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kogti [31]
2 years ago
10

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
Physics
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

#SPJ1

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san4es73 [151]

Answer:

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Explanation:

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7 0
2 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

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}

q_0=0.01166\ c

Finally

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

i_2=5\ A

3 0
3 years ago
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​
Pani-rosa [81]

Explanation:

If the stones are unloaded from the boat, the weight of the boat will decrease. Therefore, the volume of the water displaced by the boat will also decrease. Due to this, the volume of the boat immersed in the water decreases. Hence, the level of the water around the boat will decrease.

5 0
3 years ago
Identify the breed of goat in the picture.
Step2247 [10]

Answer:

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Explanation:

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Answer:

a n c

Explanation:

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