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3 years ago

10

An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runner’s average speed? (Round your answer to the ne

arest tenth of a meter per second.)

2 answers:

kati45 [8]3 years ago

6 0

Answer:

8.7

Explanation:

LUCKY_DIMON [66]3 years ago

5 0

The answer that is got 8.7 . I got that because if you divide 200 by 23 you get <span>8.69565217391 and if you round that you get 8.7</span>

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A stone was dropped off a cliff and hit the ground with a speed of 136 ft/s. what is the height of the cliff? (use 32 ft/s2 for

mina [271]

<span>(v1)^2=(v0)^2-2ad; v1= final velocity; v0= initial velocity; a=gravitational constant;d=distance Since the stone was dropped we know v0=0. (v1)^2=2ad (136)^2=2(32)d d=(136)^2/(2*32) d=289ft</span>

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2 years ago

Which separation method would be most successful in separating the components of a homogeneous mixture? screening evaporation ce

ohaa [14]

Filtering the homogeneous mixture to get the materials out and boil away any water

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2 years ago

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More current makes a ________ electromagnet.<br> longer<br> stronger<br> weaker

Taya2010 [7]

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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2 years ago

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

3 0

2 years ago