Question

directions: select the correct speed. an archery range calculated the average speed of wooden arrows shot during a competition. which is the best representation of the average speed? 5.6×10⁻³ mm/sec, 5.6×10³ mm/hr, 5.6×10⁻³ mm/hr, 5.6×10³ mm/sec

Explanation:

Step1: Analyze Units for Speed

Speed of an arrow should be a reasonable magnitude. Let's check units:

• \(5.6 \times 10^{-3} \frac{\text{mm}}{\text{sec}}\): This is very slow (0.0056 mm per second), unrealistic for an arrow.
• \(5.6 \times 10^{3} \frac{\text{mm}}{\text{hr}}\): Convert to mm/s: \(5.6 \times 10^{3} \frac{\text{mm}}{\text{hr}} \div 3600 \frac{\text{sec}}{\text{hr}} \approx 1.56 \frac{\text{mm}}{\text{sec}}\), still too slow.
• \(5.6 \times 10^{3} \frac{\text{mm}}{\text{sec}}\): This is 5600 mm per second (or 5.6 meters per second), a reasonable speed for an arrow. Wait, no—wait, the labels: Wait, the two with "sec" at the bottom: let's re - check. Wait, the top and bottom have \(\frac{\text{mm}}{\text{sec}}\) with \(10^{-3}\) and \(10^{3}\)? Wait, no, the bottom one (the lower label) is \(5.6 \times 10^{3} \frac{\text{mm}}{\text{sec}}\)? Wait, no, looking at the image: the top and bottom labels (vertical) are \(\frac{\text{mm}}{\text{sec}}\), and left/right (horizontal) are \(\frac{\text{mm}}{\text{hr}}\). Wait, the correct speed for an arrow: let's think about units. Arrows move fast, so seconds are a better unit than hours. Now, \(5.6\times 10^{-3}\) mm/sec is 0.0056 mm per second, which is extremely slow. \(5.6\times 10^{3}\) mm/sec is 5600 mm per second = 5.6 meters per second, which is a reasonable speed for an arrow (since 1 m/s is about 2.237 mph, so 5.6 m/s is about 12.4 mph, which is slow for an arrow? Wait, maybe I messed up. Wait, no—maybe the exponents: Wait, the bottom label (the one at the bottom of the target) is \(5.6\times 10^{3}\frac{\text{mm}}{\text{sec}}\)? Wait, no, the user's image: "5.6 x 10^3 mm/sec" (the bottom one) and "5.6 x 10^-3 mm/sec" (the top one). The horizontal ones are \(5.6\times 10^{3}\frac{\text{mm}}{\text{hr}}\) (left and right). So we need to pick the reasonable one. Arrows move at speeds where seconds are the appropriate time unit, and the magnitude should be reasonable. \(5.6\times 10^{3}\) mm/sec is 5.6 meters per second (since 1000 mm = 1 m), which is a plausible speed (maybe a slow arrow, but compared to the other options: \(5.6\times 10^{-3}\) mm/sec is 0.0056 mm per second, which is like a snail's pace. The \(\frac{\text{mm}}{\text{hr}}\) are way too slow (since 1 hour is 3600 seconds, so \(\frac{\text{mm}}{\text{hr}}\) is much smaller than \(\frac{\text{mm}}{\text{sec}}\) for the same coefficient magnitude). So the correct one is the bottom label (the lower vertical label) with \(5.6\times 10^{3}\frac{\text{mm}}{\text{sec}}\)? Wait, no, wait the image: the top vertical label is \(5.6\times 10^{-3}\frac{\text{mm}}{\text{sec}}\), the bottom vertical label is \(5.6\times 10^{3}\frac{\text{mm}}{\text{sec}}\), and the horizontal ones are \(5.6\times 10^{3}\frac{\text{mm}}{\text{hr}}\). So the correct speed should be the one with \(\frac{\text{mm}}{\text{sec}}\) and a reasonable magnitude. Since \(10^{-3}\) is too small, \(10^{3}\) is reasonable. So the bottom label (the lower of the two vertical labels) which is \(5.6\times 10^{3}\frac{\text{mm}}{\text{sec}}\) is the correct one.

Step2: Confirm the Correct Label

Looking at the target, the two labels with \(\frac{\text{mm}}{\text{sec}}\) are the top ( \(5.6\times 10^{-3}\)) and bottom ( \(5.6\times 10^{3}\)). The horizontal labels are \(\frac{\text{mm}}{\text{hr}}\) which are too slow. Since an arrow's speed should be measured in mm per second (or m/s, km/h, etc.) with a reasonable magnitude, \(5.6\times 10^{3}\frac{\text{mm}}{\text{sec}}\) (the bottom vertical label) is the correct representation.

Answer:

The bottom label (the one with \(5.6 \times 10^{3} \frac{\text{mm}}{\text{sec}}\))