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guajiro [1.7K]
3 years ago
6

We want to get a better, more concrete idea of the strength of the electric force. So imagine you could remove all the electrons

from 1 mL (1 cubic centimeter) of water. Dump the electrons on yourself, and put the electron-free nuclei (protons and neutrons) on the moon. Estimate how much force will there be between the positive and negative charges? (Of course you have to look up the earth-moon distance.) What fraction of your weight is that force? How many electrons do you have in 1 cc? Consider that virtually all the mass is given by protons and neutrons. Since the proton and the neutron have almost equal masses, and most atoms have equal numbers of protons and neutrons, so the water mass, divided by 2, is the mass of protons. Once you know the total mass of protons, divide by the mass of a single proton and you have the total number of protons. Knowing that, how many electrons do you then suppose you have?)

Physics
2 answers:
Flauer [41]3 years ago
6 0

Answer:

The electric force is 2.2962c + 005 times the gravity force on average human.

Explanation:

Get mass of water in gm

get the mass of protons in 1mL water

get number of electrons that should be present in 1mL of water

electric force Np (protons number), Ne (electrons number)

then get the ratio between the Electric force and weight of average human body;

see workings in attached picture.

Advocard [28]3 years ago
5 0

Answer:

Explanation:

solution answered below

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The outbreak of tornadoes that tore across the Gulf and the East Coast Tuesday and Wednesday was unusual for two reasons. For one thing, the severe weather encompassed a significant swath of the country. For another, winter is the least likely time for tornadic thunderstorms.

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A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
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Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg =  0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

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(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

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