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3 years ago

Which of these statements articulate one of Kirchhoff's laws? (Select all that apply)

Physics

1 answer:

aleksley [76]3 years ago

8 0

Answer: E) and F)

Explanation:

E) In any closed loop, the sum of all voltage drops must be zero, as it is stated in the Kirchhoff Voltage Law (KVL).

This is a direct consequence of the energy conservation principle, as if it could be possible to return to the starting point in a loop with a different voltage, this energy could be used as a infinite energy source, which is against laws of nature, specially Thermodynamic's 2nd law.

F) At any junction, the sum of the currents into it, must be equal to the sum of the currents that leave the junction, as it is stated in the Kirchhoff"s Current Law (KCL).

This is a direct consequence of the charge conservation principle, as the charge can't neither be accumulated nor be destroyed at any point.

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Suppose (for this statement only), that q is moved from the origin but is still within both the surfaces. The flux through both

o-na [289]

Answer:

-True - True - true - false -false - false

Explanation:

True The flow depends only on the charge into the surface, not on the relative position
True The two vectors are radial, so their relative direction do not changes
True It just depends on the charge inside
False, it only depends on the charge, not on the form from the integration surface
False, because if it has a load inside it can be considered in the center, but if the load is outside the flow lines change direction with respect to the surface
False The flow depends only on the load inside, not on its position

7 0

3 years ago

Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

$F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f$

7 0

4 years ago

Which term best describes the basic unit that makes up all matter?

neonofarm [45]

Answer:

C. Atoms

Explanation:

The basic unit of matter and the smallest, indivisible unit of a chemical element. It comprises a nucleus (neutrons + protons) that is surrounded by a cloud of electrons.

6 0

2 years ago

what are the 3 properties of components of the universe that can be determined using electromagnetic radiation?

stepan [7]

Your answer is electricity, light and magnetism. They can be determined usinf elecromagnetic radioation.
<span>
Even the energy can't be detected by our eyes, there are a lot of measurement instruments that can measure infrared (IR), gamma rays, radio or X-rays or ultraviolet (UV)</span>

4 0

3 years ago

a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resi

777dan777 [17]

Answer:

<h2>d₂ = 3d</h2><h2>The diameter of the second wire is 3 times that of the initial wire.</h2>

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂ and L =L₂

Equating 1 and 2;

Rπd²/4L = (Rπd₂²)/36L₂

Rπd²/4L = (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

5 0

3 years ago