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dlinn [17]
3 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. m/s
Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

v^2-u^2=2as

Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

v^2-1.96=40

v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

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7 0
3 years ago
Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horiz
steposvetlana [31]

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

v_x = 3.80 m/s

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

v_y = u_y + gt

where

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t is the time

Therefore, after t = 1.00 s, the vertical speed is

v_y = 0 + (9.8)(1.00)=9.8 m/s

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

v=\sqrt{v_x^2 +v_y^2}=\sqrt{(3.8)^2+(9.8)^2}=10.5 m/s

b) \theta = -68.8^{\circ}

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

\theta = tan^{-1} (\frac{v_y}{v_x})=tan^{-1} (\frac{-9.8}{3.8})=-68.8^{\circ}

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so v_x has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so v_y is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

\theta = -68.8^{\circ}

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3 years ago
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Answer:

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Explanation:

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Answer:

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