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garri49 [273]
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.
Physics
2 answers:
madreJ [45]3 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]3 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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Answer:

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Explanation:

Given;

mass of the cart = 2.5 kg

angle of inclination, β = 18.5⁰

length of inclined plane = 3.3m

Part(i) the time taken for this cart to reach the bottom of the inclined plane

s = ut + ¹/₂×at²

initial vertical velocity, u = 0

s = 3.3 m

s =  ¹/₂×at²

t = \sqrt{\frac{2s}{a} }

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane

V = a×t

V = 3.11 × 1.457 = 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane

KE = ¹/₂MV²

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Two identical metal spheres a and b are connected by a plastic rod. both are initially neutral. 1.0×1012 electrons are added to
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The plastic rod is made of insulator (plastic), so it does not allow charges moving from one sphere to another. This means that all the electrons given to sphere A will remain on sphere A.

The number of electrons initially given to sphere A is N=1.0 \cdot 10^{12}, and since the charge of 1 electron is e=-1.6 \cdot 10^{-19} C, the net charge left on sphere A after the removal of the rod will be
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Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i
LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

           v^{2} = u^{2} + 2as

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or,                 h = \frac{v^{2}}{2g}

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                   = 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c)  Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d)  Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.

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3 years ago
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