Assign deliveryCost with the cost (in dollars) to deliver a piece of baggage weighing baggageWeight. The baggage delivery service charges twenty dollars for the first 50 pounds and one dolllar for each additional pound. The baggage delivery service calculates delivery charge by rounding to the next pound. Assume baggageWeight is always greater than 50 pounds.

Answer:

Load to fracture = 17212.5 N

Explanation:

Firstly we need to find the flexural strength of the material. We do this by using the date obtained by the test of the circular sample and the equation shown below

σ (flexural strength) = F x L/ (π x r³), where F is the load at fracture, L is the distance between the two load points, r is the radius of the circular cross section. Substituting the values we get

σ (flexural strength) = 3000 x 40 x 10⁻³/ (π x 5 x 10⁻³)³ = 306 x 10⁶ N/m²

Now, the flexural stress for a square sample is written as

σ (flexural strength) = 3 x F x L/ 2 x b x d² , and since in a square sample breadth = width the equation becomes

σ (flexural strength) = 3 x F x L/ 2 x d³

Solving for F, we get

F = σ (flexural strength) x 2 x d³ /3 x L

For same material we will use the σ (flexural strength) as calculated above, furthermore L remains the same and d = 15 x 10⁻³. Solving for F

F = 306 x 10⁶ x 2 x (15 x 10⁻³)³ /3 x 40 x 10⁻³ = 17212.5 N

The load required for this specimen to fracture would be 17212.5 N.

Answer:

Please see attachment.

Explanation:

To calculate the downstream velocity of air in a pipe, use the principle of conservation of mass and the continuity equation, which relates the product of cross-sectional area and velocity at two points in a system. The cross-sectional areas of the pipe at both points are computed, and using the continuity equation, the downstream velocity can be found. Convert diameters to meters, calculate areas, and apply the equation.

The student's question involves calculating the velocity of air at a downstream location in a pipe system, using the principles of fluid dynamics. First, to find the velocity of gas at the downstream point, we can invoke the principle of conservation of mass, specifically the continuity equation for an incompressible fluid. This equation states that the product of the cross-sectional area (A) and velocity (V) at one point must equal the product of A and V at any other point in the system, assuming the density of the fluid remains constant.

First, we must calculate the cross-sectional areas at both points:

Using the given velocities and the continuity equation A1 x V1 = A2 x V2, we can solve for the downstream velocity, V2.

Given:

We can convert these diameters to meters, find the areas, and then apply the continuity equation to find the downstream velocity. However, it seems the student is not required to solve for changes in air density due to temperature and pressure variations, hence we're treating the air as incompressible.

The velocity of the gas at the downstream point is approximately 57.3 m/s, considering the conservation of mass flow rate and the given conditions of temperature and pressure.

To find the velocity of the gas at the downstream point, we need to apply the principle of conservation of mass for a compressible flow, which states that the mass flow rate must be constant throughout the pipe. The mass flow rate [tex](\(\dot{m}\))[/tex] can be expressed using the following equation:

[tex]\[\dot{m} = \rho \cdot A \cdot v\][/tex]

where:

- [tex]\(\rho\)[/tex] is the density of the air,

- A is the cross-sectional area of the pipe,

- v is the velocity of the air.

First, we'll calculate the density of the air at both the upstream and downstream points using the ideal gas law:

[tex]\[\rho = \frac{P}{R \cdot T}\][/tex]

where:

- P is the absolute pressure,

- R is the specific gas constant for air [tex](\(R = 287.05 \, \text{J/(kg·K)}\))[/tex],

- T is the temperature in Kelvin.

Let's start by converting the gauge pressures to absolute pressures. Since gauge pressure is the pressure relative to atmospheric pressure, we need to add atmospheric pressure (1.01325 bar) to each gauge pressure:

Upstream Conditions:

- Gauge pressure: [tex]\(1.80 \, \text{bar}\)[/tex]

- Absolute pressure: [tex]\(P_1 = 1.80 \, \text{bar} + 1.01325 \, \text{bar} = 2.81325 \, \text{bar} = 281325 \, \text{Pa}\)[/tex]

- Temperature: [tex]\(27.0^\circ \text{C} = 27.0 + 273.15 = 300.15 \, \text{K}\)[/tex]

- Pipe diameter: [tex]\(7.00 \, \text{cm} = 0.07 \, \text{m}\)[/tex]

