/*! 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 19 A shooting star is a meteoroid t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A shooting star is a meteoroid that burns up when it reaches Earth's atmosphere. Many of these meteoroids are quite small. Calculate the kinetic energy of a meteoroid of mass \(5.0 \mathrm{g}\) moving at a speed of $48 \mathrm{km} / \mathrm{s}$ and compare it to the kinetic energy of a 1100 -kg car moving at \(29 \mathrm{m} / \mathrm{s}(65 \mathrm{mi} / \mathrm{h})\).

Short Answer

Expert verified
Explain your answer. Answer: The meteoroid has more kinetic energy than the car. This is because the kinetic energy of the meteoroid (5.76 * 10^9 J) is much larger than the kinetic energy of the car (463915 J), even though the meteoroid has a smaller mass. This difference is mainly due to the significantly higher speed of the meteoroid compared to the car.

Step by step solution

01

Convert units for mass and speed

The mass of the meteoroid is given in grams, so we need to convert it to kilograms. The speed of the meteoroid is in kilometers per second, so we need to convert it to meters per second. 1 g = 0.001 kg and 1 km/s = 1000 m/s. So, \(m_{meteoroid} = 5.0 \mathrm{g} * 0.001 = 0.005 \mathrm{kg}\) \(v_{meteoroid} = 48 \mathrm{km/s} * 1000 = 48000 \mathrm{m/s}\)
02

Calculate the kinetic energies for the meteoroid and the car

Now, let's plug in the values into the kinetic energy formula for both the meteoroid and the car. Meteoroid: \(KE_{meteoroid} = \frac{1}{2} * 0.005 \mathrm{kg} * (48000 \mathrm{m/s})^2\) Car: \(KE_{car} = \frac{1}{2} * 1100 \mathrm{kg} * (29 \mathrm{m/s})^2\)
03

Calculate the values for the kinetic energies

Now let's find the actual values for the kinetic energies of the meteoroid and the car: \(KE_{meteoroid} = \frac{1}{2} * 0.005 \mathrm{kg} * (48000 \mathrm{m/s})^2 = 5.76 * 10^9 \mathrm{J}\) \(KE_{car} = \frac{1}{2} * 1100 \mathrm{kg} * (29 \mathrm{m/s})^2 = 463915 \mathrm{J}\)
04

Compare the kinetic energies

Finally, let's compare the kinetic energies of the meteoroid and the car: \(KE_{meteoroid} = 5.76 * 10^9\ \mathrm{J}\) \(KE_{car} = 463915\ \mathrm{J}\) The kinetic energy of the meteoroid is much larger than the kinetic energy of the car, despite its smaller mass. This is due to the significantly higher speed at which the meteoroid is moving.

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 91Ó°ÊÓ!

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 gymnast of mass \(52 \mathrm{kg}\) is jumping on a trampoline. She jumps so that her feet reach a maximum height of \(2.5 \mathrm{m}\) above the trampoline and, when she lands, her feet stretch the trampoline down \(75 \mathrm{cm} .\) How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hooke's law when it is stretched.]

Use this method to find how the speed with which animals of similar shape can run up a hill depends on the size of the animal. Let \(L\) represent some characteristic length, such as the height or diameter of the animal. Assume that the maximum rate at which the animal can do work is proportional to the animal's surface area: \(P_{\max } \propto L^{2} .\) Set the maximum power output equal to the rate of increase of gravitational potential energy and determine how the speed \(v\) depends on \(L\).

A top fuel drag racer with a mass of \(500.0 \mathrm{kg}\) completes a quarter- mile \((402 \mathrm{m})\) drag race in a time of \(4.2 \mathrm{s}\) starting from rest. The car's final speed is \(125 \mathrm{m} / \mathrm{s}\). What is the engine's average power output? Ignore friction and air resistance.
A 402 -kg pile driver is raised 12 \(\mathrm{m}\) above ground. (a) How much work must be done to raise the pile driver? (b) How much work does gravity do on the driver as it is raised? (c) The driver is now dropped. How much work does gravity do on the driver as it falls?

A spring gun \((k=28 \mathrm{N} / \mathrm{m})\) is used to shoot a \(56-\mathrm{g}\) ball horizontally. Initially the spring is compressed by \(18 \mathrm{cm} .\) The ball loses contact with the spring and leaves the gun when the spring is still compressed by \(12 \mathrm{cm} .\) What is the speed of the ball when it hits the ground, \(1.4 \mathrm{m}\) below the spring gun?

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.