Question

decide how the objects in the following table would be listed, if they were listed in order of increasing kinetic energy. that is, select \1\ next to the object with the lowest kinetic energy, select \2\ next to the object with the second lowest kinetic energy, and so on.

object	kinetic energy
a 9.3 g bullet moving at 545. m/s.	select <
a 9.8 g bullet moving at 605. m/s.	select <
a 9.3 g bullet moving at 605. m/s.	select <

Explanation:

Step1: Recall kinetic - energy formula

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the velocity.

Step2: Convert mass to SI units

For all bullets, convert mass from grams to kilograms. $1\ g=0.001\ kg$. So, $m_1 = 12.3\ g = 0.0123\ kg$, $m_2=m_4 = 9.3\ g=0.0093\ kg$, $m_3 = 9.8\ g = 0.0098\ kg$.

Step3: Calculate kinetic energy for each bullet

For the first bullet with $m_1 = 0.0123\ kg$ and $v_1 = 605\ m/s$, $K_1=\frac{1}{2}\times0.0123\times(605)^{2}=\frac{1}{2}\times0.0123\times366025 = 2247.46\ J$.
For the second bullet with $m_2 = 0.0093\ kg$ and $v_2 = 545\ m/s$, $K_2=\frac{1}{2}\times0.0093\times(545)^{2}=\frac{1}{2}\times0.0093\times297025 = 1381.07\ J$.
For the third bullet with $m_3 = 0.0098\ kg$ and $v_3 = 605\ m/s$, $K_3=\frac{1}{2}\times0.0098\times(605)^{2}=\frac{1}{2}\times0.0098\times366025 = 1793.53\ J$.
For the fourth bullet with $m_4 = 0.0093\ kg$ and $v_4 = 605\ m/s$, $K_4=\frac{1}{2}\times0.0093\times(605)^{2}=\frac{1}{2}\times0.0093\times366025 = 1700.6\ J$.

Step4: Rank kinetic energies

Comparing the values: $K_2

Answer:

1. A 9.3 g bullet moving at 545 m/s