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2 years ago

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An irregularly shaped rock is placed in 50 mL of water. The new reading in the container is now 74.2 mL. If the mass of the rock

is 88.53 g, what is the density of the rock

1 answer:

cricket20 [7]2 years ago

8 0

Taking into account the definition of density, the density of the rock is 3.6583 g/mL.

<h3>Definition of density</h3>

Density is defined as the property that matter has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

The expression for the calculation of density is:

density= mass÷ volume

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

<h3>Density of the rock</h3>

In this case, you know that:

Mass= 88.53 g
Volume of the rock= Volume after place the rock in the water - Volume of water= 74.2 mL - 50 mL= 24.2 mL

Replacing in the definition of density:

density of rock= 88.53 g÷ 24.2 mL

Solving:

<u><em>density of rock= 3.6583 g/mL</em></u>

In summary, the density of the rock is 3.6583 g/mL.

Learn more about density:

brainly.com/question/952755

brainly.com/question/1462554

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Answer:

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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

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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

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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.

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$q_0=0.01166\ c$

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