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3 years ago

8

The rectangular region of the xy plane shown has nonuniform surface charge density σ = σ0 (1+ y b), where σ0 is a constant. Find

the net charge.

1 answer:

sergiy2304 [10]3 years ago

8 0

Answer:

This is net charge on the surface is Q = σ₀ x (y + 2by²)

Explanation:

The surface charge density is defined as the amount of charge Q per unit area A

σ = dq / dA

dq = σ dA

Since the surface is a rectangular region we use an xy coordinate system so the area difference

dA = dxdy

dq = σ dx dy

We replace, evaluate the integral

∫ dq = ∫ σ₀ (1 + yb) dxdy

realizamos laintegral de dx

Q -0 =σ₀ ∫ (1 + yb) (x-0) dy

Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0

We perform the other integral (dy)

Q = σ₀ x (y + 2y² b)

Evaluated between Y = 0 and Y = y

Q = σ₀ x (y + 2by²)

This is net charge on the surface

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The four terrestrial planets are so called because they ?

velikii [3]

Well, you need no look further than the word "terrestrial" If you notice the beginning of the word you notice that it consists mostly of the word "terra" Terra by definition is just land. Due to the solid land of these 4 planets, they're called terrestrial planets, the other 4 aren't made of land but of gas which is why they aren't classified as terrestrial planets.

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4 years ago

Read 2 more answers

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

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4 years ago

Cosmic inflation refers to ____(A) the current period of cosmic acceleration.(B) stellar winds creating bubbles of gas.(C) the U

Brilliant_brown [7]

Answer:

(D) a brief early period of hyper rapid expansion of space-time.

Explanation:

Cosmic inflation is a theory which states that in an interval of 10⁻³⁶ seconds to around 10⁻³³ seconds after the big bang there was a massive expansion.

According to the big bang theory the universe came to be from a singularity i.e., a point of infinite gravitational field. The universe is expanding this means that according to the conservation of energy the universe was expanding from the beginning.

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3 years ago

A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d

TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

$W=mgh.................(1)$

where m is the mass, g is acceleration due to gravity taken as $9.8m/s^2$ and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

$h=132sin12^o\\h=27.44m$

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

$W=54*9.8*27.44\\W=14523.55J$

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3 years ago

A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi

asambeis [7]

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

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3 years ago