An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such defines the structure of a chemical element.

Generally, these atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

Hence, an object is most likely to become electrically charged by gaining or losing negative particles.

5 0

3 years ago

Escribe un listado de 5 acciones que promueven la interculturalidad y la inclusión

allochka39001 [22]

de mategoloalterfsqol

6 0

3 years ago

Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o

blondinia [14]

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

$Q=ne$

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

$Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C$

The potential difference V between the disks separated by a distance d is given by,

$V=Ed$

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

$V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V$

The capacitance C of the capacitor is given by,

$C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F$

The capacitance of a parallel plate capacitor is given by,

$C=\frac{\epsilon_0A}{d}$

Here, ε₀ is the permittivity of free space and A is the area of the disks.

Rewrite the expression for A.

$C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3$

the area A of the disks is given by,

$A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }$

Here, D is the diameter of the disk.

$D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm$

The diameter of each disc is found to be 1.32 cm.

7 0

3 years ago

Newton's Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force F, where r is the dista

Marrrta [24]

Answer:

W = - 5.01 10¹⁰ J

Explanation:

Work is defined by the expression

W = ∫ F.dr

Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product

W = ∫ F dr

W = G m₁m₂ ∫ 1 /r² dr

W = G m₁ m₂2(-1 / r)

We evaluate between the lower limits r = Re and upper r = ∞

W = G m₁m₂ (-1 / Re + 1 / ∞)

W = - G m₁ m₂ / Re

Let's calculate

W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶

W = - 5.01 10¹⁰ J

4 0

3 years ago

Normal faults occur from _____ stress and reverse faults occur from _____ stress.

Normal faults occur from _ stress and reverse faults occur from _ stress.