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2 years ago

malott, m. e. (2003). paradox of organizational change: engineering organizations with behavioral systems analysis.

Engineering

1 answer:

mylen [45]2 years ago

7 0

Answer:

Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.

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Your driver license will be _____ if you race another driver on a public road, commit a felony using a motor vehicle, or are fou

ser-zykov [4K]

Answer:

Revoked

Explanation:

Your driver license will be revoked if you race another driver on a public road, commit a felony using a motor vehicle, or are found guilty of reckless driving three times in one year.

4 0

2 years ago

A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W

irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

$k_p = \dfrac{1 + sin\phi}{1 - sin\phi}$

$k_p = \dfrac{1 + sin30^0}{1 - sin30^0}$

k_p = 3

Passive earth pressure

$P_p = \dfrac{1}{2}k_p \gamma H^2$

$P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2$

P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0

3 years ago

Calculate the time taken to completely empty aswimming pool 15

user100 [1]

Answer:

Time needed to empty the pool is 401.35 seconds.

Explanation:

The exit velocity of the water from the orifice is obtained from the Torricelli's law as

$V_{exit}=\sqrt{2gh}$

where

'h' is the head under which the flow of water occurs

Thus the theoretical discharge through the orifice equals

$Q_{th}=A_{orifice}\times \sqrt{2gh}$

Now we know that

$C_{d}=\frac{Q_{act}}{Q_{th}}$

Thus using this relation we obtain

$Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}$

Now we know by definition of discharge

$Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}$

Using the above relations we obtain

$Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\$

The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0

Solving for 't' we get

$\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.$

4 0

3 years ago

I am making composites of silicone rubber and copper particles; by mixing thermally conductive particles into the thermally insu

Gelneren [198K]

Answer:

The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

Explanation:

Let us assume that the total mass of composite is 100 lbm So as per the given conditions

15 lbm is copper and 85 lbm is rubber.
Density of rubber is 70 lbm/ft3
Specific gravity of Copper is 9

So As per the formula of specific gravity

$S_{cu}=\frac{\rho_{cu}}{\rho_w}$

here density of water is 62.4 lbm/ft3

Solving for Density of Copper gives

$S_{cu}=\frac{\rho_{cu}}{\rho_w}\\9=\frac{\rho_{cu}}{62.4}\\\rho_{cu}=9 \times 62.4\\\rho_{cu}=561.5 lbm/ft3$

For composition on volume basis, volume of individual components and composite are calculated as

$V_{cu}=\frac{m_{cu}}{\rho_{cu}}\\V_{cu}=\frac{15}{561.5}\\V_{cu}=0.0267 ft^3\\\\V_{r}=\frac{m_{r}}{\rho_{r}}\\V_{r}=\frac{85}{70}\\V_{r}=1.214 ft^3\\\\V_{c}=V_{r}+V_{cu}\\V_{c}=1.214+0.0267 \\V_{c}=1.2409 ft^3$

The composition is given as

$c_{cu}=\frac{V_{cu}}{V_{c}}\\c_{cu}=\frac{0.0267}{1.2409} \times 100 \%\\c_{cu}=2.154 \%\\\\c_{r}=\frac{V_{r}}{V_{c}}\\c_{r}=\frac{1.214}{1.2409} \times 100 \%\\c_{r}=97.83 \%$

So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

6 0

3 years ago

Answer the following either true (T) or false (F) (5 pts)

likoan [24]

Answer:

1. True

2. True

3. False

Explanation:

The office location is where the soil layer is not uniform. The thickness of the soil varies which could lead to doors being jammed. The engineer needs to estimate the differential in clay soil.

The inclined surface can hold less weight than a vertical surface. The capacity to hold the weight is due to the gravitational force which is exerted to the load.

6 0

3 years ago