A particular Table in a relational database contains 100,000 Data Records/rows, each of which Data Record/row requires 200 bytes. A select statement returns all Data Records/rows in the Table that satisfy an equality search on an attribute. Estimate the time in milliseconds to complete the query when each of the following Indexes on that attribute is used.
A. No Index (Heap File of Data Records)
B. A Static Hash Index (with no overflow buckets/Pages). Assume the cost of applying the hash function is H, negligible.

Answer:

The answer is 0.06027 m³

Explanation:

Solution

Given that:

The first step to take is to determine the initial state of the volume for R-134a refrigerant

Now,

v₁ =V/m

V = the volume of weighted piston cylinder device at the normal state

m = the mass of the R-134a refrigerant

Thus,

We substitute the values 0.14 for V and 0.2 kg for m

Which results in

v₁ = 0.14/0.2

v₁  = 0.7 m₃/kg

The next step is to find the saturated pressure of the R-134a refrigerant from the temperature table of saturated refrigerant R-134a which is equivalent to the normal temperature of 26.4°C.

Thus, by applying the method of interpolation we have the following

P₁ =101.73 - ((101.73-92.76) * (-26-(-26.4)/-26- (-28))

P₁ = 99.936 kPa

So,

The refrigerant in the weighted piston–cylinder device is then heated until the temperature gets to a 100°C

Hence, the temperature and pressure at a state of two becomes

P₂ = 99.936 kPa

T₂ = 100°C

The next step is to determine the specific volume of the refrigerant R-134a at a final state from the super heated refrigerant R-134a which is equivalent to the pressure of 99.936 kPa

v₂ =0.30138 m³ /kg

Now,

we now calculate the final state of  the weighted piston cylinder device

V₂ = mv₂

V₂ = 0.2 * 0.30138

V₂ = 0.06027 m³

Hence ,the final volume of the weighted cylinder piston device is 0.06027 m³

Answer:

The value of relative density is 75 % while that of degree of saturation is 54.82%.

Explanation:

Given Data:

Bulk density of Sandy Soil=[tex]\gamma_b=18.8\ kN/m^3[/tex]

Void Ratio in Densest state=[tex]e_{max}=0.88[/tex]

Void Ratio in Loosest state=[tex]e_{min}=0.48[/tex]

Water content=[tex]w=12\%[/tex]

To Find:

Relative Density=[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%[/tex]

Degree of Saturation=[tex]S=\dfrac{w\times G_s}{e}[/tex]

Now all the other values are given except e. e is calculated as follows

e is termed as In situ void ratio and is given as

[tex]e=\dfrac{\gamma_w \times G_s-\gamma_d}{\gamma_d}[/tex]

Here

γ_w is the density of water whose value is 1

G_s is the constant whose value is 2.65

γ_d is the dry density of the sandy soil which is calculated as follows:

[tex]\gamma_d=\dfrac{\gamma_b}{1+\dfrac{w}{100}}[/tex]

Putting values

[tex]\gamma_d=\dfrac{18.8}{1+\dfrac{12}{100}}\\\gamma_d=16.78\ kN/m^3=1.678 g/cc \\[/tex]

Putting this value in the equation of e gives

[tex]e=\dfrac{1 \times 2.65-1.678}{1.678}\\e=0.579=0.58[/tex]

So the value of Relative density is given as

[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%\\D_R=\dfrac{0.88-0.58}{0.88-0.48} \times 100 \%\\D_R=75 \%[/tex]

So the value of relative density is 75 %

Now the value of degree of saturation is given as

[tex]S=\dfrac{w\times G_s}{e}\\S=\dfrac{12\times 2.65}{0.58}\\S=54.82 \%[/tex]

The value of degree of saturation is 54.82%.

