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Maru [420]
3 years ago
11

An electron is isolated from an atom and exists in vacuum. A group of scientists collectively state that they can remove part of

the electron’s charge so that it may be transferred to another uncharged particle. Which of the following claims, with appropriate evidence, is correct about the removal of partial charge from the electron? Justify your selection.
Physics
1 answer:
lesya692 [45]3 years ago
5 0

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Explanation:

The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.

The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:

  • Protons: they carry electric charge of +e
  • Electrons: they carry electric charge of -e

Where "e" is the fundamental charge (e=1.6\cdot 10^{-19}C). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.

An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of e. Therefore, the correct statement is

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Learn more about particles:

brainly.com/question/2757829

#LearnwithBrainly

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A student is pedaling a stationary bicycle at the U of M Rec Center. Her alveolar PO2 = 115 mm Hg and her oxygen consumption is
polet [3.4K]

Answer:

This is an alveolar gas equation question and it is used to approximate the partial pressure of oxygen in the alveolus (PAO2):

The equation states;

PₐO₂ = P₁O₂ - (PₐCO₂/R) = [(PB - PH₂O) × F₁O₂ - (PₐCO₂/R)]   ...................Eqn 1

where

PₐO₂ = Alveolar partial pressure of O2 = 115mmHg

P₁O₂ = Inspired partial pressure of O2 = 150mmHg

PB = barometric pressure,

PH₂O = Water vapor pressure (usually 747 mmHg),

F₁O₂ = fractional concentration of inspired oxygen,

and R = gas exchange ratio. (Usually around 0.8)

FₐO₂ = Fraction of alveolar O₂

(O₂) = 1L/min = 1dm³

From eqn 1. we have

PₐCO₂ = (P₁O₂ - PₐO₂)/R

   = (150 - 115)x0.8

PₐCO₂ = 28mmHg

Similarly from Eqn 1, we have

F₁O₂ = (PₐO₂ + PₐCO₂/R)/(PB - PH₂O)

F₁O₂ = (115 + (28/.8))/(747 - 47)

F₁O₂ = 0.21

Now to find the Alveolar Ventilation A, we will use this equation;

O₂ = A(F₁0₂ - FₐO₂)                                                                   .................Eqn 2

But FₐO₂ = PₐO₂/(PₐO₂ + PₐCO₂)

FₐO₂ = 115/ (115+28) = 0.8

A = O₂/(F₁0₂ - FₐO₂)

A = 0.001/(0.21 - 0.8)

A = 0.00169m³/min

Hence, the aveolar ventilation is 0.00169m³/min

8 0
3 years ago
Which theory was first proposed by Albert Einstein
elena-s [515]
General relativity was first proposed by Einstein.

Hope this helps! Let me know if you need the definition :)
5 0
3 years ago
Read 2 more answers
Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second
zlopas [31]
Given:

10^10 electrons per second

To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:

From literature, 

1 Coulomb is equivalent to 6.242×10^18 electrons<span>.

So,

= 10^10 electrons * (1 coulomb/</span><span>6.242×10^18</span> electrons) / second
<span>= 1.602 x 10^-9 coulumbs

This value is too small to be used in an actual setting. 

</span><span>
</span>
3 0
3 years ago
What is TRUE of carbon monoxide?
Aleks04 [339]

¡Hello!

Carbon monoxide is an odorless, colorless and tasteless gas produced by the incomplete combustion of carbon in fossil fuels such as wood, propane, charcoal, oil, gas, coal or other fuel.

6 0
2 years ago
What is the difference between the B-field and the H-field?
Simora [160]
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So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA.  The units work out like 
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux.  The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field.  And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever.   I have included a picture that also shows M, the magnetization of a material along with H and B.  M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!

3 0
3 years ago
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