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3 years ago

Which option distinguishes the members of a software deployment process team most likely involved in the following scenario?

Engineering

1 answer:

Alchen [17]3 years ago

5 0

Answer:

A local bank, with several branches in three cities, requests changes to its mortgage calculation software.

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What is the mass of the same dragster body (volume of 150 cm3) if it is made of basswood instead?

dusya [7]

Answer:

the answer is 61.5

Explanation:

8 0

3 years ago

What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?

Dominik [7]

Answer:

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)= $\frac{s}{(s+3)^2}$

U(s)= $\frac{1}{s}$ for unit step function

output =H(s)×U(s)

= $\frac{s}{(s+3)^2}$ × $\frac{1}{s}$

= $\frac{1}{(s+3)^2}$

taking inverse laplace of output

output=t× $e^{-3t}$

at t=0 putting the value of t=0 in output

output =0

3 0

4 years ago

You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,

nadezda [96]

Answer:

Explanation:

cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

= 1.075 x 10⁻³

Young's modulus = stress / strain

= 83.32 x 10⁶ / 1.075 x 10⁻³

= 77.5 x 10⁹ N/m²

5 0

3 years ago

The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i

Phantasy [73]

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²) ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )

solve it we get

Eo = 340.74 GPa

8 0

3 years ago

A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowr

hram777 [196]

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = $12\ \text{m}^3/\text{s}$

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}$

Froude number is given by

$Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5$

Since $F_r>1$ the flow is super critical.

Flow is critical when $Fr=1$

Depth is given by

$d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}$

The depth of the channel will be 1.2 m for critical flow.

4 0

3 years ago