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3 years ago

How does the electric force between two charged particles change if one particle's charge is increased by a factor of 2?

Physics

2 answers:

Burka [1]3 years ago

8 0

if one of the charge is doubled, the electric potential energy would be doubled too

Ray Of Light [21]3 years ago

5 0

C. It is increased by a factor of 2. :) APEX

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In 1-2 sentences describe how your nose cleans the air you inhale.

svp [43]

Answer: Your nose inhales O2 and goes through these little nose hairs to keep stuff that doesn't belong in your airway, it goes through your nose into your lungs taking in O2 for your blood. The other substances that you have inhaled get absorbed elsewhere and it is converted into CO2 which is what you exhale.

Explanation:

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3 years ago

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

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2 years ago

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kow [346]

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2 years ago

0/2 File Limit

slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

$v_f = 46.8 m/s$

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

$v_i = 15 m/s$

$g = - 9.81 m/s^2$

now when it will reach the ground then we have

$y = v_y t + \frac{1}{2} at^2$

so we have

$-100 = 15 t - \frac{1}{2}(-9.81) t^2$

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so we have

$t = 6.3 s$

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$v_f = v_i + at$

$v_f = -15 + (9.81)(6.3)$

$v_f = 46.8 m/s$

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3 years ago

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