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garri49 [273]
2 years ago
8

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

ch. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s2 during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.
Physics
2 answers:
madreJ [45]2 years ago
7 0
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
xxTIMURxx [149]2 years ago
5 0

Answer:

v_f = 6.48 m/s

Explanation:

As we know that her speed while she walk is given as

v_o = 1.4 m/s

now here acceleration is given as

a = 0.20 m/s^2

now the distance moved by her

d = 100 m

now the final speed is given as

v_f^2 - v_0^2 = 2 a d

so we have

v_f^2 - 1.4^2 = 2(0.20)(100)

v_f = 6.48 m/s

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Explanation :  Given,

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Initial conc.    0.1                0.1          0

At eqm.         (0.1-2x)       (0.1+2x)      x

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Now put all the values in this expression, we get :

1.67\times 10^{-7}=\frac{(0.1+2x)^2\times (x)}{(0.1-2x)^2}

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x=1.67\times 10^{-7}M

Concentration of S_2 at equilibrium = x=1.67\times 10^{-7}M

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