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3 years ago

Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart are the two charges?

Physics

1 answer:

77julia77 [94]3 years ago

7 0

Answer:

The separation distance between the two charges must be 82704.2925 m

Explanation:

Given:

Two negative charges that are both q = -3.8 C

Force of 19 N

Question: How far apart are the two charges, s = ?

First, you need to get the electrostatic force of this two negative charges:

$F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }$

Here

k = electric constant of the medium = 9x10⁹N m²/C²

Substituting values:

$s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m$

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4 years ago

A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic

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Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

Cross sectional area of copper $A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}$

Cross sectional area of aluminium wire

$A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\$

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

Resistance of copper $R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega$

Resistance of aluminium $R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega$

The total resistance of the wire can be obtained as follows;

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$

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3 years ago

Read 2 more answers

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3 years ago

Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact

Westkost [7]

Answer:

-5.8868501529 m/s² or -5.8868501529g

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Explanation:

t = Time taken

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Dividing by g

$\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g$

The acceleration is -5.8868501529 m/s² or -5.8868501529g

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The time taken is 0.118909090909 s

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3 years ago

Water pollution that elevates the temperature of the water

juin [17]

Answer:

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Explanation:

Lower levels of dissolved oxygen due to the inverse relationship that exists between dissolved oxygen and temperature. As the temperature of the water increases, dissolved oxygen levels decrease.

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Increase in algal blooms (Photo of algal blooms).

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Change in the abundance and spatial distribution of coastal and marine species and decline in populations of some species.

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