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zaharov [31]
3 years ago
10

Neutral hydrogen can be modeled as a positive point charge +1.6×10^−19C surrounded by a distribution of negative charge with vol

ume density given by rhoE(r)=−Ae−2r/a0 where a0=0.53×10^−10m is called the Bohr radius, A is a constant such that the total amount of negative charge is −1.6×10^−19C, and e = 2.718 is the base of the natural log.
Required:
a. What is the net charge inside a sphere of radius a0?
b. What is the strength of the electric field at a distance a0 from the nucleus?
Physics
1 answer:
Archy [21]3 years ago
4 0

Answer:

a) 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

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Answer:

1. 0.45 s.

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Explanation:

From the question given above, the following data were obtained:

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1. Determination of the time taken for the pencil to hit the floor.

Height (h) = 1 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.8

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Take the square root of both side

t = √(1/4.9)

t = 0.45 s.

Thus, it will take 0.45 s for the pencil to hit the floor.

2. Determination of the velocity with which the pencil hit the floor.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 0.45 s.

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 0.45)

v = 0 + 4.41

v = 4.41 m/s

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A force of 10. N toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden floor. the c
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1) 5 N

The crate is initially moving, so we must calculate the force of kinetic friction, which is given by:

F_f = \mu_k mg

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\mu_k=0.2 is the coefficient of friction between the crate (made of wood) and the floor (made of wood). The coefficient of kinetic friction between wood and wood is about 0.2.

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Substituting the numbers into the formula, we find

F_f=(0.2)(25 N)=5 N


2) 5 N

There are two forces acting on the crate along the horizontal direction:

- The force that pushes the crate toward the right, of magnitude F=10 N

- The force of friction, which acts in the opposite direction (so, towards the left), of magnitude F_f = 5 N

Since the two forces are in opposite directions, the net force is given by their difference:

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3) Yes

The crate is accelerating. In fact, according to Second Newton's Law:

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where Fnet is the net force on the crate, m is its mass, a is its acceleration. We can immediately see that since Fnet is not zero, the acceleration is also non-zero, so the crate is accelerating.

We can even calculate the magnitude of the acceleration. In fact, the mass of the crate is given by:

m=\frac{Weight}{g}=\frac{25 N}{9.8 m/s^2}=2.55 kg

And by using (1) we find

a=\frac{F_{net}}{m}=\frac{5 N}{2.55 kg}=1.96 m/s^2


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