Question

1. the acceleration of free - fall is about 10 m/s². why does the seconds unit appear twice? 19. when an object is thrown upward, how much speed does it lose each second (ignore air resistance)?

Explanation:

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity and $\Delta t$ is change in time. In SI - units, velocity $v$ has units of m/s and time $t$ has units of s. When we calculate acceleration $a=\frac{\text{m/s}}{\text{s}}=\text{m/s}^2$. So the seconds unit appears twice because acceleration is the rate of change of velocity with respect to time.

Step2: Consider free - fall acceleration

The acceleration due to gravity $g = 10\ m/s^2$ near the Earth's surface. When an object is thrown upward, the acceleration acting on it is the acceleration due to gravity, which acts downward. According to the definition of acceleration $a=\frac{\Delta v}{\Delta t}$, and since $a = g= 10\ m/s^2$ and $\Delta t = 1\ s$, we have $\Delta v=a\times\Delta t$.

Step3: Calculate the change in speed

Substituting $a = 10\ m/s^2$ and $\Delta t = 1\ s$ into $\Delta v=a\times\Delta t$, we get $\Delta v=10\ m/s$.

Answer:

1. The seconds unit appears twice in the acceleration unit ($m/s^2$) because acceleration is the rate of change of velocity with respect to time, and velocity has units of m/s and time has units of s, so when calculating the rate of change $\frac{\text{m/s}}{\text{s}}=\text{m/s}^2$.
2. It loses 10 m/s of speed each second.