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enot [183]
3 years ago
9

This position-time graph describes an object's motion. Use it to determine the distance the object moves (in m) during the 10.0

seconds of motion.

Physics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

  27

Explanation:

It moves 13 meters in the first two seconds, backtracks 9 meters in the next 4 seconds, then moves forward 5 meters in the last 4 seconds. The total distance traveled is ...

  13 m + 9 m + 5 m = 27 m

_____

In the first interval, the object moves from 1 m to 14 m away from the position reference, hence a move of 13 meters.

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its
Savatey [412]

Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

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Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

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c).

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3 years ago
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