Technician A says that reinforcements may be made of plastic.

Technician B says that reinforcements may be made of nylon.

Who is right?

Answer:

Total distributed load on BE = 5 m²

Explanation:

The first process is to get the value for the Dead load (DL) on the slab;

This is determined by using the formula:

DL = ρ × t

DL = 23.6 kK/m³ × 0.2 m

DL = 4.72 kN/m²

From table 1.4 which relates to the office buildings, we derive the value for the minimum live load (LL) = 2.40 kN/m³

Hence the total load TL = Dead Load DL) + Live loaf (LL)

DL = (4.72 + 2.40) kN/m³

DL = 7.12 kN/m³

Now; from the imaginative view of the information given; the member BE which get the load from half area of the panel BEDC & half the area of BEFA panel parallel to member BE; then the tributary area on member BE can be calculated as;

[tex]A_{BE} = ( \dfrac{a}{2}+\dfrac{a}{2}) \times width[/tex]

[tex]A_{BE} = ( \dfrac{2}{2}+\dfrac{2}{2}) \times 1[/tex]

[tex]A_{BE} =2\times 1[/tex]

[tex]A_{BE} =2 m^2/m[/tex]

The total distributed load acting on BE is:

[tex]Total \ load = TL \times A_{BE}[/tex]

[tex]Total \ load = 7.12 \ \dfrac{kN}{m^2 }\times (2 \times \dfrac{m^2}{m})[/tex]

Total load = 2.5 × 2

Total load = 5 m²

Answer:

The shortest time needed for all four of them to cross the bridge = 17 minutes

Explanation:

As given ,  to cross the bridge

Person 1 takes 1 minute,

Person 2 takes 2 minutes,

Person 3 takes 5 minutes, and

Person 4 takes 10 minutes.

For the first time ,

Person 1 and Person 2 will go to cross the bridge

As Person 2 takes 2 minutes

So, total time taken by Person 1 and person 2 together will be 2 minutes

Now,

Person 1 will come back with flashlight.

He takes 1 minutes.

So, Total time becomes 2 + 1 = 3 minutes

Then,

Person 3 and Person 4 will cross the bridge

As Person 4 takes 10 minutes

So, total time taken by Person 3 and person 4 together will be 10 minutes

So, Total time becomes 3 + 10 = 13 minutes

Now,

Person 2 will come back with flashlight.

He takes 2 minutes.

So, Total time becomes 13 + 2 = 15 minutes

Then,

Person 1 and Person 2 will cross the bridge

As Person 2 takes 2 minutes

So, total time taken by Person 1 and person 2 together will be 2 minutes

So, Total time becomes 15 + 2 = 17 minutes

∴ we get

The shortest time needed for all four of them to cross the bridge = 17 minutes

Answer:

The fifth deviation is +1

Explanation:

Given

[tex]Deviations = -5, +2, +4,-2[/tex]

Required

Determine the fifth deviation

Represent the fifth deviation with x.

As a theorem, the sum of deviations from mean is always 0.

So, we have:

[tex]-5 +2 +4-2+x = 0[/tex]

[tex]-1+x = 0[/tex]

Add 1 to both sides

[tex]1 -1+x = 0 +1[/tex]

[tex]x = 0 +1[/tex]

[tex]x= +\ 1[/tex]

Hence, the fifth deviation is +1

Answer:

- Vernier thrust is 5.37 kN

- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Explanation:  

Given that;

For individual thrust chambers (vacuum);

Fc = 73.14 kN ,  Cc = 3279 m/sec

For Overall engine with Vernier (vacuum);

Foa = 297.93 kN = ,  Coa = 3197 m/sec.

- determine the Vernier thrust

Vernier thrust Fv =  Foa - ( 4 × Fc )

Vernier thrust Fv  = 297.93 - ( 4 × 73.14)

Vernier thrust Fv  = 297.93 - 292.56

Vernier thrust Fv  = 5.37 kN

Therefore, Vernier thrust is 5.37 kN

-

Vernier mass flow rate;

we know that

[tex]Co_{a}[/tex] = Fc + Fv  / mc + mv

mv = Foa/Coa - Fc/Cc

we convert kilonewton to kilograms

1 kn = 102 kg

Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg

Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg

we substitute

mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)

mv = 9.5054 - 9.1006

mv = 0.4048 kg/s

Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Answer:

electrical

computer

mechanical

and manufacturing .... I think

The answer is electrical, manufacture, computer, and chemical

Complete Question:

Throughout your job search, you’ll find that _____(the economy/education/salary) is closely related to the career of your choosing. It’s important to take the time to find out what to expect now, so you can start developing the ____ (task/communication/skill) that you need to excel in that career.

