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2 years ago

If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters.

1 answer:

5 0

Before any calculations, we need to determine first the crystal structure of the lead metal. From literature, the lead metal assumes an FCC structure. So, it would have 4 atoms per units cell where the three atoms is the sum of all the portion of an atoms in each face of the cell and the 1 atom is the sum of all the portion of the corner atoms. The volume of the unit cell is equal to the edge length raise to the power three or V = a^3. The edge length can be calculated from the radius of the atoms by the pythagorean theorem. We do as follows:

V = a^3
a^2 + a^2 = (4r)^2
2a^2 = (4r)^2
a = 2r
V = (2r)^3
V = 16r^3
V = 16 (0.175x10^-9)^3
V = 1.21 x 10^-28 m^3

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Discuss the production, transmission, and usage of electricity in the context of conservation of energy. When electricity is “us

mr Goodwill [35]

There are huge losses in the transmission, production and usage of electricity and the reduction of these losses in order to save electricity is called as conservation of energy.

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2 years ago

Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass

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Use Scoratic it works with any time of subject

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2 years ago

You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init

Lady_Fox [76]

Answer:

$T_{f}$ = 85.89 ° C

Explanation:

The linear thermal expansion process is given by

ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

ΔV = V β ΔT

Let's apply this equation to our case

ΔV / V = -0.507% = -0.507 10-2

ΔT = (ΔV / V) 1 /β

ΔT = -0.507 10⁻² 1 / 1.15 10⁻³

ΔT = -4.409

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$T_{f}$ = 90.3-4409

$T_{f}$ = 85.89 ° C

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2 years ago

Read 2 more answers

A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-

andreyandreev [35.5K]

The resultant static friction force is equal to 20 N to the left.

Why?

I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:

"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"

So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.

We can calculate the static friction force by using the following formula:

$Fr=F(appliedforce)$

Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.

Calculating we have:

$Fr=F(appliedforce)$

$Fr=20N$

Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)

So,

$FrictionForce=AppliedForce$

Since the static friction force is equal to the applied force, the crate does not start to move.

Have a nice day!

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2 years ago

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

L = W

Area

A = W W = W²

R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

Length L = D= 2W

Area A = W L

R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

R₂ / R₁ = 2W²/L

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2 years ago