Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?

Answer:

Option B

Explanation:

An evaporator along with  cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.

Hence, option B is correct

This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = [tex]\frac{1}{2}[/tex]mv²

we substitute

⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = [tex]\frac{\pi }{4}[/tex]d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

Answer:

Option B

Explanation:

Catalytic convertor with in the exhaust pipes of any vehicle converts harmful gases (such as hydrocarbons (HC), carbon monoxide (CO) and nitrogen oxides (NOx)) produced by combustion of fossil fuels (petrol, diesel etc.) into less harmful/toxic pollutants by acting as a catalyst for redox reaction. A catalytic convertor releases some mild gases from the exhaust like less harmful CO2, nitrogen and water vapour (steam)

Hence, option B is correct

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]

[tex]\rho = 7900 \ kg/m^3[/tex]

[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]

[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]

Here, we can't apply the lumped capacitance method, since Bi > 0.1

[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]

[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]

[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]

However, on a single rod, the energy extracted is:

[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]

Hence, for centerline temperature at 50 °C;

The surface temperature is:

[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]

Answer:

The temperature after a long time will return to 15°C

Explanation:

Determine the temperature of the slab after a very long time

First we calculate the heat flow for m^2 area normal to the surface

= q / A = 650°c - 15°C / ( 1 / h  + L / K )

           = 635°c  / ( 1 / 220 + 0.1 / 110 )  = 116.416 kw/m^2

Total heat content in the slab is calculated as

= m* c * ΔT

= 8530 * A * 0.1 * 380 * ( 650 - 15 )

= 205828.9  kJ/m^2

The temperature will return to 15°C after a long time

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v[tex]_f[/tex] = 0.001057 m³/kg

v[tex]_g[/tex] = 1.0037 m³/kg

u[tex]_f[/tex] = 486.82 kJ/kg

u[tex]_g[/tex] 2524.5 kJ/kg

h[tex]_g[/tex] = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v[tex]_g[/tex]

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m[tex]_{out[/tex] = m₁ - m₂

m[tex]_{out[/tex] = 1.89414  - 0.003985

m[tex]_{out[/tex] = 1.890155 kg

so, Initial internal energy will be;

U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]

U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex]  + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E[tex]_{in[/tex] -  E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]

QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁

QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

probably in it's chromosomes

Answer:

The appropriate answer is "0.0043".

Explanation:

The given values is:

Error probability,

p = 0.01

Received digits,

n = 10

and,

[tex]x\sim Binomial[/tex]

As we know,

⇒  [tex]P(x)=\binom{n}{x}p^xq^{n-x}[/tex]

Now,

⇒  [tex]P(x >1) =1- \left \{ P(x=0)+P(x=1) \right \}[/tex]

⇒                 [tex]=1-\left \{\binom{10}{0}(0.01)^0(0.99)^{10-0}+\binom{10}{0}(0.01)^1(0.99)^{10-1} \right \}[/tex]

⇒                 [tex]=1-0.9957[/tex]

⇒                 [tex]=0.0043[/tex]