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cricket20 [7]
2 years ago
13

One half of 17’ 9 3/16”

Engineering
1 answer:
Mademuasel [1]2 years ago
6 0
555.66 is the answer I think
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Given the following MATLAB statement: ( 3 + 2 ) / 5 * 4 + 5 ^ 2 In what order will these operations be done?
Andrei [34K]

Answer:

first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.

7 0
3 years ago
As the impurity concentration in solid solution of a metal is increased, the tensile strength:________.a) decreasesb) increasesc
valkas [14]

Answer:

Increases

Explanation:

By inhibiting the motion of dislocations by impurities in a solid solutions, is a strengthening mechanism. In solid solutions it is atomic level strengthening resulting from resistance to dislocation motion. Hence, the strength of the alloys can differ with respect to the precipitate's property. Example, the precipitate is stronger (ability to an obstacle to the dislocation motion) than the matrix and it shows an improvement of strength.

5 0
3 years ago
A ballistic pendulum consists of a 3.60 kg wooden block on the end of a long string. From the pivot point to the center‐of‐mass
Pavel [41]

Answer:

17.799°

Explanation:

When the bullet hits the block at that time the momentum is conserved

So, initial momentum = final momentum

P_i=P_f

So 28\times 10^{-3}\times 210=(3.6+0.028)v_f

v_f=1.6207\ m/sec

Now energy is also conserved

So \frac{1}{2}\times (3.6+0.028)\times 1.6207^2=(3.6+0.028)\times 9.81\times 2.8(1-cos\Theta )

cos\Theta =0.8521So\ \Theta =17.799^{\circ}

3 0
3 years ago
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
For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
3 years ago
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