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3 years ago

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Objects 1 and 2 attract each other with an electrostatic force of 18 units. If the charge of Object 2 is increased by a factor o

f 3 AND the distance separating Objects 1 and 2 is increased by a factor of 3, then the new electrostatic force will be _____ units.

1 answer:

kap26 [50]3 years ago

6 0

Answer:

$F'=18/3=6\ units$

Explanation:

<u>Electrostatic Force </u>

We know particles 1 and 2 attract each other with an electrostatic force of 18 units. The force between two charged particles is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where $q_1,\ q_2$ are the charges and r is the distance between them.

We are told the new charge 2 is

$q_2'=3q_2$

And the distance is three times the original

$d'=3d$

The new force is

$\displaystyle F'=\frac{k\ q_1\ (3q_2)}{(3r)^2}$

Operating

$\displaystyle F'=\frac{3k\ q_1\ q_2}{9r^2}$

$\displaystyle F'=\frac{k\ q_1\ q_2}{3r^2}$

This force is one third of the other one, so

$\boxed{F'=18/3=6\ units}$

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To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

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Replacing we have to

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

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4 years ago

A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

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Using formula for length

$\Delta L=L_{2}-L_{1}$

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$\Delta L=16.1\ m$

For an open-open tube,

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Put the value into the formula

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$f=1.065\ kHz$

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Answer:

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The impedance of the circuit is given by :

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