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3 years ago

Why are metals so much better at conducting than other solids

Physics

1 answer:

Over [174]3 years ago

5 0

Answer:

Explanation:

Metals are by far better conductors of electricity than most solids because individual atoms in a metal form a matrix which their outer electrons can move freely. These atoms form a "sea" of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

Solids are usually rigid substances with their atoms in a fixed place unable to move freely.

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Which shows the correct order of forces, from weakest to strongest?

Norma-Jean [14]

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2 years ago

Answer question and I will give brainliest

8090 [49]

D. mechanical energy is transformed into sound energy.

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3 years ago

Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 2

liubo4ka [24]

Answer:

The volume of the larger cube is 5.08 g/cm³.

Explanation:

Given that,

Mass of smaller cube = 20 g

Density of smaller cube $\rho= 7.87 g/cm^2$

Dylan has two cubes of iron.

The larger cube has twice the mass of the smaller cube.

$M_{l}=2m_{s}$

Density is same for both cubes because both cubes are same material.

The density is equal to the mass divided by the volume.

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

Where, V = volume

m = mass

$\rho=density$

We need to calculate the volume of smaller mass

The volume of smaller mass

$V_{s}=\dfrac{m_{s}}{\rho_{s}}$

$V_{s}=\dfrac{20}{7.87}$

$V_{s}=2.54\ cm^3$

Now, We need to calculate the volume of large cube

$V_{l}=\dfrac{m_{l}}{\rho_{l}}$

$V_{l}=\dfrac{2\times20}{7.87}$

$V_{l}=5.08\ g/cm^3$

Hence, The volume of the larger cube is 5.08 g/cm³.

8 0

3 years ago

At the instant shown in the diagram, the car's centripetal acceleration is directed

serg [7]

Answer:

At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.

Explanation:

The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.

3 0

3 years ago

Four long wires are each carrying 6.0 A. The wires are located

Firdavs [7]

Answer:

$B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

$B=\frac{\mu_oI}{2\pi r}$

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

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:

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

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

$r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m$

Then you have:

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

5 0

3 years ago