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polet [3.4K]
3 years ago
7

A 3-oz serving of roasted, skinless chicken breast contains 140 Cal, 27 g of protein, 3 g of fat, 13 mg of calcium, and 64 mg of

sodium. One-half cup of potato salad contains 180 Cal, 4 g of protein, 11 g of fat, 24 mg of calcium, and 662 mg of sodium. One broccoli spear contains 50 Cal, 5 g of protein, 1 g of fat, 82 mg of calcium, and 20 mg of sodium.
a) Write 1x5 matrices C, P, and B that represent the nutritional values of each food.

b) Find C+2P+3B and tell what the entries represent.
Engineering
1 answer:
Art [367]3 years ago
7 0

Answer:

Explanation:

  • a) Given C [ cal pro fat calc sod]

              [140 27 3  13  64]

  • P = [cal pro fat calc sod]

     [180 4 11 24 662]

  • B = [cal pro fat calc sod]

      [50  5    1  82   20]

To find C+2P+3B = [140 27 3  13  64] + 2[180 4 11 24 662] + 3[50  5    1  82   20]

= [650  54  28  307  1448]

The entries represent skinless chicken breast , One-half cup of potato salad and One broccoli spear.

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In multi-grade oil what is W means?
irga5000 [103]

Answer:

winter viscosity grades

Explanation:

The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.

4 0
3 years ago
An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator
MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit i_m=385mA=385\times 10^{-3}A=0.385A

Inductance of the inductor L=400mH=400\times 10^{-3}h=0.4H

Capacitance C=4.43\mu F=4.43\times 10^{-3}F

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm

Capacitive reactance will be equal to X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm

Impedance of the circuit will be Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm

So maximum voltage will be \Delta V_{max}=0.385\times 865.44=333.194volt

(B) Phase difference will be given as \Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70

So current will be leading by an angle 54.70

5 0
3 years ago
Which explanation best summarizes what went wrong during Paul’s cost analysis?
Valentin [98]

Answer:

wut is it

Explanation:

4 0
1 year ago
I want a real answer to this, not just take my points
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7 0
2 years ago
Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature
sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

\epsilon_{car}=1-\frac{T_c}{T_H}

Where T_c is the atmospheric temperature and T_H is the maximum burn temperature.

For the case (B), the efficiency we will use is:

\epsilon_{b}=0.4\epsilon_{car}

The work done by the engine can be calculated as:

W=\epsilon Q=\epsilon H_v\cdot m_{fuel} where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

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TC=m\cdot c

8 0
2 years ago
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