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Natasha2012 [34]
3 years ago
13

an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity

Physics
2 answers:
MatroZZZ [7]3 years ago
6 0
  • Total displacement=150-65m=85m
  • Time=5s

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}

\\ \rm\longrightarrow Avg\:Velocity= 17m/s

ladessa [460]3 years ago
5 0

Answer:

displacement = 150 - 65 = 85m \:  \\ time = 5sec \\ velocity = displacement/ time= \frac{85}{5}  = 17  \frac{m}{s}  \: south\\ thank \: you

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A person yells across a canyon and hears an echo 2.5 seconds later. The air temperature is 21°C. How far away is the canyon wall
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Answer: The canyon wall is 850 m away from the person

Explanation:

Well, the speed of sound V in air at 21 \° C is defined as 343.60 m/s, this can be calculated by the following equation:

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Now that we know the speed of sound, we can use the following equation to find the distance between the person and the canyon wall:

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In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
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Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

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So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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