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11Alexandr11 [23.1K]
3 years ago
8

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radi

us of the uranium nucleus is approximately 7.4×10−15m .
a. What is the electric field this nucleus produces just outside its surface?Express your answer using two significant figures.
b. What magnitude of electric field does it produce at the distance of the electrons, which is about 1.1×10-10m ?
c. The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?Express your answer using two significant figures.=____________N/C
Physics
1 answer:
guapka [62]3 years ago
7 0

Answer:

(A) E = 2.633 * 10^19 N/C

(B) E = 1.096 * 10^13 N/C

(C) E = -1.096 * 10^13 N/C

Explanation:

Parameters given:

Number of protons, N = 92

Radius of Uranium nucleus = 7.4 * 10^-15 m

Electronic charge, ẹ = 1.6023 * 10^-19 C

Electric field at a point R due to a charge Q is given as

E = (k*Q) / (R^2)

Where k =Coulombs constant

(A) Since there are 92protons,the total Electric field due to the protons will be:

E = (k*e*N) / (R * R)

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (7.4 * 10^-15)^2

E = 2.633 * 10^19 N/C

(B) At the position of the electrons, R = 1.1 * 10^-10m. Therefore, Electric field will be:

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = 1.096 * 10^13 N/C

(C) There are 92 electrons in the Uranium atom and electrons have a charge - e, hence, the Electric field due to the electrons at the nucleus will be:

E = -(k*e*N) / (R * R)

E = -(9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = -1.096 * 10^13 N/C

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5 0
3 years ago
Consider the following waves representing electromagnetic radiation: An illustration shows two waves representing electromagneti
Murljashka [212]

Answer:

a) red wave hs a longer wavelength than the green wave

b)f = 1.875 10¹¹ Hz ,  f_green = 5.45 10¹⁴Hz

c)   E = 1.24 10⁻²² J , E_green = 3.6 10⁻¹⁹ J

d) The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

Explanation:

a) The green wave are on the left in the electromagnetic spectrum so the red wave has a longer wavelength than the green wave

The green wavelength is in the range of 550 10⁻⁹ m

The speed of the wave is

            c = λ f

            f = c /λ

b) The frequency of the red wave is

            f = 3 10⁸ / 1.6 10⁻³

            f = 1.875 10¹¹ Hz

For the green wave

           f_green = 3 10⁸/550 10⁻⁹

           f_green = 5.45 10¹⁴Hz

c) The photon energy is given by the Planck equation

             E = h f

             E = 6.63 10⁻³⁴ 1.875 10¹¹

             E = 1.24 10⁻²² J

For the green wave

              E_green = 6.63 10⁻³⁴ 5.45 10¹⁴

              E_green = 3.6 10⁻¹⁹ J

d) The speed of electromagnetic waves is constant and has a value of 3 108 m / s

e)  

The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

6 0
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2 years ago
A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under
Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0
3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

5 0
3 years ago
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