Answer:

Education; skills.

Explanation:

Throughout your job search, you'll find that education is closely related to the career of your choosing. It's important to take the time to find out what to expect now, so you can start developing the skills you need to excel in that career.

This ultimately implies that, all job openings that are being advertised by various organizations usually have a minimum requirement such as academic experience or level i.e the prospective candidate must have attained a level of education. Also, it is very important for undergraduates and potential employees to ensure that they are being proactive, as well as developing the requisite skills needed to excel in their career.

Answer:

both are c

Explanation:

Answer:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Explanation:

Given

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]

[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]

[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]

[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]

[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]

[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]

[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]

[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]

[tex]0.78.[/tex]

The 0.3's is will be plotted as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

The 0.4's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}[/tex]

The 0.5's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]

The 0.6's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

Lastly, the 0.7's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

The combined steam and leaf plot is:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Answer:

The answers are below

Explanation:

a. 180 in^3 to L

1 in³ = 0.0164L

180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]

b. 750 ft-lbf to kJ

1 ft-lbf = 0.00136 kJ

750 ft-lbf  = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]

c. 75.0 hp to kW

1 hp = 0.746 kW

75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]

d. 2500.0 lb/h to kg/s

1 lb/h = 0.000126 kg/s

2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]

e. 120 psia to kPa

1 psia = 6.89 kPa

120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]

f. 120 psig to kPa

1 psig = 6.89 kPa

120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]

g. 300 ft/min to m/s

1 ft/min = 0.005 m/s

300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]

h. 125 km/h to miles/h

1 km/h = 0.62 mph

125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]

i) 6000 N to Ibf

1 N = 0.2248 lbf

6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]

j. 6000 N to ton

1 N =  0.000102 Ton-force

6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]

Answer:

no it has to be removed

Explanation:

It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.

Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.

Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.

Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.

Learn more about drum brakes here:

https://brainly.com/question/14937026

#SPJ2

Answer:

Explanation:

The given equation is :

[tex]\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f[/tex]

a) Question Completion:

INPUT HOURS OF LABOR

Country   Cookies      Milk

Atlantis       2 hours   1 hour

Neverland  4 hours   1 hour

Answer:

1. Atlantis has the absolute advantage in the production of cookies.

2. No country has the absolute advantage in the production of milk.

Explanation:

Absolute advantage refers to superior production capability.  It is determined when a country, for example, has the ability to produce a particular good or service at lower cost or more efficiently (i.e. with less resources) than the other country.  In the scenario above, Atlantis has an absolute advantage in the production of cookies because it can produce the same quantity of cookies using 2 labor hours that Neverland can produce using 4 labor hours.  But for the production of milk, Atlantis and Neverland share the same comparative advantage less they can use less labor hours to produce milk than they can produce cookies.

Answer:

This band gap also allows semiconductors to convert light into electricity in photovoltaic cells and to emit light as LEDs when made into certain types of diodes. Both these processes rely on the energy absorbed or released by electrons moving between the conduction and valence bands.

Explanation:

On the internet

Answer:

Attached below

Explanation:

PWM signal source has 1 KHz base frequency

Analog filter : with time constant = 0.01 s

low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]

PWM duty cycle is a constant block

Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

[tex]\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)[/tex]

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

[tex]\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}[/tex]

where;

[tex]Q_g[/tex] = heat generated per unit volume

[tex]\alpha[/tex] = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

[tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0[/tex]

where;

The radial directional term = [tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})[/tex] and the axial directional term is [tex]\dfrac{\partial^2 T}{\partial z^2 }[/tex]

d) The heat equation for a wire going through a furnace is:

[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]

since;

the steady-state is zero, Then:

[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

[tex]\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}[/tex]

Answer:

A) attached below

B) Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Explanation:

attached below is a detailed solution

A) attached below

B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A

Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Answer:

a) the friction angle of the sand is 36.87°

b) the deviator stress at failure is 450 kN/m³

Explanation:

