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wariber [46]
3 years ago
11

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

Physics
1 answer:
mihalych1998 [28]3 years ago
6 0
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
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Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an al
olga nikolaevna [1]

Answer:

6.67154\times 10^{-9}\ F

13.009503 C

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

k = Dielectric constant of air 3\times 10^6\ V/m

Side of plate = 0.7 km

A = Area

d = Distance = 650 m

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F

The capacitance is 6.67154\times 10^{-9}\ F

Electric field is given by

Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C

The charge on the cloud is 13.009503 C

7 0
3 years ago
what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)​
Alexxx [7]

Answer:

3.15m³

Explanation:

To solve this problem, let us first find the mass of the petrol from the given dimension.

       Mass  = density x volume

Volume of petrol  = 4.2m³

Density of petrol  = 0.3kgm⁻³  

       Mass of petrol  = 4.2 x 0.3  = 1.26kg

So;

      We can now find the volume of the alcohol

 Volume of alcohol = \frac{mass}{density}  

Mass of alcohol  = 1.26kg

Density of alcohol  = 0.4kgm⁻³  

  Volume of alcohol  = \frac{1.26}{0.4}   = 3.15m³

7 0
3 years ago
Compare the gravitational acceleration on the following objects compared to the Sun using:
arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765  x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0
3 years ago
A heat engine uses thermal energy that flows from a heat source with a temperature of 325 K to a cold sink, which has a temperat
Ilya [14]
The answer is "58.5"
5 0
3 years ago
Read 2 more answers
A boy of mass 30.0 kg is sledding down a 70.0-m slope starting from rest. The slope is angled at 15.0° below the horizontal. Aft
Studentka2010 [4]

The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

<h3>Their speed at the bottom</h3>

Apply the principle of conservation of energy,

E(up) - E(friction) = E(bottom)

mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²

v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }

where;

  • u is the speed of the after 28 m

u = √2gh

u = √(2gL sin15)

u = √(2 x 9.8 x 28 x sin 15)

u = 11.92 m/s

v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s

Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

Learn more about speed here: brainly.com/question/6504879

#SPJ1

8 0
2 years ago
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