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Fudgin [204]
3 years ago
13

The drawing shows a triple jump on a checkerboard, starting at the center of square A and ending on the center of square B. Each

side of a square measures 4.6 cm. What is the magnitude of the displacement of the colored checker during the triple jump (there is only one right corner)
Physics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

Each jump covers one center of the square to the other. If you can imagine the scenario, a single jump would cover a length of the side of one whole square. So, if it took three jumps, the total distance would be

d = 3 * 5 cm

d = 15 cm

Therefore, the magnitude or distance is 15 cm. The displacement depends on the direction of motion. 

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The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6
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Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

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Explanation:

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v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s

Average velocity is 3.93 cm/s

v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s

Root mean square velocity is 4.79 cm/s

v_p=7.68-1.4=6.28\ cm/s

Velocity peak to peak is 6.28 cm/s

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