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shtirl [24]
4 years ago
15

A spherical, non-conducting shell of inner radius r1 = 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC distri

buted uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 12 cm from the center of the shell? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)
Physics
1 answer:
Marizza181 [45]4 years ago
4 0

Answer:

Explanation: r₂ = 15 x 10⁻³ m ; r = 12 x 10⁻³m ; r₁ = 10 x 10⁻³ m

     E = Q /4π∈r²x ( r³ - r₁ ³) / ( r₂³ - r₁³ )

         = 15 x 10⁻⁶ x 8.99 x 10⁹x( 12³ - 10³ ) / ( 15³- 10³ ) X 12³

          = 23.92 N/C

       

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states that the velocity of the fluid in contact with the surface is equal to the velocity of that surface
nikklg [1K]

Answer:

The No-slip boundary condition.

Explanation:

The no-slip boundary condition states that the velocity of the fluid in contact with the surface is equal to the velocity of that surface, both the normal and tangential components of the fluid velocity field are equal to zero.

7 0
3 years ago
A new system of units is invented where the basic unit of length is the "dak." The "dak" is defined as equal to 3 inches, and th
Goryan [66]

Answer:

1 cm = 0.131 daks

Explanation:

It is given that,

1 dak = 3 inches

and

1 inch = 2.54 cm

We need to find how many "daks" are there in a centimeter. We can do the conversions as follows :

1\ dak=1\ dak\times \dfrac{3\ inches}{1\ dak}\times \dfrac{2.54\ cm}{1\ inch}

1\ dak=7.62\ cm

1\ cm=\dfrac{1}{7.62}\ dak

1 cm = 0.131 dak

So, there are 0.131 daks in 1 centimeter. Hence, this is the required solution.

5 0
4 years ago
The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

5 0
3 years ago
A block of mass 8 m can move without friction
nekit [7.7K]

Answer:

Let M1 = 8 kg and M2 = 34 kg

F = M a = (M1 + M2) a

F = M2 g     the net force accelerating the system

M2 g = (M1 + M2) a

a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2

5 0
2 years ago
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