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Zarrin [17]
2 years ago
10

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

Engineering
1 answer:
Ann [662]2 years ago
6 0

The number of tubs that each of them sold is; 24 tubs each

The number of days it will take for both of them to sell same amount of tubs is; 4 days

Number of cookies that Nicole had already sold = 4 tubs

Number of cookies sold by Josie before counting = 0 cookies

Nicole now sells 5 tubs per day and

Josie now sells 6 tubs per day.

Let the number of days it will take for them to have sold the same amount of cookies be d.

Thus;

5d + 4 = 6d + 0

6d - 5d = 4

d = 4 days

Thus, total number of cookies for both are;

Total for Nicole = 4 + 5(4) = 24 cookies

Total for Josie = 6(4) = 24 cookies

Read more about proportion at; brainly.com/question/870035

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Please help this is due today!!!!!
White raven [17]

Answer:

1:c 2:False

Explanation:

7 0
2 years ago
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati
Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0
3 years ago
The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?
bonufazy [111]

Answer:

Explained

Explanation:

Cold working: It is plastic deformation of material at temperature below   recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.

Given melting point temp of lead is 327° C and lead recrystallizes at about

0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.

6 0
3 years ago
Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)-
pickupchik [31]

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

\frac{T}{J}= \frac{\tau }{r}

where, T = torque

            J = polar moment of inertia of shaft

            τ = torsional shear stress

             r = raduis of the shaft

Therefore from the above relation we see that

            \tau = \frac{T.r}{J}

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

4 0
3 years ago
Which of the following are tips to help a speaker use their own voice?
djverab [1.8K]
D) All of the above.
4 0
3 years ago
Read 2 more answers
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