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OLEGan [10]
3 years ago
13

Evaluate to three significant figures and using appropriate prefix: (354 mg)(45 km)/(0.0356 kN)

Engineering
1 answer:
frutty [35]3 years ago
5 0

Answer:

0.447 s²

Explanation:

First, convert to SI units.

(354 mg) (45 km) / (0.0356 kN)

(0.354 g) (45000 m) / (35.6 N)

One Newton is kg m/s²:

(0.354 g) (45000 m) / (35.6 kg m/s²)

(0.000354 kg) (45000 m) / (35.6 kg m/s²)

Simplify:

0.447 s²

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BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of
Tanya [424]

(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ

<u>Explanation:</u>

<u />

Given-

Lattice parameter of Li  = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200

(a)

Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b)

Density is represented by ρ

ρ = n X (AM) / v X Nₐ

where,

Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

4 0
3 years ago
Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)
Novosadov [1.4K]

Answer:

a)  ∝  and β

   The phase compositions are :

    C_{\alpha } = 5wt% Sn - 95 wt% Pb

    C_{\beta } =  98 wt% Sn - 2wt% Pb

b)

The phase is; ∝  

The phase compositions is;   82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

The phase compositions are :

C_{\alpha } = 5wt% Sn - 95 wt% Pb

C_{\beta } =  98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝  

The phase compositions is;  82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

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3 years ago
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Both a and b

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2 years ago
A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

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Explanation:

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L is the length of the wire,

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For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

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