Question

1. a $5 \times 10^{-2}$ kg bullet has a kinetic energy of $1.6 \times 10^{3}$ j. what is the speed of the bullet?

Explanation:

Step1: Recall kinetic energy formula

The kinetic energy formula is $KE = \frac{1}{2}mv^2$, where $KE$ is kinetic energy, $m$ is mass, and $v$ is speed.

Step2: Rearrange for speed $v$

Rearrange the formula to solve for $v$:
$$v = \sqrt{\frac{2KE}{m}}$$

Step3: Substitute given values

Substitute $KE = 1.6 \times 10^3\ \text{J}$ and $m = 5 \times 10^{-2}\ \text{kg}$:
$$v = \sqrt{\frac{2 \times 1.6 \times 10^3}{5 \times 10^{-2}}}$$

Step4: Calculate numerator first

$$2 \times 1.6 \times 10^3 = 3.2 \times 10^3$$

Step5: Perform division inside root

$$\frac{3.2 \times 10^3}{5 \times 10^{-2}} = 6.4 \times 10^4$$

Step6: Compute square root

$$v = \sqrt{6.4 \times 10^4} = \sqrt{64 \times 10^3} = 80\sqrt{10} \approx 253\ \text{m/s}$$

Answer:

The speed of the bullet is approximately $\boldsymbol{253\ \text{m/s}}$ (or exactly $80\sqrt{10}\ \text{m/s}$)