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mestny [16]
3 years ago
7

A charge of 145 µC is at the center of a cube of edge 60.0 cm. No other charges are nearby. (a) Find the flux through each face

of the cube. N · m2/C (b) Find the flux through the whole surface of the cube. N · m2/C (c) Would your answers to parts (a) or (b) change if the charge were not at the center? (a) would change. (b) would change. Both (a) and (b) would change. Neither would change.
Physics
1 answer:
IrinaK [193]3 years ago
8 0

Explanation:

Given that,

Charge, q=145\ \mu C

Edge of the cube, a=60\ cm=0.6\ m

(a) The Gauss law gives the relation between the charge and the electric field. Mathematically it is given by :

\phi=\dfrac{q_{encl}}{\epsilon_o}

As there are 6 faces of a cube. It will share equal amount of flux. So,

\phi=\dfrac{q_{encl}}{6\epsilon_o}

\phi=\dfrac{145\times 10^{-6}}{6\times 8.85\times 10^{-12}}

\phi=2.73\times 10^6\ Nm^2/C

(b) For the whole surface, the flux is given by :

\phi=\dfrac{q_{encl}}{\epsilon_o}

\phi=\dfrac{145\times 10^{-6}}{8.85\times 10^{-12}}

\phi=1.63\times 10^7\ Nm^2/C

(c) If the charge is not present at the center of the cube, it  will present on another faces. As a result, the net charge doesn't change. So, the electric flux in part (b) doesn't change.

Also, if the charge is not at the center, the distribution of electric field is not uniform in each faces. Hence, the flux in part (a) changes.

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a) The x-coordinate of the center of gravity is \frac{19}{18}\cdot a.

b) The y-coordinate of the center of gravity is \frac{19}{18}\cdot a.

<h3>Determination of the coordinates of the center of gravity</h3>

Let suppose that each square has an <em>uniform</em> density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

\vec r_{1} = (0.5\cdot a, 0.5\cdot a)

\vec r_{2} = (1.5\cdot a, 0.5\cdot a)

\vec r_{3} = (0.5\cdot a, 1.5\cdot a)

\vec r_{4} = (1.5\cdot a, 1.5\cdot a)

The center of gravity of the <em>entire</em> system is found by applying definition of <em>weighted</em> averages:

\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}

Where W_{1}, W_{2}, W_{3} and W_{4} are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}

\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)

\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a  \right) \blacksquare

a) The x-coordinate of the center of gravity is \frac{19}{18}\cdot a. \blacksquare

b) The y-coordinate of the center of gravity is \frac{19}{18}\cdot a. \blacksquare

To learn more on center of gravity, we kindly invite to check this verified question: brainly.com/question/20662119

6 0
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A projectile's horizontal range on level ground is r=v20sin2θ/g. at what launch angle or angles will the projectile land at half
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As given the formula of range is

R = \frac{v^2sin2\theta}{g}

now for the maximum range

\theta = 45

R_{max} = \frac{v^2}{g}

now for half of this maximum range we will have

\frac{R_{max}}{2} = \frac{v^2sin2\theta}{g}

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now from above we have

sin2\theta = \frac{1}{2}

so two possible values for above is given as

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d_1 = 2d

Area of object 2, A_2 = \frac{\pi d^2 }{4}

Area of object 1:

A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2

Since all other parameters are still the same except the drag coefficient:

For object 1:

D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2

For object 2:

D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2

Since the drag force for the two objects are the same:

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