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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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I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]

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Then you have:

B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T

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