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2 years ago

9

Statement true about fats in food

2 answers:

Elodia [21]2 years ago

7 0

Answer:

What that means please explain

NNADVOKAT [17]2 years ago

4 0

Answer:

Fats are important for good health and proper functioning of the body. They are a source of energy, essential fats and enhance the absorption of fat soluble vitamins. However, too much fat and/or the wrong type of fat may negatively affect our health. Fats also give foods a particular texture, appearance and flavour.

Explanation:

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A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second

Andreyy89

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.

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How wold you classify the earliest examples of S.T.E.M discoveries provided in this lesson?

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2 years ago

Which basic principle influences how all HVACR systems work?

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Answer:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

Explanation:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

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3 years ago

Which traditional subject is part of construction management or construction science syllabi?

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2 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

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3 years ago