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Mrrafil [7]
3 years ago
8

I really need help ASAP!!!

Engineering
1 answer:
ValentinkaMS [17]3 years ago
8 0

Explanation:

He would work on the thing like in the method you work on your question.

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A boy eat a energy of a sandwich to run a race because when they eat a sandwich it helps them to help it mid workout and real nutritions of NYC and bring extra fuel and eating the right thing
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List and describe three classifications of burns to the body.
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AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.

Explanation:

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If you are a mechanical engineer answer these questions:
Natasha_Volkova [10]

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1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

3 0
3 years ago
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
Dafna11 [192]

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
3 years ago
A 600 MW power plant has an efficiency of 36 percent with 15
ololo11 [35]

Answer:

401.3 kg/s

Explanation:

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

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