Answer:

The relative density = 0.83 which is equivalent to 83%

The degree of saturation, S = 0.58 which gives 58% saturation

Explanation:

The parameters given are;

Water content W% = 12%

Bulk unit weight, γ = 18.8 kN/m³

Void ratio of  [tex]e_{min}[/tex]  = 0.48

Void ratio of  [tex]e_{max}[/tex] = 0.88

[tex]G_S[/tex] = Constant (As learnt from an answer to the question on the current page) = 2.65 for Sandy soil

[tex]\gamma =\dfrac{W}{V} = \dfrac{W_{w}+W_{s}}{V}[/tex]

Where, V = 1 m³

W = 18.8 KN

Bulk unit weight, γ = [tex]\gamma_d[/tex] × (1 + W)

∴ 18.8 =  [tex]\gamma_d[/tex] × (1 + 0.12)

[tex]\gamma_d[/tex] = 18.8/ (1.12) = 16.79 kN/m³

[tex]\gamma_d =\dfrac{W_s}{V} = \dfrac{W_{s}}{1} = 16.79 \, kN/m^3[/tex]

[tex]W_s[/tex] = 16.79 kN

∴ [tex]W_w[/tex] = 18.8 kN - 16.79 kN = 2.01 kN

[tex]m_w = 2.01/9.81 = 0.205 \, kg[/tex]

Volume of water = 0.205 m³

[tex]\gamma = \dfrac{GS \times \gamma _{w}\times \left (1+w \right )}{1 + e} = \dfrac{GS \times 9.81\times \left (1+0.12 \right )}{1 + e} =18.8[/tex]

e + 1 = 0.58×GS = 0.58×2.65 =

e = 1.54 - 1 = 0.55

The relative density is given by the relation;

[tex]Relative \ density, Dr=\dfrac{e_{max} - e}{e_{max} - e_{min}}[/tex]

[tex]Relative \ density, Dr=\dfrac{0.88 - e}{0.88 - 0.48} = \dfrac{0.88 - 0.55}{0.4} = 0.83[/tex]

The relative density = 0.83

The relative density in percentage = 0.83×100 = 83%

S·e = GS×w = 0.12·2.65

S×0.55 = 0.318

The degree of saturation, S = 0.58

The degree of saturation, S in percentage = 58%.

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

ANSWER:

1) The car is not accelerating, because for the the car to turn while accelerating down the street, it has to deaccelerate (reduce speed) due to CENTRIPETAL FORCE that is acting on the car during the turn. This force acts to bring the car in the direction where it has turned to.

2) You will feel the movement as everyone in the car will tend to fall to the left, while the car makes a right turn. This is because of CENTRIFUGAL FORCE, which is acting to take the car to a left direction, while the car is turning to the right direction.

Answer:

(a) 6.91 mm (b) 160 MPa

Explanation:

Solution

Given that:

E = 200 GPa

The rod length = 48 mm

P =P¹ = 6 kN

Recall that,

1 kN = 10^3 N

1 m =10^3 mm

I GPa = 10^9 N/m²

Thus

The rod deformation is stated as follows:

δ = PL/AE-------(1)

σ = P/A----------(2)

Now,

(a) We substitute the values in equation and obtain the following:

48 * 10 ^⁻3 m =  (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]

Thus, we simplify

A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²

A =0.0375 * 10 ^⁻3 m²

A =37.5 mm²

A = π/4 d²

Thus,

d² = 4A /π

After inserting the values we have,

d = √37.5 * 4/3.14 mm

= 6.9116 mm

or d = 6.91 mm

Therefore, the smallest that should be used is 6.91 mm

(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)

Thus,

σ = P/A

σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²

σ= 160 MPa

Note: I MPa = 10^6 N/m²

Hence the the corresponding normal stress is σ= 160 MPa

Answer is given below

Explanation:

we know that some common types of throttling devices are

so here throttling devices commonly used in refrigeration and air-conditioning because

Answer:

a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s

b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2

c) m' = 2.915 kg/s

Explanation:

Given:-

- The inlet pressure, Pi = 1.0 MPa

- The inlet temperature, Ti = 750 K

- Inlet velocity is negligible

- Steady, Isentropic Flow

- The specific heat ratio of air, k = 1.4

- Exit Mach number, Mae = 2

- The throat area, Ath = 20 cm^2

- Gas constant of air, R = 0.287 KJ / kg.K

Find:-

(a) the throat conditions (static pressure, temperature, density, and mach number)

b) the exit plane conditions

c) the mass flow rate

Solution:-

- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).

- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).

- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:

                             [tex]P_o = 1.0 MPa\\\\T_o = 750 K[/tex]

- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:

                             [tex]p_o = \frac{P_o}{RT_o} = \frac{1000}{0.287*750} = 4.64576\frac{kg}{m^3}[/tex]

- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:

                            [tex]\frac{P^*}{P_o} = 0.5283\\\\P^* = 0.5283*1 = 0.5283 MPa\\\\\frac{T^*}{T_o} = 0.8333\\\\T^* = 0.8333*750 = 624.75 K\\\\ \frac{p^*}{p_o} = 0.6339\\\\p^* = 0.6339*4.64576 = 2.945 \frac{kg}{m^3} \\\\[/tex]

                            [tex]V^* = \sqrt{kRT^*} =\sqrt{1.4*287*624.75} = 501.023 \frac{m}{s}[/tex]

- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:

                           [tex]\frac{P_e}{P_o} = 0.1278 \\\\P_e = 0.1278*1.0 = 0.1278 MPa\\\\\frac{T_e}{T_o} = 0.5556 \\\\T_e = 0.5556*750 = 416.7 K\\\\\frac{p_e}{p_o} = 0.23 \\\\p_e = 0.23*4.64576 = 1.069 \frac{kg}{m^3} \\\\\frac{A_e}{A_t_h} = 1.6875 \\\\A_e =1.6875*20 = 33.75 cm^2\\[/tex]

- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:

                        [tex]V_e = Ma_e*\sqrt{kRT_e} = 2*\sqrt{1.4*287*416.7} = 818.36 \frac{m}{s}[/tex]  

- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows:

                         [tex]m' = p^* A_t_h V^*\\\\m' = 2.945*20*10^-^4*501.023\\\\m' = 2.951 \frac{kg}{s}[/tex]

Answer: Provided in the explanation section

Explanation:

A precise explanation is provided here to make it easy for understanding

(a). The transfer option shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a functoon of frequency.

H(jw) = V₀ (jw) / Vin(jw)

(b).  To determine the transfunction of a filter, we connect a sinusoidal source to be the input part, measure the amplitude and phases of both the input signal and resulting output signal using voltmeters.

Oscilloscope or other instruments, and divide the Output phasor by the input phasor. This is is repeated for all frequencies.

cheers i hope this helped !!

Answer:

a co mam zroic!!

Explanation:

Answer:

1.50 miles/hour is the correct answer to the given question  

Explanation:

As mention in the question engine power = 190 horsepower engine.......eq(1)

we have to converted into the watt as we know that 1 horse power=746 watt

putting this value in eq(1) we get

[tex]Engine \ power\ =190 *746\\\ Engine \ power =146216\ NM/S[/tex]

Drag coefficient = 0.33

Projected area = 26.36 [tex]ft^{2}[/tex]............eq(2)

we have to converted into the [tex]m^{2}[/tex]  as we know that [tex]1m^{2}\ =\ 10.7639\ ft^{2}[/tex]

putting this value in eq(2) we get

Projected area =  [tex]2.448921\ m^{2}[/tex]

Now we to calculated the Drag Force

[tex]F\ =\frac{\ Drag \ coefficient\ *P*\ V^{2}\ *A }{2}[/tex]....................eq(3)

Putting the above value in eq(3) we get

[tex]F=\ \frac{0.33\ * 1.2 * \ V^{2 \ * 2.44821\ *N} }{2} \\F=\ 0.4848V^{2}[/tex]

As we know that

[tex]F=0.4848\ V^{2}\ *V \\F= \frac{\ 0.4848V^{3} NM}{S}[/tex]

As mention in the question

Power of Engine =Drag free force

[tex]146216\ =\ 0.4848\ V^{3}\\ V^{3} =301550.889\\V=67.05M/S[/tex]

We have to converted this value into the miles/hour

1 M/S =2.22369 M/H

Putting this value into the V we get

V=67.05 * 2.2234 M/H

V=150 M/H

Answer:

  4.94 kW

Explanation:

The heat energy produced by the fuel in one hour is ...

