What are the initial questions that a systems analyst must answer to build an initial prototype of the system output.

An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [

crimeas [40]

Answer:

Modulus of resilience will be $3216942.308j/m^3$

Explanation:

We have given yield strength $\sigma _y=818MPa$

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience $=\frac{\sigma _y^2}{2E}$ , here $\sigma _y$ is yield strength and E is elastic modulus

Modulus of resilience $=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3$

5 0

3 years ago

You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall

choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are $\rightarrow 120ft5in+2(10ft)$

The two long are $\rightarrow 122ft1in=122.08ft$

The two shors are $\rightarrow 86ft4.5in = 86.375ft$

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

$Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3$

That in cubic yards is equal to $180.15 (1cy=27ft^3)$

Hence, we need order 5% plus that represent with the quantity

$Q_{ordered}=1.05*180.15=189.15cy$

8 0

3 years ago

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio

saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

$K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}$

Put the value into the formula

$K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}$

$K_{H}=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}$

Put the value into the formula

$K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}$

$K_{V}=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

3 0

3 years ago

Aaron needs to create a building design for a restaurant with colors that depict excitement and vibrancy. Which color can Aaron

zloy xaker [14]

Answer:

I'm no engineer, but blue and purple are cool colors and white is every color so I'd go with orange

7 0

3 years ago

Read 2 more answers

Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa

DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l ) = ∝ * ΔT

( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )

∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

8 0

3 years ago