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3 years ago

6 use superposition to find voltage v(t) across the 100 ohm resistor

1 answer:

6 0

It'll be my pleasure to analyze the circuit, describe my analysis in detail,
and give you a clear, precise, and accurate answer.

As soon as you let me see the circuit diagram, with values marked on
all of its components and power sources.

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Ken wants to jump and have some fun too! Barbie loans Ken her bungee cord. Is this a good idea? Explain with evidence and reason

skelet666 [1.2K]

Answer:

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4 0

3 years ago

Read 2 more answers

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

3 0

3 years ago

Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri

Romashka-Z-Leto [24]

The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

$E = \frac{Q}{4\pi \varepsilon_o r^2}$

where;

E is the electric field
Q is the charge
r is the radius

The electric field reduces by a factor of $\frac{1}{r^2}$

<h3>Surface area of a Gaussian surface;</h3>

The surface area of a sphere is given as;

$A = 4\pi r^2$

<h3>Change in area with r</h3>

$\frac{dA}{dr} = 8\pi r$

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446

7 0

2 years ago

A stone is thrown in the upwards direction at the velocity of 4 m/s. It attains a certain height and then it falls back. During

Sati [7]

Answer: 0.8 m

Explanation:

Velocity of throw = 4m/s

Maximum Height attained(h) =?

Downward acceleration experienced = 10m/s^2

Using the relation:

v^2 = u^2 + 2aS

v = final Velocity = 0 (at maximum height)

u = Initial Velocity = 4

a = g downward acceleration = - 10

0 = 4^2 + 2(-10)(S)

0 = 16 - 20S

20S = 16

S = 16 / 20

S = 0.8m

Maximum Height attained = 0.8m

3 0

3 years ago