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2 years ago

This is Formation of the Solar System Lab Report

Physics

1 answer:

Ainat [17]2 years ago

3 0

Answer:

the answer is B it is right

Explanation:

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An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

2 years ago

Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol

Tju [1.3M]

Answer:

$R \approx 2.418\times 10^{9}\,km$

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

$T = \frac{2\pi}{\omega}$

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

$\omega = \sqrt{\frac{a_{r}}{R} }$

Ahora se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }$

La aceleración experimentada por el planeta es:

$a_{r} = G\cdot \frac{M_{sun}}{R^{2}}$

Se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }$

La distancia del planeta con respecto al sol es finalmente despejada:

$R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}$

$R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}$

Finalmente, se sustituyen las variables y se determina la distancia:

$R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}$

$R \approx 2.418\times 10^{12}\,m$

$R \approx 2.418\times 10^{9}\,km$

4 0

3 years ago

Does it take more force to accelerate a dump truck or a car from 0 m/s/s to 10 m/s/s? Explain.

soldier1979 [14.2K]

Answer:

It takes more force to accelerate a dump truck

Explanation:

Because a dump truck has more mass so it would also have more friction and accelerate slower.

3 0

2 years ago

Read 2 more answers

What is a car's acceleration if it increases its speed from 5 m/s to 20 m/s in 3 s?

liberstina [14]

the correct answer would be C. 5m/s^2

20-5 = 15

15/3 = 5

8 0

2 years ago

Read 2 more answers

What is long straight tube that moves food by persistalsis to stomach

Katyanochek1 [597]

Esophagus.

The esophagus is a long straight tube about 10 inches long that arises from the pharynx, passes through the diaphragm and continues into the stomach. The muscular walls of the esophagus move food into the stomach by peristalsis.

3 0

2 years ago