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3 years ago

1. What happens when like magnetic poles are brought close together?

Physics

2 answers:

VashaNatasha [74]3 years ago

8 0

Answer:

They repel from each other

Explanation:

nata0808 [166]3 years ago

8 0

Answer:

It is also observed that, magnets attract as well as repel. We can explain this dual nature of magnetic force by proposing that each magnet has two poles, north pole (N) and south pole (S). You will observe two things during the activities:

1) When two magnets are brought near each other, like poles repel; opposite poles attract.

2) When a magnet is brought near a piece of iron, the iron also gets attracted to the magnet, and it acquires the same ability to attract other pieces of iron.

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You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-

Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

$E_P=\int dE \cos$

$E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}$

$\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ$

$\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}$

If we put an electron on point P, then force on point e is :

$\vec{F}=-|e|\vec{E_P}$

$F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}$

If r >> x , then $\frac{x^2}{r^2} \approx 0$

Then, $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0

3 years ago

The linear expansivity of metal P is twice that of another metal Q. When these materials are heated through the same temperature

kirza4 [7]

It’s p as it’s to the 19th century

5 0

3 years ago

Can you please solve this for me urgently want to make sure if my answers are correct?

MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

$\sum d_{x} = - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\ { \underline{d _{x} = - 5.702 \: m}} \\ \\ \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0 \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y} = - 11.476 \: m}}$