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evablogger [386]
3 years ago
7

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Engineering
1 answer:
Kamila [148]3 years ago
7 0

Answer:

(b)False

Explanation:

I_{xy} defined as

      I_{xy} =\int \left (x\cdot y\right )dA

Where x is the distance from centroidal x-axis

           y is the distance from centroidal y-axis

          dA is the elemental area.

The product of x and y can be positive or negative ,so the value of  I_{xy} can be positive as well as negative .

So from the above expressions we can say that the product of I_{x},I_y is different from I_{xy} .

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Two gases—neon and air—are expanded from P1 to P2 in a closed-system polytropic process with n = 1.2. _____ produces more work w
Makovka662 [10]

Answer:

Note that Air requires lesser work

Explanation:

Calculate  for general work done

SInce Gas constant 'R' for:  Neon = 0.4119KJ/kg.k , and Air = 0.287 kJ/kg·K

Calculate for work done of NEON

Calculate for work done of Air

See solution attached.

7 0
3 years ago
Can I get an answer to this question please
crimeas [40]

Answer:

  (i) 12 V in series with 18 Ω.

  (ii) 0.4 A; 1.92 W

  (iii) 1,152 J

  (iv) 18Ω — maximum power transfer theorem

Explanation:

<h3>(i)</h3>

As seen by the load, the equivalent source impedance is ...

  10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

  (36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

<h3>(ii)</h3>

The load current is ...

  (12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

  P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

<h3>(iii)</h3>

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

  (600 s)(1.92 J/s) = 1,152 J

__

<h3>(iv)</h3>

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

  ((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0
2 years ago
Can i use two shunts and one meter
Lyrx [107]

Answer:

no

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph
kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.  

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0
3 years ago
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si
alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

<u>A) Classifying the soil according to USCS system</u>

 ( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0
3 years ago
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