**Problem Statement:**A neutron with a mass of \(1.67 \times 10^{-27}\) kg is traveling at a speed of \(8.0 \times 10^{6}\) m/s. How many Joules of kinetic energy does it have?**Note:**Your answer is assumed to be reduced to the highest power possible.**Solution:**To determine the kinetic energy (\(KE\)) of the neutron, the following formula is used:\[ KE = \frac{1}{2} mv^2 \]Where:* \( m \) is the mass of the neutron (\(1.67 \times 10^{-27}\) kg)* \( v \) is the velocity of the neutron (\(8.0 \times 10^{6}\) m/s)Plugging in the given values into the formula:\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \text{ kg} \times \left(8.0 \times 10^{6} \text{ m/s}\right)^2 \]Calculating the velocity squared:\[ \left(8.0 \times 10^{6}\right)^2 = 64 \times 10^{12} \]Here, we used the property of exponents: \( (a \times 10^b)^2 = a^2 \times 10^{2b} \).Next, we multiply this result by the mass of the neutron and divide by 2:\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 64 \times 10^{12} \]\[ KE = \frac{1}{2} \times 106.88 \times 10^{-15} \]\[ KE = 53.44 \times 10^{-15} \]Since \( 53.44 \times 10^{-15} \) can be rewritten to the highest power:\[ KE = 5.344 \times 10^{-14} \text{ Joules} \]Hence, the kinetic energy of the neutron is \( 5.344 \times 10^{-14} \) Joules.

Given: The mass of the neutron is 1.67x10-27 kg. The speed of the neutron is 8x106 m/s.

Answered: A 1.67*10^-27 kg neutron is traveling…

A 1.67*10^-27 kg neutron is traveling at (8.0x10^6) m/s. How many Joules of kinetic energy does it have? Note: Your answer is assumed to be reduced to the highest power possible.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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Question

**Problem Statement:**
A neutron with a mass of \(1.67 \times 10^{-27}\) kg is traveling at a speed of \(8.0 \times 10^{6}\) m/s. How many Joules of kinetic energy does it have?

**Note:**
Your answer is assumed to be reduced to the highest power possible.

**Solution:**

To determine the kinetic energy (\(KE\)) of the neutron, the following formula is used:

\[ KE = \frac{1}{2} mv^2 \]

Where:
* \( m \) is the mass of the neutron (\(1.67 \times 10^{-27}\) kg)
* \( v \) is the velocity of the neutron (\(8.0 \times 10^{6}\) m/s)

Plugging in the given values into the formula:

\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \text{ kg} \times \left(8.0 \times 10^{6} \text{ m/s}\right)^2 \]

Calculating the velocity squared:

\[ \left(8.0 \times 10^{6}\right)^2 = 64 \times 10^{12} \]

Here, we used the property of exponents: \( (a \times 10^b)^2 = a^2 \times 10^{2b} \).

Next, we multiply this result by the mass of the neutron and divide by 2:

\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 64 \times 10^{12} \]

\[ KE = \frac{1}{2} \times 106.88 \times 10^{-15} \]

\[ KE = 53.44 \times 10^{-15} \]

Since \( 53.44 \times 10^{-15} \) can be rewritten to the highest power:

\[ KE = 5.344 \times 10^{-14} \text{ Joules} \]

Hence, the kinetic energy of the neutron is \( 5.344 \times 10^{-14} \) Joules.

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