Book edition 7th

Author(s) Douglas C. Giancoli

Pages 978 pages

ISBN 978-0321625922

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Short Answer

Astronomers estimate that a 2.0-km-diameter asteroid collides with the Earth once every million years. The collision could pose a threat to life on Earth. (a) Assume a spherical asteroid has a mass of 3200 kg for each cubic meter of volume and moves toward the Earth at \(15\;{\rm{km/s}}\). How much destructive energy could be released when it embeds itself in the Earth? (b) For comparison, a nuclear bomb could release about \(4.0 \times {10^{16}}\;{\rm{J}}\). How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

a) The destructive energy released when the asteroid embeds itself in the Earth is \(1.50 \times {10^{21}}\;{\rm{J}}\).

b) The number of equivalent nuclear bombs is 37500.

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TABLE OF CONTENTS

Identification of the given data

The diameter of the asteroid is \(d = 2\;{\rm{km}} = 2000\;{\rm{m}}\).

The density of the asteroid is \(D = 3200\;{\rm{kg/}}{{\rm{m}}^3}\).

The speed of the asteroid is \(v = 15\;{\rm{km/s}} = 15000\;{\rm{m/s}}\).

The energy released by a nuclear bomb is \({E_{{\rm{nuclear}}}} = 4 \times {10^{16}}\;{\rm{J}}\).

(a) Definition of kinetic energy

The energy possessed by a body by virtue of its motion is called kinetic energy. Mathematically, it is given by:

\(KE = \frac{1}{2}m{v^2}\) … (i)

Here, m is the mass and v is the velocity of the body.

(a) Determination of the mass of the asteroid

The mass of the asteroid is determined from the formula of density. The volume of the asteroid is given as:

\(V = \frac{4}{3}\pi {r^3}\)

Since the asteroid is 2 km wide, its radius will be \(r = \frac{d}{2} = 1000\;{\rm{m}}\).

\(V = \frac{4}{3} \times \frac{{22}}{7} \times {\left( {1000\;{\rm{m}}} \right)^3}\) … (ii)

Therefore, the mass of the asteroid is:

\(\begin{aligned}{c}{\rm{Mass}} = {\rm{Density}} \times {\rm{Volume}}\\m = \left( {3200\;{\rm{kg/}}{{\rm{m}}^3}} \right) \times \frac{4}{3} \times \frac{{22}}{7} \times {\left( {1000\;{\rm{m}}} \right)^3}\\ = 1.34 \times {10^{13}}\;{\rm{kg}}\end{aligned}\)

(a) Determination of the destructive energy released

Suppose the kinetic energy of the asteroid is the destructive energy released when it embeds itself in the Earth.

From equation (i), the kinetic energy of the asteroid is given by:

\(\begin{aligned}{c}KE = \frac{1}{2}\left( {1.34 \times {{10}^{13}}\;{\rm{kg}}} \right){\left( {15000\;{\rm{m/s}}} \right)^2}\\ = 1.50 \times {10^{21}}\;{\rm{J}}\end{aligned}\)

(b) Determination of the number of equivalent nuclear bombs

The number of equivalent nuclear bombs that have to explode simultaneously to release the destructive energy of the asteroid’s collision with the Earth is determined as:

\(\begin{aligned}{c}N = 1.50 \times {10^{21}}\;{\rm{J}} \times \frac{{1\;{\rm{bomb}}}}{{4 \times {{10}^{16}}\;{\rm{J}}}}\\ = 37500\;{\rm{bombs}}\end{aligned}\)

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Physics Principles with Applications

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Short Answer

Step by Step Solution

Identification of the given data

(a) Definition of kinetic energy

(a) Determination of the mass of the asteroid

(a) Determination of the destructive energy released

(b) Determination of the number of equivalent nuclear bombs

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A ball of mass 0.220 kg that is moving with a speed of 5.5 m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.8 m/s. Calculate

(a) the velocity of the target ball after the collision, and

(b) the mass of the target ball.

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Physics Principles with Applications

Answers without the blur.

Short Answer

Step by Step Solution

Identification of the given data

(a) Definition of kinetic energy

(a) Determination of the mass of the asteroid

(a) Determination of the destructive energy released

(b) Determination of the number of equivalent nuclear bombs

Most popular questions for Physics Textbooks

A ball of mass 0.220 kg that is moving with a speed of 5.5 m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.8 m/s. Calculate

(a) the velocity of the target ball after the collision, and

(b) the mass of the target ball.

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