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____ [38]
3 years ago
7

Some of our appliances have a magnet in it.yes or no​

Physics
2 answers:
lilavasa [31]3 years ago
7 0
Yes

Magnets are used for all types of stuff that we use on a regular basis
Montano1993 [528]3 years ago
6 0

Answer:

yes !!

Explanation:

Magnets can be found in the simplest or most complex devices you use every day. From home appliances such as the refrigerator, microwave oven and electric fan, to your company's office equipment such as computers and printers. All these devices use magnets.

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If a sprinter accelerates from rest to 12 m/s north , what is their change in velocity ?
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Answer: It would be 12 m/s.

Explanation: It would be this because If you go from rest to sprint it would be 12 m/s. Also, I did this the other day.

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The center on a high school basketball team wants to increase the maximum height of his jump. What statement correctly describes
Katen [24]

Jumping with more acceleration

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In which type of collision is no kinetic energy converted to heat or sound<br> energy?
mezya [45]

Answer:

elastic collision

An elastic collision occurs when the two objects "bounce" apart when they collide. Two rubber balls are a good example. In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation.

I hope it's helpful!

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3 years ago
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.

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3 years ago
Convert the following.
Alexandra [31]

Answer:

a-

50°C whn converted into K - 323 K

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