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Chapter 26: Problem 47

An object of mass \(0.12 \mathrm{kg}\) is moving at $1.80 \times 10^{8} \mathrm{m} / \mathrm{s}$ What is its kinetic energy in joules?

Short Answer

Expert verified
Answer: The kinetic energy of the object is 1.944 x 10^15 joules.

Step by step solution

01

Write down the kinetic energy formula

Firstly, we must write down the formula we need to calculate the kinetic energy: \(K.E. = \frac{1}{2}mv^2\)
02

Plug the given values into the formula

Now that we have the formula, we can plug in the given values for mass and velocity: \(K.E. = \frac{1}{2}(0.12 kg)(1.80 \times 10^8 m/s)^2\)
03

Calculate the square of velocity

Before we proceed, we will calculate the square of the given velocity: \((1.80 \times 10^8 m/s)^2 = (1.80^2) \times (10^8)^2 = 3.24 \times 10^{16} m^2/s^2\)
04

Finish the calculation

Now, we can substitute the value calculated in the previous step into the kinetic energy formula and find the answer: \(K.E. = \frac{1}{2}(0.12 kg)(3.24 \times 10^{16} m^2/s^2)\) \(K.E. = 0.06 kg \times 3.24 \times 10^{16} m^2/s^2\) Finally, we can calculate the kinetic energy: \(K.E. = 1.944 \times 10^{15} J\) The kinetic energy of the object is \(1.944 \times 10^{15}\) joules.

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Most popular questions from this chapter

A charged particle is observed to have a total energy of \(0.638 \mathrm{MeV}\) when it is moving at \(0.600 c .\) If this particle enters a linear accelerator and its speed is increased to \(0.980 c,\) what is the new value of the particle's total energy?
A plane trip lasts \(8.0 \mathrm{h} ;\) the average speed of the plane during the flight relative to Earth is \(220 \mathrm{m} / \mathrm{s}\). What is the time difference between an atomic clock on board the plane and one on the ground, assuming they were synchronized before the flight? (Ignore general relativistic complications due to gravity and the acceleration of the plane.)
An electron has a total energy of 6.5 MeV. What is its momentum (in MeV/c)?
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An electron has momentum of magnitude $2.4 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .$ What is the electron's speed?
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