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?)
2 answers:
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.
You might be interested in
<h2>Answer:
</h2>
The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting
the radius of the orbit (measured from the center of the planet to the satellite).
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>
In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.
Answer:
(a). The tension in the cable is 658.33 N.
(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.
(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.
Explanation:
Given that,
Weight of beam= 190 N
Here, The center of gravity is at its center
According to figure,
The angle is
The horizontal component is
The vertical component is
(a). We need to calculate the tension in the cable
Using formula of net torque acting on the pivot
Put the value into the formula
(b). We need to calculate the horizontal components of the force exerted on the beam at the wall
Using formula of horizontal component
Put the value into the formula
(c). We need to calculate the vertical components of the force exerted on the beam at the wall
Using formula of vertical component
Put the value into the formula
Hence, (a). The tension in the cable is 658.33 N.
(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.
(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.
