Given F1 > F2, the net force will be in the direction of F₁, which is towards the west.

$$E_{k1} = \frac{1}{2}m(v_1)^2$$ $$E_{k1} = \frac{1}{2}m(2 \text{ m}\cdot\text{s}^{-1})^2$$ $$E_{k1} = \frac{1}{2}m(4 \text{ m}^2\cdot\text{s}^{-2})$$ $$E_{k1} = 2m$$

$$E_{k2} = \frac{1}{2}m(v_2)^2$$ $$E_{k2} = \frac{1}{2}m(8 \text{ m}\cdot\text{s}^{-1})^2$$ $$E_{k2} = \frac{1}{2}m(64 \text{ m}^2\cdot\text{s}^{-2})$$ $$E_{k2} = 32m$$

The factor of increase is the ratio of the final kinetic energy to the initial kinetic energy.

$$\text{Factor} = \frac{E_{k2}}{E_{k1}}$$ $$\text{Factor} = \frac{32m}{2m}$$ $$\text{Factor} = 16$$

The correct option is D.
D) 16

48. The pressure versus time graphs represent sound waves. We need to identify the graph that best represents a sound wave with the highest frequency and the loudest sound.

• Frequency is the number of cycles per unit time. A higher frequency means more oscillations (waves) occur within the same time interval.
• Loudness is related to the amplitude of the sound wave. A louder sound corresponds to a larger amplitude, which means a greater variation in pressure from the equilibrium.

Let's analyze the graphs:

• Graph A shows many oscillations within the given time, indicating high frequency. It also shows the largest vertical displacement (amplitude), indicating a loud sound.
• Graph B shows fewer oscillations than A (lower frequency) and a smaller amplitude than A (quieter sound).
• Graph C shows the fewest oscillations (lowest frequency) and a smaller amplitude than A (quieter sound).
• Graph D shows fewer oscillations than A (lower frequency) and a smaller amplitude than A (quieter sound).

Comparing all options, Graph A has both the highest frequency (most waves in the given time) and the largest amplitude (loudest sound).

The correct option is A.
A)