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yan [13]
3 years ago
15

Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maxi

mum rate of charge flow in a lightning bolt is about 20,000 C/s this lasts for 100 μs or less.
1.How much charge flows between the ground and the cloud in this time? (Q= ? C)
2.How many electrons flow during this time? (n_e = ? )
Physics
1 answer:
V125BC [204]3 years ago
6 0

Answer:

1.25\times 10^{19}\ electrons

Explanation:

r = Rate of flow of charge = 20000 C/s

t = Time taken = 100 μS

Charge is given by

Q=rt\\\Rightarrow Q=20000\times 100\times 10^{-6}\\\Rightarrow Q=2\ C

The charge that flows between the ground and the cloud in this time is 2 C

Number of electrons is given by

n_e=\dfrac{Q}{e}\\\Rightarrow n_e=\dfrac{2}{1.6\times 10^{-19}}\\\Rightarrow n_e=1.25\times 10^{19}\ electrons

The number of electrons is 1.25\times 10^{19}\ electrons

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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

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