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3 years ago

How are magnetic poles and electrical charges similar?

Physics

2 answers:

pshichka [43]3 years ago

7 0

Answer: The correct answer is "Their same forces repel each other, and their opposite forces attract each other".

Explanation:

Magnet has two poles: South pole and North pole.

There is magnetic force of repulsion between the same poles as the same poles repel each other. There is magnetic force of attraction between the opposite poles as the opposite poles attract each other.

The charges can be negative and positive.

There is electrostatic force of repulsion between the same charges as the same charges repel each other. There is electrostatic force of attraction between the opposite charges as the opposite charges attract each other.

Therefore, the correct option is (A).

exis [7]3 years ago

3 0

I believe the answer is a

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A cube with sides of area 22 cm2 contains a 21.2 nanoCoulomb charge. Find the flux of the electric field through the surface of

VladimirAG [237]

Answer:

Explanation:

According to Gausses theorem, electric flux coming out of charge on all sides q is equal to

q / ε₀

q is charge , ε₀ is permittivity of air whose value is given by

8.85 x 10⁻¹²; q = 21.2 x 10⁻⁹ C

Putting the values we get

FLUX = $\frac{21.2\times10^{-9}}{8.85\times10^{-12}}$

2.4 x 10³ Nm²/C

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3 years ago

What is another unit for momentum besides kg-m/s?<br> a. N<br> b. N-s<br> c. N-s2<br> d. N/s

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'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.

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3 years ago

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$

Therefore, the particle's charge-to-mass ratio is $958.1\times10^5C/kg$

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

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4 years ago

Select all of the answers that apply. Jenna dives 20meters into the ocean. How much pressure does she feel?

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4 0

4 years ago

Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of

Mars2501 [29]

Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved

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3 years ago