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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.

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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Artyom0805 [142]

Answer:

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Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

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EARTH SCIENCE PLEASE ANSWER

Rzqust [24]

The Earth Science answers are shown below.

Explanation:

1. The movement of the sun will change the angle it has on the sky in 30 minutes, it is always moving from the east to the west, so in 30 minutes it would move more west, no matter at what time you make the experiment. From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles

2. No, both marks are the same distance from the ground. the amount of stick above the mark will not affect the distance that the shadow of the mark moves at all. The Sun's clockwise motion is an apparent motion caused by the rotation of the Earth. The counterclockwise rotation of the Earth in the Sun's light causes the shadow of the gnomon to move clockwise. As the Sun appears to move higher above the horizon before solar noon, the shadow grows shorter and shorter.

3. In the summer the shadows are shorter, and in the winter the shadows are longer. In the morning your shadow will point west and in the afternoon it will point east. If your shadow is long, it is near sunrise or sunset. Your shadow is shortest around noon.

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The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

So

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

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fgiga [73]

Your order is correct here.

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3 years ago

Read 2 more answers

A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonis

icang [17]

Answer:

t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

$y = u t + \dfrac{1}{2}at^2$

$-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2$

$4.9 t^2 - 2.51 t - 3.16 = 0$

using quadratic formula to solve the equation

$t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}$

t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

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3 years ago