  (2.29 L/h)(0.749 kg/L)(43.7 MJ/kg) = 74.954677 MJ/h

Then the power output is ...

  (74.954677 MJ/h)(1 h)/(3600 s) = 20.8207 kJ/s

Multiplying this heat energy by the efficiencies of the processes involved, the output power is ...

  (20.8207 kW)(0.25)(0.95) = 4.94 kW

Answer:

Delta star transformer

Secondary side

[tex]V_L[/tex] = 415 V

[tex]V_P[/tex] = 239.6 V

[tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 800 A

Primary side

[tex]V_L[/tex] = Phase voltage, [tex]V_P[/tex] = 3300 V

Primary phase current = 58.08 A

[tex]I_L[/tex]  = 100.6 A

Turns ratio = 83/(660·√3)

Star-star transformer

Secondary side

[tex]V_L[/tex] = 3300 V

[tex]V_P[/tex] = 1905.26 V

[tex]I_P[/tex] = Line current, [tex]I_L[/tex] =  100.6 A

Primary side

[tex]V_L[/tex] = 11000 V

[tex]V_P[/tex] = 6350.9 V

Line current, [tex]I_L[/tex]   = Phase current, [tex]I_P[/tex]   = 3.018 A

Turns ratio = 3/100.

Explanation:

Delta star transformer

For the star Connection to the 3-Phase load,

Secondary line voltage, [tex]V_L[/tex] = 415 V

Secondary phase voltage, [tex]V_P[/tex] =  [tex]V_L[/tex] ÷ √3 = 415 ÷ √3 = 239.6 V

Secondary phase current, [tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 800 A

For the primary side of the delta star transformer

Line voltage, [tex]V_L[/tex] = Phase voltage, [tex]V_P[/tex] = 3300 V

K =  415/(3300×√3) = 83/(660·√3)

Primary phase current = 83/(660·√3)× 800 A = 58.08 A

Line current, [tex]I_L[/tex]   = Phase current, [tex]I_P[/tex] × √3  = 100.6 A

The turns ratio = 83/(660·√3)

For the star-star transformer, we have;

Secondary line voltage, [tex]V_L[/tex] = 3300 V

Secondary phase voltage, [tex]V_P[/tex] =  [tex]V_L[/tex] ÷ √3 = 3300 ÷ √3 = 1905.26 V

Secondary phase current, [tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 100.6 A

For the primary side of the star-star transformer

Line voltage, [tex]V_L[/tex] = 11000 V

Phase voltage, [tex]V_P[/tex] = 11000 V ÷ √3 = 6350.9 V

K =  3300/11000 = 3/100

Primary phase current = 3/100× 100.6 = 3.018 A

Line current, [tex]I_L[/tex]   = Phase current, [tex]I_P[/tex]   = 3.018 A

The turns ratio = k = 3/100.

Answer:

a) [tex]L = 1.5[/tex]

[tex]L_q = 0.9[/tex]

[tex]W = \dfrac{1 }{8 } \, hour[/tex]

[tex]W_q = \dfrac{3}{40 } \, hour[/tex]

[tex]P = \dfrac{3}{5 }[/tex]

b) The new backacter should be recommended

c) The additional backacter should not be deployed

Explanation:

a) The required parameters are;

L = The number of customers available

[tex]L = \dfrac{\lambda }{\mu -\lambda }[/tex]

μ = Service rate

[tex]L_q[/tex] = The number of customers waiting in line

[tex]L_q = p\times L[/tex]

W = The time spent waiting including being served

[tex]W = \dfrac{1 }{\mu -\lambda }[/tex]

[tex]W_q[/tex] = The time spent waiting in line

[tex]W_q = P \times W[/tex]