- Velocity: [tex]\(v_1 = 30.0 \, \text{m/s}\)[/tex]

Downstream Conditions:

- Gauge pressure: [tex]\(1.63 \, \text{bar}\)[/tex]

- Absolute pressure: [tex]\(P_2 = 1.63 \, \text{bar} + 1.01325 \, \text{bar} = 2.64325 \, \text{bar} = 264325 \, \text{Pa}\)[/tex]

- Temperature: [tex]\(60.0^\circ \text{C} = 60.0 + 273.15 = 333.15 \, \text{K}\)[/tex]

- Pipe diameter: [tex]\(5.50 \, \text{cm} = 0.055 \, \text{m}\)[/tex]

Calculate the density at both points:

[tex]\[\rho_1 = \frac{P_1}{R \cdot T_1} = \frac{281325 \, \text{Pa}}{287.05 \, \text{J/(kg·K)} \times 300.15 \, \text{K}} \approx 3.268 \, \text{kg/m}^3\][/tex]

[tex]\[\rho_2 = \frac{P_2}{R \cdot T_2} = \frac{264325 \, \text{Pa}}{287.05 \, \text{J/(kg·K)} \times 333.15 \, \text{K}} \approx 2.769 \, \text{kg/m}^3\][/tex]

Calculate the cross-sectional area of the pipes:

[tex]\[A_1 = \pi \left(\frac{0.07 \, \text{m}}{2}\right)^2 \approx 0.00385 \, \text{m}^2\][/tex]

[tex]\[A_2 = \pi \left(\frac{0.055 \, \text{m}}{2}\right)^2 \approx 0.00238 \, \text{m}^2\][/tex]

Mass flow rate at the upstream point:

[tex]\[\dot{m} = \rho_1 \cdot A_1 \cdot v_1 = 3.268 \, \text{kg/m}^3 \times 0.00385 \, \text{m}^2 \times 30.0 \, \text{m/s} \approx 0.378 \, \text{kg/s}\][/tex]

Velocity at the downstream point:

Using the conservation of mass flow rate:

[tex]\[\dot{m} = \rho_2 \cdot A_2 \cdot v_2\][/tex]

Solving for [tex]\(v_2\):[/tex]

[tex]\[v_2 = \frac{\dot{m}}{\rho_2 \cdot A_2} = \frac{0.378 \, \text{kg/s}}{2.769 \, \text{kg/m}^3 \times 0.00238 \, \text{m}^2} \approx 57.3 \, \text{m/s}\][/tex]

So, the velocity of the gas at the downstream point is approximately [tex]\(57.3 \, \text{m/s}\).[/tex]

Answer:

a. Park should be adjacent to 48th Street, b. Difference in noise level = 707dBa, c. Yes

Explanation:

Data given for Avenue A

Cars = 496, Medium Truck = 52, Heavy Truck = 19, Buses = 10

Data given for 48th Street

Cars = 822, Medium Truck = 22, Heavy Truck = 8, Buses = 3

Consider the PCEs to be Cars = 1, Medium Truck = 13, Heavy Truck = 47, Buses = 18

a. Noise Level = Number of vehicles x PCE

For Avenue A

Noise level = (496 x 1) + (52 x 13) + (19 x 47) + (10 x 18) = 2245dBa

For 48th Street

Noise level = (822 x 1) + (22 x 13  + (8 x 47) + (3 x 18) = 1538dBa

The park should adjacent to 48th street as it is quieter than Avenue A

b. Let the Setback be 50ft. We know that the reduction of noise for 100ft = 5-8 dBa, hence

For Avenue A Noise Reduction due to 50 ft = (8/100) x 50 = 4dBa

Noise at Setback distance = 2245 - 4 = 2241dBa

Considering the same setback the noise at 48th street would be = 1538 - 4 = 1534 dBa

The difference is noise level between the two sides would be = 2241 - 1534 = 707 dBa

c. Yes the developer concerns are valid because there is a clear difference in noise levels of the two sites. This can be seen even after setting the same Setback. Locating the park next to Avenue A will cause serious noise problems.

The output DC voltage after using a full-wave rectifier on an AC source with a Vrms of 60V, considering a forward voltage drop of 0.7V and negligible ripple, is approximately 53.7V.

