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2 years ago

True or False: Galvanized steel pipe can withstand moderate freezing.

Engineering

1 answer:

yaroslaw [1]2 years ago

5 0

Answer:

True

Explanation:

Older galvanized steel pipes, which have a tendency to freeze, are a bit more forgiving and will likely not burst. They can withstand extreme cold and warm temperatures.

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4. A banking system provides users with several services:

VLD [36.1K]

A diagram showing a use case diagrams for these requirements is given in the image attached.

<h3>What is system Case diagram?</h3>

A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.

A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.

Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.

Learn more about Case diagram from

brainly.com/question/12975184

#SPJ1

4 0

1 year ago

Write 83,120 in expanded form using powers of 10.

maksim [4K]

Answer:

8*10000+3*1000+1*00+2*10+2

Explanation:

8 0

2 years ago

To solve the problem, make assumptions for missing data and justify. Given:

finlep [7]

Answer:

5,4,1, this is a explication

6 0

2 years ago

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25 °C,

sashaice [31]

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity $= \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}$

theraml conductivity = k = 22 W/m.K

h = 300 W/ m^2.K

$T_i$ = 25 degree C = 298 k

$T_o$ = 60 degree C = 333 k

$T_{\infty}$ = 75 degree C = 348 L

diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

$Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226$

$\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}$

$\tau = 0.036 t$

$\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}$

at Bi = 0.226

Ai = 0.982

$\lambda = 0.876$

$\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}$

$0.3 = 0.982 e^{-0.2t}$

$0.305 = e^{-0.2t}$

-1.187 = - 0.02t

t = 59.37 s

7 0

2 years ago

For beam design, the cross section dimensions of the beam are determined: A. Based on allowable normal stress, but the allowable

noname [10]

Answer:

B. Based on the allowable shear stress, but the allowable normal stress should always be checked to be sure it is not exceeded

Explanation:

Shear stress is analyzed to determine the shear forces along the lenght of the beam. This is represented in a shear force diagram. The beam cross sectional design is determined in such a way as to minimize the shear stress. Allowable normal stress should always be checked in a structure if failure is to be prevented.

8 0

3 years ago