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4 years ago

Which statement descThe image shows the right-hand rule being used for a current-carrying wire.

Physics

2 answers:

Nat2105 [25]4 years ago

8 0

b and heres the image fam! im on edge my self!

Makovka662 [10]4 years ago

4 0

Answer:

The second option.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.

Explanation:

The first figure that I copy here with is the figure corresponding to this question.

The thumb is pointing upward.

The rule is that the thumb aims to the direction of the flow of current and the other fingers give the field lines.

The second figure that I attach is a free image from internet and it shows the direction of both the current and the fiedl lines.

So, the conclusion is that the current goes upward the wire and the field lines go out of the paper (screen) for the points to the left of the wire and in on the right side of the wire.

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Answer:

Date: N/A

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Explanation: Mark me as brainiest hope this helps I just did it

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4 years ago

A teacher will never give a student any additional information about a test in a one-on-one meeting because it would not be fair

laila [671]

Answer:

False

Explanation:

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3 years ago

Read 2 more answers

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

The frequency changes by a factor of 0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = 0.27

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3 years ago

An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associat

-BARSIC- [3]

Answer:

either +z direction or -z direction.

Explanation:

The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.

You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction

In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.

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3 years ago

An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin

fgiga [73]

Given that,

The electric field is given by,

$\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}$