To convert an AC voltage to DC voltage using a full-wave rectifier, you first need to understand the relationship between the root mean square (RMS) value of the AC voltage and its peak value. The RMS value is given as Vrms = 60V. For a full-wave rectified sine wave, the peak voltage (Vp) is Vrms × √2. Therefore, Vp = 60V × √2 ≈ 84.85V. However, because of the rectifier's forward voltage drop (Vf), we have to subtract this from the peak voltage to get the peak DC voltage (Vdc_peak). Therefore, Vdc_peak = Vp - Vf ≈ 84.85V - 0.7V ≈ 84.15V.

However, to find the average or output DC voltage (Vdc_avg) from a full-wave rectifier, you'd multiply the peak DC voltage by 2/π (since it's a sinusoidal wave), so Vdc_avg ≈ (2/π) × 84.15V ≈ 53.7V. Thus, the output DC voltage after using a full-wave rectifier directly on the line voltage, assuming negligible ripple, is approximately 53.7V.

Answer:

y  ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

[tex]v =1/n R^{2/3} s^{1/2}[/tex]

where R is hydraulic radius

S = bed slope

[tex]Q = Av =A 1/n R^{2/3} s^{1/2}[/tex]

[tex]A = 1/2 \times (B+B+4y) \times y =(B+2y) y[/tex]

[tex]R =\frac{A}{P}[/tex]

P is perimeter [tex]=  (B+2\sqrt{5} y)[/tex]

[tex]R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}[/tex]

[tex]Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}[/tex]

solving for y[tex]100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}[/tex]

solving for y value by using iteration method ,we get

y  ≈ 2.5

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

[tex]V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}[/tex]

V = 7.07  m/s

from bernou;ii's theorem we have

[tex]\frac{p_1}{\gamma}  +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma}  +\frac{V_2^2}{2g} + z_2 + h_l[/tex]

as we know pipe is horizontal and with constant velocity so we have

[tex]\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}[/tex]

[tex]P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma[/tex]

[tex]135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81[/tex]

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have [tex]\epsilon  = 0.15 mm[/tex]

[tex]\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025[/tex]

reynold number is

[tex]Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}[/tex]

Re = 378750

from moody chart

[tex]For Re = 378750 and \frac{\epsilon}{d} = 0.0025[/tex]

[tex]f_{new} = 0.025[/tex]

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have [tex]\frac{\epsilon}{d} =0.006[/tex]

[tex]\epsilon = 0.006 \times 60[/tex]

[tex]\epsilon = 0.36 mm[/tex]

thus pipe is old one with increased roughness

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

Answer:

5778.86W

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2. find the specific weights of the water and its density using T = 20C, using thermodynamic tables

note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

3. use the Bernoulli equation to find the height of the pump's power, taking into account that the potential and kinetic energy changes are insignificant

4. find the ideal pump power

5. find the real power of the pump using the efficiency equation

Answer:

generalizing

Explanation:

We all have a generalization system that operates as an autopilot, allowing us to be fast and consistent with our own identity. Thanks to these we package and label all the information with which we are bombed every second, to immediately think and act.

Otherwise, if we pay attention to each data individually, however tiny, every minute of our lives would become an exhausting and extremely slow process of analyzing and digesting, leaving us so overloaded to the point of collapse and not being able to function more mentally.

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

[tex]\rho = \frac{m}{V}[/tex]

Where

m = Mass

V = Volume

For state one we know that

[tex]\rho_1 = \frac{m_1}{V}[/tex]

[tex]m_1 = \rho_1 V[/tex]

[tex]m_1 = 1.18*1[/tex]

[tex]m_1 = 1.18Kg[/tex]

For state two we have to

[tex]\rho_2 = \frac{m_2}{V}[/tex]

[tex]m_2 = \rho_2 V[/tex]

[tex]m_1 = 7.2*1[/tex]

[tex]m_1 = 7.2Kg[/tex]

Therefore the total change of mass would be

[tex]\Delta m = m_2-m_1[/tex]

[tex]\Delta m = 7.2-1.18[/tex]

[tex]\Delta m = 6.02Kg[/tex]

Therefore the mass of air that has entered to the tank is 6.02Kg

Answer:

% od sidelap is [tex]P_w = 0.251\ or\ 25%[/tex]

Explanation:

