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vitfil [10]
3 years ago
11

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

lp please.​
Physics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

E = 2k  \frac{\lambda}{ r}

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

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1. How do energy and matter recycle in a closed-loop system?
Makovka662 [10]

Answer:

In a closed-loop system, matter is used to generate energy, and the energy generated is used to produce matter, and the cycle goes on without end.

Explanation:

A closed-loop system is one in which materials or energy is recycled without end through a production cycle. This means that a raw material is used to produce a finished product, and the finished product at the end of its use cycle is converted back and used as a raw material to produce more of it again. Energy and matter can also be cycled in the same way in an energy and matter closed-loop system, converting matter to energy, and the energy is put back into the production of more of the matter.

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3 years ago
In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6
Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

            E = 2.5 x 10⁻¹⁴ J

8 0
3 years ago
Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 g of carbon-14, how much carbon-14 m
MakcuM [25]

Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

Explanation:

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Let the sample present 11,430 years(t) ago = N_o

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ln[N]=ln[N]_o-\lambda t

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There are no true statements on that list.
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