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deff fn [24]
3 years ago
9

The Thomson model of a hydrogen atom is a sphere of positive charge with an electron (a point charge) at its center. The total p

ositive charge equals the electronic charge e. Prove that when the electron is at a distance r from the center of the sphere of positive charge, it is attracted with a force F=\frac{e^2r}{4\pi\varepsilon_oR^3} where R is the radius of the sphere.
Physics
1 answer:
Shtirlitz [24]3 years ago
4 0

Answer:

E = (1 / 4π ε₀ )  q r / R³

Explanation:

Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law

       F = K q₁ q₂ / r²

we used law Gauss

Ф = ∫ E .dA = q_{int} /ε₀

E 4π r² = q_{int} /ε₀  

E = q_{int} / 4π ε₀ r²

we replace the charge inside  

E = (1 / 4π ε₀ r²) ρ 4/3 π r³  

E = ρ r / 3 ε₀

the density for the entire atom is  

ρ = Q / V  

V = 4/3 π R³  

we substitute  

E = (r / 3ε₀ ) Q 3/4π R³  

E = (1 / 4π ε₀ ) q r / R³

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Design a rectangular milk carton box of width ww, length ll, and height hh which holds 474 cm3474 cm3 of milk. The sides of the
Hunter-Best [27]

Answer:

L = W = 6.810 cm

H =  10.22 cm

Explanation:

given data

volume L W H = 474 cm³

sides of the box cost = 2 cent/cm²

top and bottom cost 3 cent/cm²

to find out

dimensions of the box that minimize the total cost of material use

solution

we know here L W H = 474

here L is length and B is width and H is height

so when we minimize the cost function

C( L, W, H ) = (2) 2 H ( L + W ) + (3) 2 L W

so put here H

substitute H = \frac{474}{LW}

we get

C( L, W ) = 1896 ( \frac{1}{W} + \frac{1}{L} ) + 6 LW

so

minimum cost will be when the two partial derivatives is 0

so

\frac{dC}{dL} = 6W - \frac{1896}{L^2} = 0

so  

\frac{dC}{dW} = 6L - \frac{1896}{W^2} = 0

L = \frac{316}{W^2}

so

by solving above equation we get

L = W = ({{316})^{1/3} = 6.810 cm

and

H = \frac{474}{6.810^2}

H =  10.22 cm

7 0
3 years ago
A 3.00-kg box that is several hundred meters above the earth’s surface is suspended from the end of a short vertical rope of neg
marshall27 [118]

Answer:

A) 3.48m/s

B) 3.92m

C) 2.32m

D 23.33m/s

Explanation:

ma(t)=mg-At

a(t)= g - (At/m)

V(t)= integrala(T)dT = gt- (At^2/2m)

Initial x coordinator of the box is zero

X(t)= integralV(t)dt= 1/2gt^2-(At^3/6m)

a) V =( 9.81×1) -(38×1^2/2×3)

V= 9.81-6.33= 3.48m/s

b)-AT^2/2m + gT= 0

T=2mg/A= (2×3×9.81)/38

T= 1.549m

X(T)= (1/2×9.81×1.549^3)- (38×1.549^3/6×3)

X(T)= 11.768- (141.23/18) = 11.768 - 7.85= 3.92m

C) 1/2gT''^2 - AT''^3/6m =0

The only non trivial solution is T''= 3mg/A

T=(3×3×9.81)/38 = 2.32m

D) V = 9.81×3) - (38×3^2/6)

V= 29 - 5.667= 23.33m/s

6 0
3 years ago
What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?
polet [3.4K]

Answer:

0.044 V

Explanation:

E = Electric field = 5.5\times 10^6\ V/m

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V

The voltage across the membrane is 0.044 V

4 0
3 years ago
What is the second law of thermodynamics
andriy [413]
It states that the total entropy of an isolated system can never decrease over time
5 0
3 years ago
What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds
Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N

The net power needed to change the speed of the vehicle is calculated as;

P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W

3 0
3 years ago
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