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2 years ago

State two advantages of sautoing

1 answer:

4 0

Sautéing is advantageous over certain other methods of cooking as it is a very fast process, and the amount of fat required is lesser as compared to deep-frying. However, sautéed dishes do contain a certain amount of fat, and are less healthy as compared to boiled or baked dishes.-google

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Please solve part two

Burka [1]

Answer:

Wat part 2

Explanation:

7 0

2 years ago

By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer

Mademuasel [1]

Answer: Attached below is the well written question and solution

answer:

i) Attached below

ii) similar parameter = $\frac{V}{VoL } = 1 / Re$

Explanation:

Using ; L as characteristic length and Vo as reference velocity

i) Nondimensionalize the equations

ii) Identifying similarity parameters

the similar parameters are = $\frac{V}{VoL } = 1 / Re$

Attached below is the detailed solution

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2 years ago

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

Ymorist [56]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

the half life of the given circuit is given by

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

$t'_{1\2} =R'Cln2$

Divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

3 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$