Given data:

flying height is 7000 ft

focal dimension is 9× 9

focal length f = 88.90 mm

side width is x = 13,500 ft

we know side width is given as

[tex]x = (1- P_w) \frac{w}{s}[/tex]

where s is scale and given as [tex]s = \frac{f}{h}[/tex]

[tex]13,500 = (1- P_w) \times \frac{\frac{9}{12}}{\frac{0.29}{7000}}[/tex]

[tex](1- P_w) = 0.748[/tex]

[tex]P_w = 0.251\ or\ 25%[/tex]

Answer:

[tex]R_{min} = 4.84\times 10^{-3} m[/tex]

Explanation:

Given data:

Applied force 750 N

Flexural strength is 105 MPa

separation is 50 mm = 0.05 m

flexural strength is given as

[tex]\sigma_f = \frac{FL}{\pi R_{min}^3}[/tex]

solving for R so we have

[tex]R_{min} = [\frac{FL}{\pi \sigma_f}]^{1/3}[/tex]

plugging all value to get minimum radius

[tex]R_{min} = [\frac{750 \times 0.05}{\pi 105 \times 10^6}]^{1/3}[/tex]

[tex]R_{min} = 4.84\times 10^{-3} m[/tex]

Answer:

Q=20.1kW

recirculation mode=Q=9.045kW

Explanation:

Hi!

To solve this problem you must use the first law of thermodynamics that establishes the energy that enters a system is the same that must come out. Given the above we have the following equation

Q = mCp (T2-T1)

WHERE

Q = Heat removed to air

m = mass air flow = 1kg / s

Cp = specific heat of air = 1.005KJ / kg °C

T2 = air inlet temperature

T1 = air outlet temperature=15°C.

for the first case

Q=(1kg/s)(1.005KJ/kg°C)(35°C-15°C)

Q=20.1kW

for the second case( recirculation mode)

Q=(1kg/s)(1.005KJ/kg°C)(24°C-15°C)

Q=9.045kW

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

[tex]C=\frac {2E\gamma}{\pi \sigma_c^{2}}[/tex] where E is the modulus  of elasticity, [tex]\gamma[/tex] is surface energy and [tex]\sigma_c[/tex] is tensile stress

Substituting the given values

[tex]C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm[/tex]

The maximum allowable surface crack is 1.44 mm

Answer:

0.3

Explanation:

Please see attachment.

Answer:

The area at the base should be 2.91cm²

Explanation:

The separation of the molten metal may occur when the pressure of the flowing liquid falls below the atmospheric pressure and the air entrapped inside the mold tries to escape but rather gets rapped inside the casting (which is solidifying) causing the liquid to separate and the casting to be porous.

The velocity at the bottom of the sprue is written as V = √(2gh), where g is the acceleration due to gravity and h is the height of the sprue

V = √(2 x 9.81 x 100 x 15) = 171.6 cm² (Multiplying by 100 to convert 9.81 m/s to cm/s)

The expression for flow rate is Q = VA where V is the velocity as calculated above and A is the area

500 = 171.6 x A, solving for A we get A = 500/171.6 = 2.91cm²

The cross - sectional Area at the base of the sprue should be 2.91cm² to avoid separation of the molten metal.

Answer:

51.77 GPa

Explanation:

For elasticity for a continous and aligned hybrid composite

[tex]E_{ci}=E_{poly}  (1-V_{A} -V_{G} )+E(V_{A})+E(V_{G} )\\=(2.5GPa)(1-0.22-0.3)+(131GPa)(0.22)+(72.5GPa)(0.3)\\=1.2+28.82+21.75\\=51.77GPa[/tex]

Answer: No, not in favor of the idea.

This is because, an increase in the refrigerator's work output is equal to work input. In real live application, the work input of the refrigerator is always greater than the additional work output. This will result to reducing the efficiency of the power plant.

Refrigerating cooling water before the condenser in a power plant is impractical due to high energy costs, added complexity, and inefficiencies, outweighing the potential efficiency gains.

While the idea of refrigerating the cooling water before it enters the condenser in a power plant to increase efficiency might seem theoretically sound, in practice it is not advisable due to several reasons:

1.Energy Cost of Refrigeration: Refrigeration itself is an energy-intensive process. The energy required to cool the water to a significantly lower temperature would likely be greater than the energy savings achieved by the increased efficiency of the heat engine. This means that the net energy gain would be negative.

