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Rudik [331]
3 years ago
13

Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical

surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 2.0 cm from the center of the two surfaces?
Physics
1 answer:
N76 [4]3 years ago
6 0

Answer:

E = K Q / R^2     by Gauss' Law where Q is the  charge enclosed by the surface of revolution and R is the distance from the enclosed charge

Since Q = d 4 pi r^2   where d is the charge density and r the radius of the inner sphere

E = K / R^2 * (4 d pi r^2) = 4 K pi d (r / R)^2 = 4 K pi d * 1/4

E =  9 * 10E9 * 3.14 * 40 * 10E-12 = 1.13 N / C

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On a planet with different gravity, would the molarity of water be different? explain your reasoning.
ruslelena [56]

On a planet with different gravity, the molarity of water won't be different as water produces regular natural gravity.

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5 0
2 years ago
What isthe correct 1/4=20%
anzhelika [568]

Answer:

25

Explanation:

5 0
3 years ago
Can you solve it descriptively . thanks
Solnce55 [7]

Answer:

|M_y| = 170.82 \ N.mm

Explanation:

From the diagram affixed below completes the question

Now from the diagram; We need to resolve the force at point  A into (3) components ; i.e x.y. & z directions which are equivalent to F_x \ , F_y \ ,  F_z

So;

F_x = positive x axis

F_y = Negative y axis

F_z = positive z axis

Then;

|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ?  ---- equation (3)

From equation (1); Let's make F_y the subject of the formula ; then :

F_y = \frac{77+11F_z}{27}

Substituting  the value for F_y into equation (2) ; we have:

(\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since  \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm

4 0
3 years ago
A wave is passing through a uniform medium. As the frequency of this wave increases, its wavelength
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B. Decreases

Why: Because the longer the wavelength the lower the frquency. The shorter the wavelength the higher the frequency.

5 0
4 years ago
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