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3 years ago

Summarize kirchoff’s laws in your own words

1 answer:

7 0

Kirchhoff's voltage law has nothing to do with gain or loss of energy by electronic components (resistors, capacitors, etc.). It is a law referring to the potential field generated by voltage sources. In this potential field, regardless of what electronic components are present, the gain or loss in "energy given by the potential field" must be zero when a charge completes a closed loop.

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Chris races his Audi north down a road for 1000 meters in 20 seconds, what is his velocity?

liq [111]

The answer is 1000/20. Or that’s what I’m guessing. Lol

5 0

4 years ago

Read 2 more answers

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

3 years ago

In electronic circuits:______.

trapecia [35]

Answer:

(a), (c) and (e) s correct.

Explanation:

a. the power used by a circuit is the resistance times the current squared.

The power is given by P = I^2 R, so the statement is correct.

b. electric and magnetic fields are transporting the energy.

false

c. electrons are transporting the energy.

The energy is transferred by flow of electrons. It is correct.

d. the power used by a circuit is the voltage times the current squared.

The power is given by P = V I, the statement is wrong.

e. the power used by a circuit is the current times the voltage.

The power is given by P = V I, the statement is correct.

4 0

3 years ago

A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy

snow_lady [41]

Answer:

The distance required = 16.97 cm

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

e = 16.97 cm

Thus the distance required = 16.97 cm

6 0

4 years ago

According to the rayleigh criterion, what is the shortest object we could resolve at the 25.0 cm near point with light of wavele

Blizzard [7]

Answer:

$6.71 *10^{-5}$ rad

Explanation:

∅ = $\frac{1.22*wavelength}{D}$ = $\frac{1.22*550 * 10^{-9} }{25 * 10^{-2} }$

∅ = $6.71 *10^{-5}$ rad

The minimum resolvable angle = $6.71 *10^{-5}$ rad

7 0

3 years ago