Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

Answer:

a)  Vr = - a^2/r cosθ  + aß / r

    Vθ = 1/r [ -a^2/r * sinθ ]

b) attached below

Explanation:

potential function

Ø= a^2 /r  cosØ + aßlnr ----- ( 1 )

a = radius ,  ß = constant

a) Expressions for Vr and Vθ

Vr =  dØ / dr  ----- ( 2 )

hence expression : Vr = - a^2/r cosθ  + aß / r

Vθ = 1/r dØ / dθ ------ ( 3 )

back to equation 1

dØ / dr = - a^2/r sinθ + 0  --- ( 4 )

Resolving equations 3 and 4

Vθ = 1/r [ -a^2/r * sinθ ]

b) expression for stream function

attached below

Answer:

Walls Built Using Bricks and Wood

The brick house is more energy-efficient than the one built with wood.

Explanation:

Because of their high thermal mass, which gives bricks the ability to absorb heat and release it over time, bricks remain more energy-efficient than other building materials, including wood.  In summer, bricks leave your home cool.  In winter, they make it warm.  With these two advantages provided by bricks over other building materials, bricks are the most energy-efficient building material.

Answer:

a) 1.9%

b) T2s = 505.5 k = 232.5°C

Explanation:

P1 = 95 kPa

T1 = 27°C  = 300 k

P2 = 600 kPa

T1 = 277°c  = 550 k

Table used : Table ( A - 17 ) Ideal gas properties of air

a) determining the isentropic efficiency of the compressor

Л = ( h2s - h1 ) / ( h2a -  h1 ) ---- ( 1 )

where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k

To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)

 Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306  hence; h2s = 500.72 kJ/kg

back to equation1

Л = 0.019 = 1.9%

b) Calculate the exit temperature of the air if compressor is reversible

if compressor is reversible the corresponding exit temperature

T2s = 505.5 k = 232.5°C

given that h2s = 500.72 kJ/kg

Answer:

D. Turn on the light source and the spectrophotometer.

Explanation:

A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.

Answer:

[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]

Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):

[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]

Now, the isentropic efficiency of the turbine can be given as follows:

[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]

Answer:

Hence the given statement is false.

Explanation:

For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.

The speed within the test section is decided by the planning of the tunnel.  

Thus by adjusting the pressure difference won't change the worth of test section velocity.

Answer:

The given statement is false .

Answer:

The right answer is "2.2099 m³".

Explanation:

Given:

Mass,

m = 5 kg

Temperature,

T = 35℃

or,

  = 35 + 273

Pressure,

P = 200 kPa

Gas constant,

R = 0.2870 kj/kgK

By using the ideal gas equation,

The volume will be:

⇒ [tex]PV=mRT[/tex]

or,

⇒    [tex]V=\frac{mRT}{P}[/tex]

By substituting the values, we get

          [tex]=\frac{5(0.2870)(35+273)}{200}[/tex]

          [tex]=\frac{441.98}{200}[/tex]

          [tex]=2.2099 \ m^3[/tex]

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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:

[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:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

Determine the design flow ins ft^3/s

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

    = 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 )  = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

                  = 1709.07 ft^3/s

Answer:

Interval:  x∈ ( 0, ∞ )

There are no transient terms

Explanation:

x (dy/dx) – y= x^2sinx

Attached below is the detailed solution of the Given problem

There are no transient terms found in the general solution

Interval:  x∈ ( 0, ∞ )

Answer:

b. 828

Explanation:

UDL = 2.6 - 0.3 [5.5] - 0.02 [ 21 ] = 0.53

USR = 0.7 - 0.3 [5.5 / 2 ] - 0.04 [ 21 ] = -0.69

UB = -0.3 [ 1.25 ] - 0.01 = -0.645

Psr = [tex]\frac{e^{-0.53} }{e^{-0.53} + e^{-0.69} + e^{-0.645} }[/tex]

Solving the equation we get 0.184.

Number of people who will take shared ride is:

0.184 * 4500 = 828 approximately.

Answer:

Explanation:

The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory

Answer:

True

Explanation:

Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.

We have this cylinder equation as

[T1-T2 / ln(r1-r2)]2πKL

The radial distance is r2-r1

The gradient of temperature is T1-T2

From the equation,

The temperature gradient has a direct and proportional relationship to radial distance

T1-T2 ∝ ln(r2-r1)

1/T1-T2 = k(r2-r1)

This inverse relationship above confirms that the statement is true

Answer:

H(s) = 20 / [ 1 + s / 10^5 ]^2

Explanation:

Given data:

cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20 dB

new nominal passband gain at cutoff = 14 dB

Represent the transfer function H(s)

The attenuation rate show that there are two(2) poles

H(s) = k / [ 1 + s/Wc ]^2  ----- ( 1 )

where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20

Input values into equation 1

H(s) = 20 / [ 1 + s / 10^5 ]^2

Answer:

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Explanation:

j

Answer:

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

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain =  1.855 *10^-5   or  18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

                   L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82²  = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

                           Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

                                 = 0.1855 * 10³  * 66 / 10000 * 10³

                                 = 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

                          =0.00001855 = 1.855 *10^-5

                         = 18.55μ

∴ Change in rod   Δd/ d = μ ΔL/L

                           = (0.33) 1.855 *10^-5 * 0.82

                           = 5.02 * 10^ -6 in

Answer:

the current consumed is 3.3 A

Explanation:

Given;

resistance, R = 30 ohms

inductance, L = 200 mH

Voltage supply, V = 230 V

frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

The current consumed is calculated as;

[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]

Therefore, the current consumed is 3.3 A

Answer: Can these nuts fit in your mouth?

Explanation:

im just here for the points >:)

Answer:

Input power = 465.63 W

current = 1.94 A

Explanation:

we have the following data to answer this question

V = 9130

i = 0.051

the input power = VI

I = 51.0 mA = 0.051

= 9130 * 0.051

= 465.63 watts

the current = 465.63/240

= 1.94A

therefore the input power is 465.63 wwatts

while the current is 1.94A

the input power is the same thing as the output power.

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:

(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:

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: 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