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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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Consider the following three statements: (i) For any electro-magnetic radiation, the product of the wavelength and the frequency
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Answer:

A and B

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Wavelength=\frac{c}{Frequency}

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Also, 1 m = 3\times 10^{-10} Å

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<u>Wavelength = 3.0 Å</u>

<u>Option B is correct.</u>

As stated above, the speed of electromagnetic radiation is constant. Hence, each radiation of the spectrum travels with same speed.

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a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
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The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

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a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

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learn more about acceleration here :

brainly.com/question/11560829

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8 0
1 year ago
A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.
wolverine [178]

Answer:

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(B) We know that work done is given by

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