Given the data in the question;

For a consolidated drained test

The effective major principle stresses

σ₃ = σ₃' = 80 kN/m²

and

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m²  = 320 kN/m²

now

a) friction angle of the sand

σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )

for sand; c' = 0

so

σ₁' = σ₃'tan²( 45° + Ф/2' )

we substitute

320 = 80 tan²( 45° + Ф/2' )

Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]

Ф' = 2 × [ 63.4349° - 45° ]

Ф' = 2 × 18.4349

Ф' = 36.87°

Therefore,  the friction angle of the sand is 36.87°

b)  deviator stress

σ₁' = σ₃'tan²( 45° + Ф/2' )

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )

σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )   3.597

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex]  = 450 kN/m³

Therefore, the deviator stress at failure is 450 kN/m³

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

Answer:

False

Explanation:

I got it wrong picking true

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

Determine the maximum speed for safe vehicle operation

firstly calculate the stopping sight distance

m = R ( 1 - cos [tex]\frac{28.655*S}{R}[/tex] )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos [tex]\frac{28.65 *s }{678}[/tex]  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + [tex]\frac{u^2}{30(\frac{a}{3.2} )-G1}[/tex]  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + [tex]\frac{u^2}{30[(11.2/32.2)-0 ]}[/tex]

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

Answer: Why is even here then.

Explanation:

Answer:

a. molecular interactions.

Explanation:

Conduction is thermal energy transfer by molecular interactions. Therefore, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Some examples of conductors include metal, steel, aluminum, copper, graphite, etc.

Hence, conduction is thermal energy transfer as a result of the movement of electrons and collision between the molecules of an object.

Answer:

vehicle density =  28.205 veh/mile

flow rate =  0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance    ...............1

vehicle density = ( 21 + 1 )  ÷ 0.78

vehicle density  k =  28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs    .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate =  0.909 veh/hr

Answer:

a) attached below

b) The worst case in tension is case ( 9-7 ) b which is

  = -65k

  The worst case in compression is case ( 9-5 ) a which is

 = 670 k  

Explanation:

Given data :

D = 150k , L = 280k , Lr = 40k , s = 50k , w = ± 120k

E = ± 200k

attached below is a detailed solution to the given problem ( problem 1 )

A)  attached below

b) The worst case in tension is case ( 9-7 ) b which is

  = -65k

  The worst case in compression is case ( 9-5 ) a which is

 = 670 k  

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter [tex]d_{o}[/tex] = 12.8 mm

Final diameter [tex]d_{f}[/tex] = 10.7

Engineering stress  [tex]\alpha _{E}[/tex] = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4([tex]d_{o}[/tex] )²  ; Ag = π/4([tex]d_{f}[/tex] )²

% = π/4 [ ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ( π/4  ([tex]d_{o}[/tex] )²) ]

= ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ([tex]d_{o}[/tex] )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  [tex]\alpha _{T}[/tex] = [tex]\alpha _{E}[/tex] ( 1 +  [tex]E_{E}[/tex] )

[tex]E_{E}[/tex]  is engineering strain

[tex]E_{E}[/tex]  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

[tex]E_{E}[/tex] = 0.431

so we substitute the value of [tex]E_{E}[/tex]  into our initial equation;

True stress  [tex]\alpha _{T}[/tex] = 460 ( 1 +  0.431)

True stress  [tex]\alpha _{T}[/tex] = 460 (1.431)

True stress  [tex]\alpha _{T}[/tex] = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

Answer:

30 mm is the minimum thickness that must be applied.

Explanation:

Given the data in the question;

Using Fourier's equation. the heat rate is  

q = kA(ΔT/Δx)

where

A is the surface area, we must consider all surfaces through which the heat can dissipate through

i.e 2×2 for one wall gives you 4m²,

there are 5 walls, so we will  have 20m² for surface area.

k is thermal conductivity of the styrofoam ( 0.030 W/m K)    

q is the heat loss (500 W  )

ΔT is the Temperature difference ( 35 - 10) = 25°C

Δx  = ?

So we substitute

500 = (0.030)(20)(25/Δx)

500 = 0.6 (25/Δx)

500 = 15 / Δx

Δx = 15 / 500

Δx = 0.03 m = 30 mm

Therefore, 30 mm is the minimum thickness that must be applied.