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Dvinal [7]
3 years ago
12

Can you list three ways that real life earthquake conditions may differ from those made by a shaking table

Engineering
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

No

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.

4 0
3 years ago
Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th
Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

PV=mRT

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

T_1 = 38\°C

T_2 = 14\°C

\eta = 97\%

\dot{v} = 510m^3/kg

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

PV = mRT

\dot{m} = \frac{PV}{RT}

\dot{m} = \frac{0.6626*10^{5}*510}{287*311}

\dot{m} = 37.85kg/min

Therfore the mass flow rate at which water condenses, then

\eta = \frac{\dot{m_v}}{\dot{m}}

Re-arrange to find \dot{m_v}

\dot{m_v} = \eta*\dot{m}

\dot{m_v} = 0.97*37.85

\dot{m_v} = 36.72 kg/min

PART B) Enthalpy is given by definition as,

H= H_a +H_v

Where,

H_a= Enthalpy of dry air

H_v= Enthalpy of water vapor

Replacing with our values we have that

H=m*0.0291(38-25)+2500m_v

H = 37.85*0.0291(38-25)-2500*36.72

H = 91814.318kJ/min

In the conversion system 1 ton is equal to 210kJ / min

H = 91814.318kJ/min(\frac{1ton}{210kJ/min})

H = 437.2tons

The cooling requeriment in tons of cooling is 437.2.

3 0
3 years ago
If you want to obtain a Class E license before you turn 18, you must______. A. be in compliance with school attendance requireme
Makovka662 [10]

Answer:

The correct option is;

A. be in compliance with school attendance requirements

Explanation:

The requirements to acquire a special restricted driver's license for driver's license for under under 18 that have had a beginner's for up to 180 days include a minimum of 40 hours driving practice 10 of which should be in the dark. The application for the conditional drivers license is to be signed by the parent or guardian and the application is to be accompanied with proof of acceptable school attendance

At 17, after holding the special restricted drivers licence for a year without issues you can obtain the full drivers license.

7 0
3 years ago
Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with
Scilla [17]

Answer:

DIAMETER  = 9.797 m

POWER = \dot W = 28.6 kW

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

v =\frac{RT}{P}

  where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

v =  \frac{0.287\times 293}{100}

v = 0.8409 m^3/ kg

from continuity equation

A_1 V_1 = A_2 V_2

\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2

D_2 = D_1 \sqrt{\frac{V_1}{V_2}}

D_2 = 8 \times \sqrt{\frac{12}{8}}

D_2 = 9.797 m

mass flow rate is given as

\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}

\dot m = 717.309 kg/s

the power produced \dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]

\dot W = 28.6 kW

8 0
3 years ago
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
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