2.Second Law of Thermodynamics: According to the second law of thermodynamics, any energy conversion process, including refrigeration, involves irreversibilities and losses. The additional step of refrigeration would introduce more inefficiencies and would not be 100% efficient. Thus, the overall efficiency of the system would decrease.

3.Complexity and Cost: Adding a refrigeration system to cool the condenser water would significantly increase the complexity and cost of the power plant. This includes the initial capital cost, operational costs, and maintenance costs. The benefits gained from a marginal increase in thermal efficiency would likely not justify these additional expenses.

4. Practical Alternatives: There are more practical and cost-effective ways to improve the efficiency of a power plant. These include optimizing the thermodynamic cycle, improving the insulation, using better quality fuels, and upgrading to more efficient machinery and technology. Improving the heat exchange process itself, rather than cooling the water, is generally more practical.

5.Environmental Considerations: The energy used for refrigeration would typically come from the power plant itself or the grid, potentially leading to increased fuel consumption and higher greenhouse gas emissions. This is counterproductive to the goal of improving efficiency and sustainability.

In summary, while lowering the temperature at which heat is rejected can theoretically improve the thermal efficiency of a heat engine, refrigerating the cooling water is not a practical or economical solution due to the high energy cost, increased complexity, and overall inefficiency of the refrigeration process itself. Instead, other methods to enhance efficiency should be explored.

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

[tex]W = \dot{m}c_p \Delta T[/tex]

Where

[tex]C_p =[/tex] Specific heat

[tex]\dot{m}[/tex]= Mass flow rate

[tex]\Delta T[/tex] = Change in Temperature

Our values are given as

[tex]C_p = 1.005kJ/kgK \rightarrow[/tex] Specific heat of air

[tex]T_1 = 50\°C[/tex]

[tex]\dot{m} = 2kg/s[/tex]

[tex]W = 8kW[/tex]

From our equation we have that

[tex]W = \dot{m}c_p \Delta T[/tex]

[tex]W = \dot{m}c_p (T_2-T_1)[/tex]

Rearrange to find [tex]T_2[/tex]

[tex]T_2 = \frac{W}{\dot{m}c_p}+T_1[/tex]

Replacing

[tex]T_2 = \frac{8}{2*1.005}+(50+273)[/tex]

[tex]T_2 = 326.98K \approx 53.98\°C[/tex]

Therefore the exit temperature of air is 53.98°C

Answer:

books = []

   fp = open("bookTitles.txt")

   for line in fp.readlines():

       title = line.strip()

       if title not in books:

           books.append(title)

   fp.close()

   fout = open("noDuplicates.txt", "w")

   for title in books:

       print(tile, file=fout)

   fout.close()

except FileNotFoundError:

   print("Unable to open bookTitles.txt")

To solve this problem it is necessary to apply the concepts related to Arc Length.

From practical terms we know that the Angle can be calculated based on the arc length and the radius of the circle, in other words

[tex]\theta = \frac{S}{r}[/tex] or

[tex]S = \theta r[/tex]

Where,

S = Length of the arc

r = Radius

From the information given, the object to bending has a millimeter thick and a radius of 12 mm. This way your net radio is given by

[tex]R_1 = r_B+\frac{t}{2}[/tex]

After Springback the radius increases and the thickness ratio is therefore maintained.

[tex]R_2 = r_S+\frac{t}{2}[/tex]

For both cases we have different angles but that maintains the electron-magnetic proportions, therefore the arc length is maintained:

[tex]S_B = S_S[/tex]

[tex]\theta_B R_1 = \theta_S R_2[/tex]

[tex]\theta_B (r_B+\frac{t}{2})=\theta_S(r_S+\frac{t}{2})[/tex]

Re-arrange to find [tex]\theta_S[/tex]

[tex]\theta_S = \frac{\theta_B (r_B+\frac{t}{2})}{(r_S+\frac{t}{2})}[/tex]

[tex]\theta_S= \frac{90\°*(12+\frac{1}{2})}{13+\frac{1}{2}}[/tex]

[tex]\theta_S= 83.33\°[/tex]

Therefore the angle that should be bent before springback is 83.33°

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

[tex]Y = 1 - e^{-kt^n}[/tex]

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

[tex]0.50 = 1 - e^{-k^146^2.1}[/tex]

