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lidiya [134]
2 years ago
5

The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

08 m is N, rounded to the tenths place.
Physics
2 answers:
solong [7]2 years ago
8 0
To answer this question, we will simply use Coulomb's which is stated as follows:
F = k*q1*q2 / r^2 where:
F is the magnitude of the force between the two charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 =  <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r is the distance between the two charges = </span><span>0.008 m

Substituting in the equation, we get:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075 which is approximately 6.1 Newtons

</span>
brilliants [131]2 years ago
8 0

On Ed. its 6.1 N. Just took the test and got it right soooo.

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The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0
2 years ago
Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea
Semmy [17]

Answer:

{\rm K} = 2.4\times 10^{-19}~J

Explanation:

The electric field inside a parallel plate capacitor is

E = \frac{Q}{2\epsilon_0 A}

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

F = qE = \frac{qQ}{2\epsilon_0 A}

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60

Plugging this identity into the force equation above gives

F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is 1.6 \times 10^{-19}

Therefore, the kinetic energy is 2.4\times 10^{-19}

8 0
2 years ago
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