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2 years ago

Note that common skills are listed toward the top, and less common skills are listed toward the bottom.

Engineering

1 answer:

Sladkaya [172]2 years ago

5 0

Answer:

BDEG

Explanation:

got it right on the test on edge because i used my b r a i n

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Learning the key concepts of each approach is essential to successful management of a project. What type of unpredictability is

Levart [38]

Answer:

lemme write it down

Explanation:

hold down okay

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3 years ago

Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the approp

Anvisha [2.4K]

Answer:

The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

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3 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

3 0

3 years ago

A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni

Dvinal [7]

Answer:

Tc = = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

$\rho = 8000 kg/m^3$

k = 50 w/m . k

c = 500 j/kg.k

$h_{\infty} = 1000 w/m^2$

$t_{\infity} = 750 k$

$t_w = 500 K$

$surface area = As = \pi dL$

$\frac{As}{L} = \pi D = \pi \timeS 0.06$

HEAT FLOW Q is

$Q = h_{\infty} As (T_[\infty} - Tw)$

$= 1000 \pi\times 0.06 (750-500)$

= 47123.88 w per unit length of rod

volumetric heat rate

$q = \frac{Q}{LAs}$

$= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}$

$q = 1.66\times 10^{7} w/m^3$

$Tc = \frac{- qR^2}{4K} + Tw$

$= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} + 500$

= 424.85 K

7 0

3 years ago

Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str

alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

= 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

= 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

ρ_avg= mass of unit cell/volume of unit cell

ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

λ^3 = 4.37

6 0

3 years ago