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Elanso [62]
3 years ago
7

a soccer player is running at 6m/s. he then stumbles over an opponents foot falling, and rolling to a stop. this took 4 seconds.

what was his acceleration?
Physics
2 answers:
Natali [406]3 years ago
8 0

a =  \frac{(vf - vi)}{t}  \\ a =  \frac{0 - 6}{4} \\ a =  - 1.5 \frac{m}{ {s}^{2} }
top one is the formula
mid- i plug in the numbers
bottom- answer
marishachu [46]3 years ago
5 0

Answer:

The acceleration of the soccer player is 1.5\ m/s^2 and it is decelerating.

Explanation:

Given that,

Initial speed of a soccer player, u = 6 m/s

Finally, it stops, v = 0

Time, t = 4 s

We need to find the acceleration of a soccer player. We know that the rate of change of velocity is called acceleration of an object. It is given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-6}{4}\\\\a=-1.5\ m/s^2

So, the acceleration of the soccer player is 1.5\ m/s^2 and it is decelerating.

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Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

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The distance of longitudinal wave

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From the first equation

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d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

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