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3 years ago

10

Why is it important to have knowledge on abbreviations, sign and symbols in masonry servicing job in the construction business s

ector?

1 answer:

marta [7]3 years ago

5 0

Maybe To know short cuts on the long run. And it is a way of coding

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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress

zvonat [6]

Answer:

a) $\mathbf{\sigma _ 1 = 4800 psi}$

$\mathbf{ \sigma _2 = 0}$

b) $\mathbf{\sigma _ 1 = 6000 psi}$

$\mathbf{ \sigma _2 = 3000 psi}$

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = 0$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 4800 psi}$

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = \frac{Pd}{4t}$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 6000 psi}$

$\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}$

$\mathbf{ \sigma _2 = 3000 psi}$

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0

3 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

"A communication between two devices is over the maximum limit of an ethernet frame size. The Transmission Control Protocol (TCP

Sliva [168]

COmmunication Devices

Explanation:

TCP/IP, or the Transmission Control Protocol/Internet Protocol, is a suite of communication protocols used to interconnect network devices on the internet. TCP/IP can also be used as a communications protocol in a private network (an intranet or an extranet).

The sequence number in a header is used to keep track of which segment out of many this particular segment might be. The next field, the acknowledgment number, is a lot like the sequence number. The acknowledgment number is the number of the next expected segment.

The 32 bit sequence number field defines the number assigned to the first byte of data contained in this segment. TCP is a stream transport protocol. To ensure connectivity, each byte to be transmitted is numbered.

3 0

3 years ago

The in-situ dry density of a sand is 1.72Mg/m3. The maximum and minimum drydensities, determined by standard laboratory tests, a

Stells [14]

Answer:

Relative density = 0.7 or 70%

Explanation:

The following information was provided by this question

Pd = 1.72mg/mg³

Pd max = 1.81 mg/mg³

Pd min = 1.54 mg/mg³

We substitute into the formula. This formula is contained in the attachment.

[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]

= 0.649350 - 0.581395 / 0.649350 - 0.552486

= 0.067955/0.096864

= 0.7015

= 0.7

The relative density is Therefore 0.7 or 70% when converted to percentage

8 0

2 years ago

¿Que piensas respecto al hecho de que María y Efraín rara vez se comunican directamente?

AleksandrR [38]

Answer:

Crean un código usando la naturaleza.

8 0

3 years ago