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2 years ago

How would a small bar magnet be oriented when placed at position X

Physics

2 answers:

kotykmax [81]2 years ago

6 0

Answer:

Explanation:

Well opposites attract, don't they?

Alex Ar [27]2 years ago

4 0

The answer is right here

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If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W

Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

$P=\frac{W}{t}$

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

$P=\frac{3600 J}{3.0 s}=1200 W$

5 0

3 years ago

Read 2 more answers

If an insulator replaces a conductor in an electrical circuit, the flow of electrons in the circuit will be

Degger [83]

The flow of electrons in the circuit will be less.

6 0

3 years ago

BEST ANSWER GETS BRAINLIEST!!!!!!!!!!

denis-greek [22]

Answer:

There are Microwaves, the type of electro magnetic radiation is a Micro-wave. We use x-rays, the type of electro magnetic radiation is a gamma wave. We also use radios, the type of electro magnetic radiation is a radio wave.

Explanation:

I remember doing this assignment too

8 0

2 years ago

1/012=1/0.05+1/d' hiiiiiiiiii

klasskru [66]

Correct question is;

1/0.12 = (1/0.05) + (1/d')

Answer:

d' = -1/700

Explanation:

1/0.12 = (1/0.05) + (1/d')

Let's rearrange to get;

(1/d') = (1/0.12) - (1/0.05)

(1/d') = (1/(12/100)) - (1/(5/100))

(1/d') = 100/12 - 100/5

Let's multiply through by 60 to get rid of the denominators on the right side;

> (1/d') = 500 - 1200

> (1/d') = -700

> d' = -1/700

8 0

2 years ago

A lead nucleus is spherical with a radius of about 7 ✕ 10⁻¹⁵ m. The nucleus contains 82 protons (and typically 126 neutrons). Be

Komok [63]

Answer:

∈= $3.1584x10^{26} \frac{V}{m}$

Explanation:

Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus

∈ $=\frac{P}{E_{o}}$

The P is the charge density and 'Eo' is the constant of permittivity in free space

to find P

$P=\frac{q}{V}$

$V=\frac{4}{3}*\pi*r^3$

$V=\frac{4}{3}\pi*(7x10^{-15})^3$

$V=2.932x10^{-14} m^3$

$P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}$

So replacing

∈ $=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}$

∈= $3.1584x10^{26} \frac{V}{m}$

3 0

3 years ago

How would a small bar magnet be oriented when placed at position X​

How would a small bar magnet be oriented when placed at position X