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3 years ago

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The electrostatic attractions in crystalline salts, such as sodium chloride, are formed by __________ between positive and negat

ive ions. A) sharing electrons B) a sea of electrons C) dipole attractions D) transferring electrons

2 answers:

nikklg [1K]3 years ago

8 0

D transferring electrons because that causes electricity

mamaluj [8]3 years ago

7 0

<h3><u>Answer;</u></h3>

<em>D) transferring electrons</em>

The electrostatic attractions in crystalline salts, such as sodium chloride, are formed by<em><u> </u></em><em><u>transfer of electrons</u></em> between positive and negative ions.

<h3><u>Explanation;</u></h3>

<u><em>Crystalline salts such as sodium chloride are formed as a result of ionic bonds. </em></u>Ionic bonds are types of bonds that are formed between non metals and non metals as a result of transfer of electrons.
<em><u>Transfer of electrons occurs when one atom involved in ionic bond formation looses electrons and the other gains electrons.</u></em> The result is that both atoms form ions, where one is a positively charged ion (cation) while the other is a negatively charged ion (anion). T<em><u>he attraction between the cation and anion results to the formation of ionic bond which results to formation of crystalline structures such as sodium chloride.</u></em>

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So, the fundamental frequency will halve.

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In this case, the mass per unit length is doubled:

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Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}$

So, the fundamental frequency will decrease by a factor $\sqrt{2}$ .

(c) the fundamental frequency will increase by a factor $\sqrt{2}$

In this case, the tension is doubled:

$T'=2T$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f$

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3 years ago

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Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

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$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

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3 years ago