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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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Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a sta
yuradex [85]

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

7 0
3 years ago
An equilateral triangle has a height of 3.32 cm. Draw the picture and use this information to determine the length of the sides
Westkost [7]

Answer:

The answer is  a=b=c=3.833 cm

Explanation:

Lets call the variables a=side a b=side b c=side c

We have that the formula of the equilateral triangle for its height is:

1)h=(1/2)*root(3)*a

2) If we resolve the equation we have

 2.1)2h=root(3)*a

 2.2)(2h/root(3))=a

3) After the replacement of each value we have that

a=2*3.32/1.73205

a=3.833 cm

And we know that the equilateral triangle has the same value for each side so a=b=c=3.833 cm

7 0
3 years ago
what will be the gravitational force between two bodies if the mass of each is doubled and the distance between them is halved?
katen-ka-za [31]

Answer:

Gravitational force will be 16 times more.

Explanation:

we know;

Gravitational force (F) = (Gm1m2)/d^2

when mass of each is doubled and distance between them is halved;

F= (G2m1×2m2)/(d/2)^2

=(4Gm1m2)/(d^2/4)

=4×4(Gm1m2)/d^2

=16(Gm1m2)/d^2

=16F

3 0
2 years ago
Light is incident normally on the short face of a 30∘−60∘−90∘ prism (Figure 1). A drop of liquid is placed on the hypotenuse of
ZanzabumX [31]

1.06 is the <u>maximum</u> refractive index that the liquid may have for the light to be totally reflected.

Only when a light source passes from a denser to a rarer medium can it completely reflect.

When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.

sin i_{c} = μ_{r} / μ_{d

From the diagram

Angle of incidence = 60°

sin60°  ≥ sini_{c} = μ_{r}/μ_{d}

μ_{r} ≤ μ_{d} sin60°

μ_{r} ≤ √1.5 × √3/2

   = 1.06

Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06

Learn more about refractive index here brainly.com/question/10729741

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2 years ago
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2 years ago
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