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Sophie [7]
3 years ago
5

HELP I NEED THIS ANSWERED AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
WITCHER [35]3 years ago
8 0

Displacement is d  


Vf² = Vi² + 2 g d  


(-20²) = (+10²) + 2 (-9.8) d  


-19.6 d = 300  


d = -15.3 m  


negative means lower


time is t  


d = Vi t + 1/2 g t²



-15.3 = 10 t + (-4.9) t²



4.9 t² - 10 t -15.3 = 0  


t = 3.06 s

Hope this helps -John

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A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then
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Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

v^{2}-u^{2}= 2as

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

v^{2}-(0)^{2}= 2(1.44)(30)

v= 9.3 m^{-1}

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

s=ut+\frac{(1)}{2}at^2

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

30=(0)t+\frac{(1)}{2}(1.44)t^2

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

T=6.45+\frac{s'}{v}

Put s= 30 m, v= 9.3 m^{-1} and s'= 100 m.

T=6.45+\frac{(100)}{9.3}

T= 17.20 s

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Explanation:

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What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i
CaHeK987 [17]

Answer:

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Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

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