A micromechanical resonator is to be designed to have a Q factor of 1000 and a natural frequency of 2 kHz. Determine the system-damping factor and the system bandwidth.

Answer:

≈ 0.516 m^3/hr

Explanation:

Inlet of compressor = 200 SCMH

sheer standard conditions = 1 atm and 288.5 K

For oxygen :

critical pressure(Pc) = 49.8 atm

critical temperature Tc = 154.6 K

hence at compressor inlet

Tr = T / Tc = 288.5/154.6 = 1.866

Pr = P / Pc = 1 / 49.8 = 0.0204

Z1 ( from compressibility chart ) = 0.98

at compressor outlet

P2 = 500 bar = 500*0.9869 = 493.45 atm  , T2 = 360 k

hence : Pr = P / Pc = 493.45 / 49.8 = 9.91

            Tr = T / Tc = 360 / 154.6 = 2.33

Z2 ( from compressibility chart ) ≈ 1

V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)

                                         = 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)

                                         = 0.516 m^3/hr

Answer:

International Film Festival

Judging the best best film:

a. The probability that out of five judges (random sample),  three are in favor of film FA is:

= 33%.

b. The demerits of classical probability are:

1. Classical probability can only be used with events that have definite numbers of possible outcomes.  

2. Classical probability can only handle events where each outcome is equally likely.

3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.

Explanation:

a) Number of judges = 11

Number of judges in favor of FA film = 6

Number of judges in favor of FB film = 5

Probability of judges in favor of FA film = 6/11

Probability of judges in favor of FB film = 5/11

Random sample of judges = 5

Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11

= 18/55

= 33%

b) Classical probability is the simple probability showing that each event has equal chance of happening.  It can be contrasted with empirical probability that is obtained from experiments.

Answer:

1.887 m

Explanation:

(15 *pi)/180

= 0.2618 rad

Polar moment

= Pi*d⁴/32

= (22/7*20⁴)/32

= 15707.96

Torque on shaft

= ((22/7)*20³*110)/16

= 172857.14

= 172.8nm

Shear modulus

G = 79.3

L = Gjθ/T

= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8

= 1.887 m

The length of the bar is therefore 1.887 meters

Answer:

The Space Needle is a cut away with minimal residual deflection due to load transfer.

Answer:

Explanation:

[tex]496=\frac{480\times 60}{demand}[/tex]

demand per day = 58 pans

Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.

Hence the cycle time would be the greater time of the two operations.

cycle time = 450 seconds

[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]

[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]

[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)

Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.

b)

Required daily production is 58 pans

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

Answer:

[tex]T_o=82.1sec[/tex]

Explanation:

From the question we are told that:

Lost Time [tex]t=12secs[/tex]

Sum of critical flow ratios [tex]X=0.72[/tex]

Generally the Webster Method's equation for Optimum cycle time is is mathematically given by

 [tex]T_o=\frac{1.5t+5}{1-x}[/tex]

 [tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]

 [tex]T_o=82.1sec[/tex]

Answer:

There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. ... Pie charts to show you how a whole is divided into different parts.

#carryonlearning

Answer: the attached picture is the answer.

Explanation:

Assuming:

the switch position connect to 1, hence 4.5V exist at across lamp1

the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2

the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.

Answer:

1.49 mm

Explanation:

The modulus of elasticity, Y = stress/strain = σ/ε

σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.

ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,

So, Y = σ/ε

Y = F/A ÷ ΔL/L

Y = FL/AΔL

making the change in length ΔL subject of the formula, we have

ΔL = FL/AY

substituting the value of A into the equation, we have

So, ΔL = FL/(πd²/4)Y

ΔL = 4FL/πd²Y

Since Y = 200 GPa = 200 × 10⁹ Pa

Substituting the values of the variables into the equation, we have

ΔL = 4FL/πd²Y

ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]

ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]

ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]

ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N

ΔL = 0.149 × 10⁻² m

ΔL = 1.49 × 10⁻³ m

ΔL = 1.49 mm

The change in length of the circular rod is 1.49 mm

Answer: Plot of  [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope  of 4.

Explanation:

Given

Equation is [tex]y=10^{4x}[/tex]

Taking log both sides

[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]

It resembles with linear equation [tex]y=mx+c[/tex]

Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.

