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ladessa [460]
3 years ago
5

Examine the pressure-measuring device shown in the figure below. (a) What is the gauge pressure reading in psi at point A? (b) W

hat is the pressure difference between point A and B? (c) Which is higher, pressure at point A or at point B? (d) What is the absolute pressure in psi at point D? State any assumptions you made

Engineering
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer: 45

Explanation:just cuase I need to

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In this assignment, you will demonstrate your ability to write simple shell scripts. This is a cumulative assignment that will c
nevsk [136]

Answer:

Explanation:

Usage: flip [-t|-u|-d|-m] filename[s]

  Converts ASCII files between Unix, MS-DOS/Windows, or Macintosh newline formats

  Options:

     -u  =  convert file(s) to Unix newline format (newline)

     -d  =  convert file(s) to MS-DOS/Windows newline format (linefeed + newline)

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8 0
3 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
Các đặc điểm chính của đường dây dài siêu cao áp .
rodikova [14]

Answer:

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Explanation:

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3 years ago
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
goldenfox [79]

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

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3 years ago
Gtjffs
grandymaker [24]

the required documents is 3000

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