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umka21 [38]
3 years ago
8

How long...you...novels? A. Have/write B. Do/write C. Have/written D. Did/go

Engineering
2 answers:
deff fn [24]3 years ago
7 0

Answer: How long have you written novels

Explanation:

How long ____ you ____ novels

It's C (Might be "b"

Conclusion: All of them don't make sense except for C

soldi70 [24.7K]3 years ago
3 0

Answer:

Answer is B.

Explanation:

Subject verb agreement

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If you deposit today 11,613 in an account earning 8% compound interest, for how long should you invest the money in order to ear
Elanso [62]
Y = a (b)^t/p

y is total money

a is original amount

b is growth / decay factor

t is time

p is the frequency of every growth or decay

15131.76 = 11613 x 1.08^x

15131.76 / 11613 = 1.08^x

1.303… = 1.08^x

log1.303…. = xlog1.08

x = 3.43902165741 years
3 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 fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.
Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

S_{ut}=K \left(\cfrac ne \right)^n

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}

And we can replace values

K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87

Thus we get that the strength coefficient is K = 591.87 MPa

6 0
3 years ago
Select the correct answer.
FrozenT [24]

Answer:

B. Italy

Explanation:

Hope this helps :)

6 0
2 years ago
Read 2 more answers
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