Answer:

what about it?

Explanation:

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

Answer:

a. Rockwell              3. hardness

b. Instron                 2. stress vs strain

c. Charpy                 1. impact strength

d. Fatigue                4. Endurance Limit

e. Brinell                  3. hardness

f. Izod                      1. impact strength

Explanation:

Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.

Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.

Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.

Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.

Answer:

Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.

Answer:

(a) the reversible power output of turbine is 5810 kw

(b) The second-law efficiency of he turbine = 86.05%

Explanation:

In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.

h1 = 3658 kJ/kg s1=7.167 kJ/kgK

In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state

h2 = 2682kl/kg S2= 7.694 kJ/kg

Assuming that the energy balance equation given  

Wout=m [h1-h2+(v1²-v2²) /2]

Let

W =5 MW

V1= 80 m/s  V2= 140 m/s

h1 = 3658kJ/kg  h2 = 2682 kJ/kg

∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)

m = 5.158kg/s

Consider the energy balance equation given  

Wrev,out =Wout-mT0(s1-s2)

Substitute Wout =5 MW m = 5.158kg/s 7

s1=  7.167 kJ/kg-K            s2= 7.694kJ/kg-K and 25°C .

Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)

= 5810 kW

(a) Therefore, the reversible power output of turbine is 5810 kw.

The given values of quantities were substituted and the reversible power output are calculated.

(b) Calculating the second law efficiency of the turbine:  

η=Wout/W rev,out

Let Wout =  5 MW and Wrev,out = 5810 kW  

η=(5 MW x 1000 kW)/(1 MW *5810)  

η= 86.05%

Answer:

[tex]v_2=549.2 m/s\\[/tex]

Explanation:

Given:

[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]

Solution:

For [tex]Co_2[/tex] y=1.4

Since Nozzle is adiababic

So,

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]

Now,

[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]

Answer:

Explanation:

[tex]r_2=[/tex]∞

[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]

[tex]q=2\pi kD.[/tex]ΔT--------(1)

[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]

[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]

By equation (1) and (2)

[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT

[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)

From equation (3) and (4)

So for sphere [tex]R_a[/tex]→0

Answer:

35 miles

Explanation:

When you are driving a truck that has an air brake system you have to keep in mind that when driving down a steep downgrade, the truck will automatically accelerate due to the inclination of the road, so in order to keep the speed to a controllable situation, you need to activate the service brake until you've reached the 35 miles per hour mark.

Answer: [tex]10.631\times 10^3\ N/m^2[/tex]

Explanation:

Given

Discharge is [tex]Q=12.5\ L[/tex]

Diameter of pipe [tex]d=150\ mm[/tex]

Distance between two ends of pipe [tex]L=800\ m[/tex]

friction factor [tex]f=0.008[/tex]

Average velocity is given by

[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]

Pressure difference is given by

[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]

Answer:

1.c

2.c

3.c

Explanation:

James is a  pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.

When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.

The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.

The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.

Thus, option C, C, and C are correct.

For more information about point of a flight suit, click here:

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Answer:

there are generally two types of toilet bowl types- round and elongated.

Answer:

embroidery scissor

Explanation:

is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.

hope this helps

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Where,

is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ =  

A₂ = Area at the throat

A₂ =  

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

or

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Hope This Helps :D

Answer:

1. Drawing grid.

2. Orthogonal.

3. Geometries.

4. CTRL+N.

5. Thirteen (13).

Explanation:

CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.

Some of the features of an AutoCAD software are;

1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.

2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.

3. Tangent is a point where two geometries meet at just a single point.

4. If you want to create a new drawing, simply press CTRL+N for the short cut key.

5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.

Answer:

Explanation:

B. you would grab the plug closest to the outlet

If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.

Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.

Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.

Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket

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Answer:

-50 W

Explanation:

The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m

Since Q = kA(T₂ - T₁)/d ,

Q = kπdL(T₂ - T₁)/d

substituting the values of the variables into the equation, we have

Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C  - 46 °C )/0.0024 m

Q = 0.01275π Wm/K(-3 K )/0.0024 m

Q = -0.03825π Wm/0.0024 m

Q = -0.1202 Wm/0.0024 m

Q = -50.07 W

Q = -50 W

So, the heat transfer rate is -50 W meaning heat transfer out of the tube.