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Tatiana [17]
3 years ago
15

A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

Avogadro’s number is 6.022 × 1023 /mol, and the elemental charge is 1.602 × 10−19 C.If each water molecule donates one electron, how much water is ionized in the lightning? One mole of water has a mass of 18 g/mol. Answer in units of g.
Physics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

m=5.78\times 10^{-3}\ g

n_e=1.935\times 10^{20} is the no. of electrons

Explanation:

Given:

  • quantity of charge transferred, Q=31\ C

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron 1.602\times 10^{-19}\ C

so,

n_e=\frac{Q}{e}

n_e=\frac{31}{1.602\times 10^{-19}}

n_e=1.935\times 10^{20} is the no. of electrons

  • Now if each water molecules donates one electron:

Then we require n=1.935\times 10^{20} molecules.

<u>Now the no. of moles in this many molecules:</u>

n_m=\frac{n}{N_A}

where

N_A=6.022\times 10^{23} Avogadro No.

n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}

n_m=3.213\times 10^{-4}\ moles

  • We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

n_m=\frac{m}{M}

where:

m = mass in gram

3.213\times 10^{-4}=\frac{m}{18}

m=5.78\times 10^{-3}\ g

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Particles 1 and 2 each mass m fixed to the ends of a rigid massless rod of length L1 + l2 with L1 = 20cm and l2 = 80 cm. The rod
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3 years ago
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a
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Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0
3 years ago
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