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3 years ago

Two students below described a part of Earth's innermost layer:

2 answers:

6 0

The answer to your question is B- They both described the inner part otherwise known as the inner core.

The inner core is hotter than the surface of the sun. The inner core is made out of iron and nickel. The inner core is extremely hot and is the last layer. It is a solid because of all the pressure from the other layers putting there weight onto this layer.

expeople1 [14]3 years ago

5 0

Answer:B

Explanation: because the inner part has the highest pressures and its actually dance inside

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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

3 years ago

Which statement is true for a sound wave entering an area of warmer air?

kherson [118]

The appropriate response is letter D. The wave ventures slower and with an expanded wavelength when a sound wave entering a range of hotter air. Hotter air implies less thick, so the wave ought to back off.

I hope the answer will help you, feel free to ask more in brainly.

4 0

3 years ago

Read 2 more answers

Which is correct? I need to know!!

Mnenie [13.5K]

What website are you using?

6 0

3 years ago

An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False

Umnica [9.8K]

Hi There! :)

An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P

7 0