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Zielflug [23.3K]
3 years ago
12

Discuss the impact of the changing urban center. Include the impacts on political, economic, and social roles and opportunities.

The changing role of the family and the diversity of class should also be introduced and discussed. (Your essay should be at least 200 words in length.)
Engineering
1 answer:
KengaRu [80]3 years ago
6 0

Answer:

The 21st century world have been earmarked with great influx of people to the urban centre,the notion of gender equality and female education have also made most traditional roles in the family changing.Before now,wives we're known for their full independence on their husband who is considered the bread winner.

Inspite of the growth of of the urban centre the availability of resources have dwindled,resulting in the surge of unemployment in many urban centre,the political entity of the society which is the government have serious challenging in managing the various threat posed by overpopulation, unemployment results in the decrease of standard of living of person and family,to cater for this family have to change their roles,wives now work to support the husband.

Explanation:

You might be interested in
A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

5 0
4 years ago
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan
hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

R = \dfrac{u^2}{15(e+f_s)}

where;

R= radius

f_s = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

f_s = 0.20

So;

R = \dfrac{30^2}{15(0.08+0.20)}

R = \dfrac{900}{15(0.28)}

R = \dfrac{900}{4.2}

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0
3 years ago
What is 90 to the power of 46
Mnenie [13.5K]

Answer:Just multiply 90 by itself 46 times

Explanation:

do it

6 0
4 years ago
If pure oxygen is fed in excess by 25%, what would the fractional conversion of methane be for the final concentration of CO2 in
inessss [21]

Answer:the fractional conversion of methane is 12.5%

Explanation:The reaction represent the combustion of methane to produce Co2 and steam.

CH4 +2O2_CO2 + 2H2O

From gay lussac law of proportionality

1mol of CH4 requires 2mol of Oxygen to produce 1 mol of CO2 and 2mol of H2O

So from the combining ratio,25% of O2 will fractional produce 25×1/2% of CH4.

While 12.5% of CO2 and 25% of steam is also produced .so in essence 2.5% of CO2 was lost in the reaction.

7 0
4 years ago
Creating vacancies in ceramics. Consider doping ZrO2 with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb
Murljashka [212]

Answer:

Creating vacancies in ceramics. Consider doping ZrO₂ with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb ions sit in Zr ion sites

a. A substitutional defect will be produced.

b. With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.

Explanation:

(a) The defect produced by dopping a little concentration of Nb₂O5 with Nb in the +5 charge state is known as a substitutional defect.

(b) With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.

Moreover, if oxygen vacancies are rate-limiting defect, the corrosion of ZrO₂ decreases and if electrons are rate-limiting then the corrosion of ZrO₂ is accelerated.

7 0
4 years ago
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