P = The system utilization

[tex]P = \dfrac{\lambda }{\mu }[/tex]

From the information given;

λ = 12 trucks/hour

μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour

Plugging in the above values, we have;

[tex]L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5[/tex]

[tex]P = \dfrac{12 }{20 } = \dfrac{3}{5 }[/tex]

[tex]L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9[/tex]

[tex]W = \dfrac{1 }{20 -12 } = \dfrac{1 }{8 } \ hour[/tex]

[tex]W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour[/tex]

(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;

μ = 60/1.5 trucks/hour = 40 trucks/hour

[tex]P = \dfrac{12 }{40 } = \dfrac{3}{10}[/tex]

[tex]W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour[/tex]

[tex]W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour[/tex]

λ = 12 trucks/hour

Total cost = [tex]mC_s + \lambda WC_w[/tex]

m = 1

[tex]C_s[/tex] = GH¢ = 1300

[tex]C_w[/tex] = 400

Total cost with the old backacter is given as follows;

[tex]1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00[/tex]

Total cost with the new backacter is given as follows;

[tex]1 \times 1300 + 12 \times \dfrac{1}{38} \times 400 = \$ 1,426.32[/tex]

The new backacter will reduce the total costs, therefore, the new backacter is recommended.

c)

Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour

[tex]\therefore W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour[/tex]

Total cost with the one backacter is given as follows;

[tex]1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00[/tex]

Total cost with two backacters is given as follows;

[tex]2 \times 1000 + 12 \times \dfrac{1}{38} \times 400 = \$ 2,126.32[/tex]

The additional backacter will increase the total costs, therefore, it should not be deployed.

Answer:

41.015

Explanation:

Solution

Given that:

Th first step to take is to find  the longitudinal stress in the cylinder

σl = PD/4t

P = the pressure

D = the diameter

t = the thickness

Thus,

σl = 1.25 * ^ 6 * 0.45 / 4 * 6 * 10 ^ ⁻3

=23.475Mpa

Now. we find the hoop stress in the cylinder∠

σh = PD/2t

σh = 1.25 * ^ 6 * 0.45/ 2 * 6 * 10 ^ ⁻3

σh  =46.875 Mpa

Then

we find the normal stress in the line of the 30° angle with the longitudinal axis stated below:

σab = σh + σl/2 + ( σh - σl/2) cos 2θ + t sin 2θ

So,

σab  =46.875 + 23.4375/2 + ( 46.875 - 23.4375/2) cos 2(30°) + 0

σab= 41.015

Therefore the normal stress to line A-B is 41.015

Answer: The summation of the reaction forces is equal to the applied load.

Explanation:

When solving an indeterminate structure, it is important for one to satisfy the force-displacement requirements, compatibility, and the equilibrium of the structure.

With regards to the above question, for a statically indeterminate axially loaded member, the summation of the reaction forces will be equal to applied load. Here, the summation of the reactive forces will then be zero. This is a condition that is necessary to solve the unknown reaction forces.

Answer:

No. of Machines Required = 47

Explanation:

First, we calculate the number of parts that can be manufactured by one machine during the shift. For that purpose, we use following formula:

No. of Parts Manufactured by 1 Machine = Total Operating Time/Time taken to perform Operation

where,

Total Operating Time = 8 h - 1 h = (7 h)(60 min/h) = 420 min

Time taken to perform operation = 20 min

Therefore,

No. of Parts Manufactured by 1 Machine = 420 min/20 min

No. of Parts Manufactured by 1 Machine = 21

Now, we will calculate the no. of non-defective parts manufactured by 1 machine. Since, it is given that one-fifth of the parts manufactured by machine are defective. Therefore, the non-defective parts will be: 1 - 1/5 = 4/5 (four-fifth).

No. of Non- Defective Parts Manufactured by 1 Machine = N = (4/5)(21)

N = 16.8 = 16 (Since, the 17th part will not be able to complete in time)

So, the no. of machines required to produce 750 non-defective parts is given by:

No. of Machines Required = No. of non-defective parts required/N

No. of Machines Required = 750/16

No. of Machines Required = 46.9

No. of Machines Required = 47 (Since, 46 machines will not be able complete the job)

Answer:

Continuous insulation helps in eliminating the  necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.

Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.

Explanation:

Solution:

The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.

Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.

By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building to withstand the test of time without the need to upgrade and repair

Now inside the placement of continuous insulation in the envelope is needed as follows:

Answer:

The volume of sludge after filtration is 0.914 m

Explanation:

Solution

Given that:

We have to find the Volume of sludge after filtration

Now,

The  Sludge concentration is = 32%,

The sludge volume = 100 m,

The sludge concentration after filtration = 35%

Then,

The mass balance equation is stated below

Cin∀in = Cout∀out

Now,

We Solve for ∀out

∀out =Cin∀in/Cout

By substituting the values

∀out = (0.32)(100 m)/(0.35) = 0.914 m

Answer:

ESTABLISHED

Explanation:

What is TCP?

A Transmission Control Protocol (TCP) is a communication protocol which allows the exchange of data between computers in a network.  

When a Transmission Control Protocol connection is up and running meaning that both sides can send and receive data then the corresponding TCP socket states is known as "ESTABLISHED".

The most common socket states are:

LISTEN:

Before a TCP connection is made, there needs to be a server with a listener that will listen on incoming connection request.

ESTABLISHED:

When a TCP connection is up and running meaning that both sides can send and receive data.

CLOSED:

The CLOSED state means that there is no TCP connection.

There are a total of 11 TCP socket states:

1. LISTEN

2. SYN-SENT

3. SYN-RECEIVED

4. ESTABLISHED

5. FIN-WAIT-1

6. FIN-WAIT-2

7. CLOSE-WAIT

8. CLOSING

9. LAST-ACK

10. TIME-WAIT

11. CLOSED

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

Answer: Fossil remains of the same land-dwelling animal.

Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.

Answer:

Fossil remains of the same land-dwelling animal

Explanation:

Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.

Answer:

Check below for answers

Explanation:

Matlab code:

function[Pa] = Psi-ToPa(psi)

        Pa = psi * 6894.75728;

end

a) To convert 120 psi to units of Pa, just call the function Pa using the command:

          Psi-ToPa(120)

ans =

       8.2737e+05

b) To convert 3000 psi to units of Pa, just call the function Pa using the command:

          Psi-ToPa(3000)

ans =

       2.0684e+07

Answer:

Pillar drill are small and used for woodworking while Radial arm pillar is mounted on a very large column.

Answer:

[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]

Explanation:

Given that :

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)

i.e

k = 0.14 W/mK

∝ = 6.35 × 10⁻⁸ m²/s

L = 0.01 m

[tex]B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857[/tex]

We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number [tex]F_o > 0.2[/tex]

⇒  [tex]\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)[/tex]

From Table 5.1 ; at [tex]B_1[/tex] = 14.2857

[tex]C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad[/tex]

[tex]\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}][/tex]

[tex]In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}[/tex]

[tex]-1.9399=-0.001350 *t_f[/tex]

[tex]t_f = \dfrac{-1.9399}{-0.001350}[/tex]

[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]

Answer: Provided in the explanation section

Explanation:

Transactional Leadership - This leadership style is mainly focused on the transactions between the leader and employees. If the employees work hard, achieve the objectives and deliver the results, they are rewarded through bonuses, hikes, promotions etc. If the employees fail to achieve the desired results, they are punished by awarding lower ratings in the performance appraisal, denying opportunities etc.

In this style, leader lays emphasis on the relation with the followers.

It is a reactive style where the growth of the employee in the organization completely depends on the performance with respect to the activities and deliverables.

It is best suited for regular operations and for a settled environment by developing the existing organizational culture which is not too challenging.

It is a bureaucratic style of leadership where the leader concentrates on planning and execution rather than innovation and creation.

A transactional leader is short-term focused and result oriented. He/she doesn't consider long-term strategic objectives regarding the organization's future.

cheers i hope this helped !!