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3 years ago

Which law describes the interactions between charged particles when they are not in contact?

Physics

2 answers:

natulia [17]3 years ago

7 0

The answer would be the coulombs law.

mojhsa [17]3 years ago

7 0

Answer:

B) Coulomb's law

Explanation:

As per coulombs law if two point charges are placed at some distance from each other then the force of interaction between them is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 , q_2$ = two charges between which force of interaction is required

d = distance between two charges

$k = \frac{1}{4\pi \epsilon_0}$

Faraday law is used to find the induced EMF due to changing magnetic flux

Ohm's law is used to relate potential difference between the ends of a conductor with its resistance and its current

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Firdavs [7]

Answer:

Explanation:

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2 years ago

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A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex

Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

$v_i = v_o$

now at the end of 80 s let the speed is

$v_f = v_1$

after another 120 s let the speed will be

$v_f' = v_2$

now we know that

$d = \frac{v_i + v_f}{2} (t)$

$d = \frac{v_o + v_1}{2}(80)$

$1000 = 40(v_o + v_1)$

also we know that

$v_1 - v_o = a(80)$

also we have

$1000 = \frac{v_1 + v_2}{2}(120)$

$1000 = 60(v_1 + v_2)$

now we can say

$(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}$

also we know

$v_2 - v_o = a(120 + 80)$

$-8.33 = 200 a$

$a = -0.042 m/s^2$

Part b)

now we have

$v_1 + v_o = 25$

$v_1 - v_o = (-0.042)(80)$

$v_1 = 10.83 m/s$

so the starting velocity of the trip is

$v_o = 25 - 10.83 = 14.17 m/s$

now speed after t = 200 s is given as

$v_2 = v_o + at$

$v_2 = 14.17 - (0.042)(200)$

$v_2 = 5.77 m/s$

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2 years ago

Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t

Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

$B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}$

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

$B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\ T$

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

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2 years ago

How much work will be done on a 30kg object traveling at 20 m/s from the rest?

Fofino [41]

Yes it will 90 i am ver vwery sureee!!!!!!

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2 years ago

A ball has a kinetic energy of 106 J. If the ball is moving at a velocity of 2.01 m/s, what is the mass of the ball, in kilogram

dem82 [27]

Answer:

Explanation:

KE= ½mv²

m = 2KE/v²

m = 2(106)/2.01²

m = 52.47394...

m = 52.5 kg

4 0

2 years ago