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3 years ago

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According to Coulomb's Law, if the distance between the nucleus and electron were doubled, the force would be__. A. 1/4 as much.

. . B. 1/2 as much.. . C. two times as much.. . D. four times as much.

1 answer:

Naya [18.7K]3 years ago

3 0

The answer is <span>A. 1/4 as much.

</span>

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

<span>where F is repulsive force, k is constant, Q is charge, r is distance between charges.</span>

k = 5.0 × 10⁹ N*m/C²

<span><span> <span> <span> <span> </span></span> </span></span> </span>

Let's take one example where:

Q1 = Q2 = 1.2 × 10⁻⁶ C²

r₁ = 0.5 m

So, F1 = k*Q1*Q2/r₁² = 5.0 × 10⁹ * 1.2 × 10⁻⁶ * 1.2 × 10⁻⁶ /0.5² = 12.96 × 10⁻²/0.25 = 51.84 × 10⁻² = 0.5184 ≈ 0.52

Let's now double r:

r₂ = 2r₁ = 2*0.5 = 1

And let's calculate force:

F2 = k*Q1*Q2/r₂² = 5.0 × 10⁹ * 1.2 × 10⁻⁶ * 1.2 × 10⁻⁶ /1² = 12.96 × 10⁻²/1 = 12.96 × 10⁻² = 0.1296 ≈ 0.13

And now, let's see how much is the second force (F2) differ from the first force (F1):
F2/F1 = 0.13/0.52 = 1/4

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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

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At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

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So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

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That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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Explanation:

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