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andrezito [222]
3 years ago
5

Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

ght near B1, the bars repel.
Which conclusion is best supported by the data?

A) A1 is the north pole of a magnet, and B1 is the south pole of a magnet.
B) A1 is the north pole of a magnet, and B1 is the north pole of a magnet.
C) A1 and B1 are opposite poles, but there is not enough information to tell which ones.
D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.
Physics
2 answers:
olga2289 [7]3 years ago
3 0

The conclusion that is best supported by the data is;

D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

Anarel [89]3 years ago
3 0

Answer:

D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

Explanation:

When A1 is brought near B1 the bars will repel each other

As we know by the properties of magnet that when two bars are of similar nature then they will repel each other

so here A1 and B1 are of same type

so correct option must be

D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

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Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0
3 years ago
A student has a displacement of 739 m north in 162 s. What was the student’s average velocity?A. 0.22 m/sB. 119,718 m/sC. 162 m/
user100 [1]

Answer:

answer below

Explanation:

Displacement of the student is 739 m due North and it takes 162 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

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v= 4.56

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A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg
iogann1982 [59]

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