[tex]0.50 = e^{-k306.6}[/tex]

taking natural log on both side

ln(0.5) = -k(306.6)

[tex]k = 2.26\times 10^{-3}[/tex]

for 86 % completion

[tex]0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}[/tex]

[tex]e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14[/tex]

[tex]-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)[/tex]

[tex]t^{2.1} = 869.96[/tex]

t = 25.10 sec

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

[tex]N_F=n_F\times \frac{N_S}{n_S}[/tex] here,

[tex]n_F[/tex] = the number of unseeded faults = 6

[tex]N_S[/tex] = number of seeded faults = 30

[tex]n_s[/tex] = number of seeded faults found = 15

So NF will be calculated as,

[tex]N_F=6\times \frac{30}{15}=12[/tex]

And the estimate of faults remaining is  [tex]N_F-n_F[/tex] = 12 - 6 = 6

https://brainly.com/question/14000001

The average heat transfer coefficient would be; 4.42 W/m gK

The convective heat transfer loss to the air from both surfaces would be 4.87 W/m K.

Net radiation to a a very large environment at 35°C would be; -10.06 W

To determine the average heat transfer, the Rayleighs number would be used. According to the formula;

Ra = gβΔTL³/Vα

= 1.827 * 10⁷

GrL = RaL/Pr

= 0.501

The correlation for estimating hL

Nu L= hL*L/k

= 4.42 W/m gK

Net radiation using the Stefan-Boltzmann equation of Q_rad = ε * σ * A * (Tplate₄ - Tsurroundings⁴)

= Q_rad_total = 2 * (-5.03 W)

= -10.06 W

Learn more about heat transfer here:

https://brainly.com/question/16055406

#SPJ2

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

Answer:

speed of water flow in the test is 27 m/s

Explanation:

given data

cross section radius r1 = 1.8 m

radius constricts r2 = 0.60 m

speed of water flow = 3.0 m/s

to find out

speed of water flow in the test section

solution

we using the continuity equation here

A1 × V1 = A2 × V2     ...............1

put here value we get speed

[tex](\frac{\pi }{4} d1^2) v1 = (\frac{\pi }{4} d2^2) v2[/tex]

1.8²×3 = 0.6² × v2

v2 = 27 m/s

speed of water flow in the test is 27 m/s

To solve this problem it is necessary to apply the concepts related to the normal effort and the shear effort due to failure. The shear stress can be defined based on normal stress and effective friction angle. Mathematically it can be defined as

[tex]\tau_f = \sigma_n tan\phi[/tex]

Where

[tex]\tau_f =[/tex] Shear stress at failure

[tex]\sigma_n =[/tex] Normal stress

[tex]\phi =[/tex] Effective friction angle

PART A) Our values are given as ,

[tex]\sigma_n = 192kPa[/tex]

[tex]\tau_f  = 120kPa[/tex]

Replacing at the previous equation we have,

[tex]\tau_f = \sigma_n tan\phi[/tex]

[tex]120 = 192tan\phi[/tex]

[tex]\phi = tan^{-1}(\frac{120}{192})[/tex]

[tex]\phi = 32.27°[/tex]

Therefore the effective friction angle of the sand is 32.27°

B) Using the maximum possible effort and the angle previously given we can calculate the maximum shear force, from which from its definition it is possible to find the force.

[tex]\tau_f = \sigma_n tan\phi[/tex]

[tex]\tau_f = 200tan(32\°)[/tex]

[tex]\tau_f = 124.97kPa[/tex]

With the value of the shear stress from its basic definition we can find the force. By definition the shear stress is given by

[tex]\tau_f = \frac{F}{A}[/tex]

Re-arrante to find the Force

[tex]F = A\tau_f[/tex]

The shear stress is a function of the Force and the Area, therefore, the area would be the square of the sides (50mm)

Replacing in the equation we have to,

[tex]F = 124.97*10^{-3}*(50*50)[/tex]

[tex]F = 312.425N[/tex]

Therefore the shear force required to cause failure of the specimen is 312.425N

Answer:

Q=339.5W

T2=805.3K

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.

3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature

Answer:

reduce traction

Explanation:

Snow, ice, rain, and flooding all reduce traction, making it harder to steer and harder to stop in time. In these conditions, you might even experience hydroplaning.