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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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Friends who are your age and share the same interests are called

Pie

The answer is C, a peer group.

6 0

3 years ago

Read 2 more answers

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43

kiruha [24]

Answer:

The magnification is $m = 0.3674$

Explanation:

From the question we are told that

The power of the lens is $P = -4.00 D(dioptre)$

Generally $1 dioptre = 1 \ meter$

The object distance is $u = -43 \ cm$ the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

Generally the focal length is mathematically represented as

$f = \frac{1}{P}$

=> $f = \frac{1}{4.00 }$

=> $f = 0.25 \ m$

converting to cm

=> $f = 0.25 \ m = 0.25 * 100 = 25 \ cm$

Generally from lens equation we have that

$\frac{1}{f} +\frac{1}{v} -\frac{1}{u}$

=> $\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}$

=> $v = -15.8 \ cm$

Generally the magnification is mathematically represented as

$m = \frac{v}{u}$

=> $m = \frac{- 15.8}{-43}$

=> $m = 0.3674$

6 0

3 years ago

If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,

jonny [76]

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J

5 0

3 years ago

What type of energy does gasoline have? How could you relate it to the movement of a vehicle that uses it?

Licemer1 [7]

Answer:

Chemical potential energy is the energy stored in the chemical bonds of a substance. The various chemicals that make up gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car.

Explanation:

5 0

3 years ago

The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle

Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

Re = ρ v D /μ

The units of each term are

ρ = [kg / m³]

v = [m / s]

D = [m]

μ = [Pa s]

The pressure

Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

Re = [kg / m³] [m / s] [m] / [kg / m s]

Re = [kg / m s] / [m s / kg]

RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity μ = 11.2 10⁻⁶ Pa s

the density ρ = 0.656 kg / m³

D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

Re = 1.19 10⁴

4 0

3 years ago