times 10^{-5} \) times that of a 1.0-megaton bomb.

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!

Key Concepts

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

Meteor Impact
When a meteor hits the Earth, it creates a magnificent event known as a meteor impact. Imagine a giant rock hurtling from space at an immense speed towards our planet. This is what occurred around 50,000 years ago near what is now Flagstaff, Arizona. This particular meteor was about 1.4 x 10鲁 kilograms in mass and zoomed in at 12 kilometers per second.
The tremendous speed and mass combine to create a vast amount of energy called kinetic energy. With such a high velocity, even a relatively small object can deliver a lot of energy upon impact.
Meteor impacts are fascinating because they show us how space rocks can influence the Earth. They can cause craters, massive explosions, and even have long-lasting environmental effects. Understanding the energies involved helps scientists study these impacts more comprehensively, assessing potential risks or discovering historical events in Earth's geology.
Nuclear Energy Comparison
To appreciate the power of a meteor impact, we often compare it to nuclear energy values. Let's take a look at a one-megaton nuclear bomb, famous for its destructive capabilities. A one-megaton bomb produces energy equal to a million tons of TNT. Since 1 ton of TNT releases 4.184 x 10鈦 joules, the whole bomb releases around 4.184 x 10鹿鈦 joules.
When we carry out such comparisons, it gives us a frame of reference or a scale to comprehend the events in natural disasters, such as meteor impacts.
  • This understanding helps convey the significance of the impact energy in tangible terms.
  • It narrates how even significant natural occurrences can either dwarf or reflect the potent energies unlocked by human technology.
  • Although a meteor impact seems colossal, the meteor that struck Arizona had kinetic energy a fraction of a 1-megaton nuclear bomb.
This proportion suggests that while we equate them to man-made explosions, nature still operates at different magnitudes. Such comparisons can aid in research and preparedness strategies.
Physics Problem Solving
Physics problems, like comparing kinetic energy from a meteor impact to nuclear energy, help students apply theoretical knowledge to real-life scenarios. This problem requires using the kinetic energy formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
Here鈥檚 how you solve it:
  • Determine the meteor鈥檚 kinetic energy by inserting given values into the formula.
  • Then, compare this energy with that released by a known energy source, like a nuclear bomb.
  • Finally, understanding these principles allows for problem-solving skills application to a variety of physics-related scenarios.
Whether calculating energies, solving complex equations, or comparing results, mastering these principles enhances our understanding of the physical world. It allows future scientists or engineers to tackle questions around energy, motion, and results of impact in contexts ranging from space exploration to safety planning on Earth.

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

A moving electron has kinetic energy \(K_{1}\) . After a net amount of work \(W\) has been done on it, the electron is moving one-quarter as fast in the opposite direction. (a) Find \(W\) in terms of \(K_{1}\) . (b) Does your answer depend on the final direction of the electron's motion?

Springs in Parallel. Two springs are in parallel if they are parallel to each other and are connected at their ends (Figure 6.33\()\) . We can think of this combination as being equivalent to a single spring. The force constant of the equivalent single spring is called the effective forceconstant, \(k_{\text { eff }},\) of the combination. (a) Show that the effective force constant of this combination is \(k_{\mathrm{eff}}=k_{1}+k_{2} .\) (b) Generalize this result for \(N\) springs in parallel.

A batter hits a baseball with mass 0.145 \(\mathrm{kg}\) straight upward with an initial speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) How much work has gravity done on the baseball when it reaches a height of 20.0 \(\mathrm{m}\) above the bat? (b) Use the work-energy theorem to calculate the speed of the baseball at a height of 20.0 \(\mathrm{m}\) m above the bat. You can ignore air resistance. (c) Does the answer to part (b) depend on whether the baseball is moving upward or downward at a height of 20.0 \(\mathrm{m}\) ? Explain.

A 12 -pack of Omni-Cola (mass 4.30 \(\mathrm{kg} )\) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 \(\mathrm{m}\) by a trained dog that exerts a horizontal force with magnitude 36.0 \(\mathrm{N}\). Use the work-energy theorem to find the final speed of the 12 -pack if (a) there is no friction between the 12 -pack and the floor, and (b) the coefficient of kinetic friction between the 12 -pack and the floor is 0.30.

You are a member of an Alpine Rescue Team. You must project a box of supphes up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathbf{k}}\) . Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha .\)

See all solutions

Recommended explanations on Physics Textbooks

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.