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3 years ago

14

Find the distance from a point charge q=100nC where the field intensity is equal to E=6kN/C. please include description of the r

esult

1 answer:

madam [21]3 years ago

4 0

Answer: 0.3872m

Explanation:

q= 100nC --> 100x10^-9 C

k= 9x10^9 Nm^2/C^2

E= 6kN/C --> 600 N/C

r=?

$E= K\frac{q}{r^{2} }$ --> $r=\sqrt{\frac{kq}{E} }$ Despejas "r"

Resuelves

<h3> $r=\sqrt{\frac{(9x10^9 Nm^2/C^2)(100x10x^{-9} C)}{6000N/C} }$ (la x es por, no es una variable)</h3><h3>r= 0.3872983346m</h3>

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C. Lucid Dreaming

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

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Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

<em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>

<em>The relationship is no longer linear and v will be more for the same value of x</em>

<em>The relationship is still linear, with lesser value of v</em>

<em>The relationship is still linear, with higher value of v</em>

<em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

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2 years ago

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Answer:

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We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

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3 years ago