Engine oil flows at a rate of 1 kg/s through a 5-mmdiameter straight tube. The oil has an inlet temperature of 45°C and it is desired to heat the oil to a mean temperature of 80°C at the exit of the tube. The surface of the tube is maintained at 150°C. Determine the required length of the tube. Hint: Calculate the Reynolds numbers at the entrance and

Answer:

y'(x) = 7 ;

y'(x) = 15x²

Explanation:

Differentiation from first principle :

y = 7x

Formula to obtain derivative from first principle:

y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h

Obtain : y(x + h)

y(x) = 7x

y(x + h) = 7(x + h)

Substitute into formula :

y'(x) = lim(h - - > 0) : [7(x + h) - 7x] / h

Expand :

y'(x) = lim(h - - > 0) : [7x + 7h - 7x] / h

y'(x) = lim(h - - > 0) : 7h/ h

y'(x) = 7

2.)

y = 5x³

y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h

Obtain : y(x + h)

y(x) = 5x³

y(x + h) = 5(x + h)³

Substitute into formula :

y'(x) = lim(h - - > 0) : [5(x + h)³ - 5x³] / h

Expand :

y'(x) = lim(h - - > 0) : [5(x³ + 3x²h + 3xh² + h³) - 5x³] / h

y'(x) = lim(h - - > 0) : [5x³ + 15x²h + 15xh² + 5h³ - 5x³] / h

y'(x) = lim(h - - > 0) : [15x²h + 15xh² + 5h³] / h

y'(x) = lim(h - - > 0) : h(15x² + 15xh + 5h²) / h

y'(x) = lim(h - - > 0) : (15x² + 15xh + 5h²)

y'(x) = 15x²

Answer:

3.19 days

Explanation:

Given data :

stream flow rate = 50 kg/min

stream contains ; 0.22% sand by mass

Cross sectional area of filter = 9 m^2

Filter successfully filters out 90% of the input sand by mass

SG = 2.25

thickness of cake formed = 0.25 meters

Determine how long a new filter can be used before replacement

Given that for every 1 minute  5.43*10^-5 m thickness of sand layer(cake) forms on the filter and the replacement of filter is done once the cake thickness = 0.25 meters

To determine the number of days ( X ) before replacing filter we apply the relationship below

5.43 * 10^-5 = 1 min

0.25 m = X

hence ; X = 0.25 / (5.43 * 10^-5 )  = 4604.051 minutes ≈ 3.19 days

Attached below is the beginning part of the detailed solution

Answer:

evaluate the problem

Explanation:

An engineering design process is a series of steps that engineers use to create products.

Therefore, if a client requests for the battery life for a tablet, the step in the design process the request is associated with is "evaluate the problem".

This is because, by finding out the battery life, you're already evaluating the problem.

Answer:

2727000 Liters

Explanation:

Determine the daily methane production in liters

For waste cooking oil ;

Delivered amount in liters  = 757.082 liters per week

daily = 757.082 / 7 = 108.154 L/day

waste vs concentration = 25,000 mg/L

Its volatile solid  = 25000 * 10^-3 (g/l) * 108.154 L/day

                           = 2703.864  g/day

hence methane production in liters

= 600 L/g * 2703.864 g/day

= 1622318.57 L

For cow manure

Delivered amount in liters = 500 liters / day

Vs concentration = 5000 mg/l

Hence amount of Vs = 5000 * 10^-3 * 500  = 2500 ( g/day)

methane production in liters

= 150 * 2500 = 375000 liters

For glass clippings

Delivered amount in liters per day  = 1135.62 / 7 = 162.23 L/day

amount of Vs = 15000 * 10^-3 (g/L) * 162.23 L/day = 2438.45 ( g/day )

hence the methane production in liters

= 300 * 2433.45 = 730035 L

Hence ∑ daily methane production of each waste

=  1622318.57 + 375000 + 730035 = 2727000 Liters

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

B) The physical properties are copper

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

Answer: 25 ft

Explanation:

For a triangle, the area is given by

A = (1/2)base*height

Rearranging,

base = 2A/height

Substituting the given values,

base = 2(350 ft^2)/(28 ft) = 25 ft

4.8 unit however you can think as 48/10 but you're reversing which 4.8. is .048 Ohms.

Answer:

a

Explanation:

Answer:

a scale model each size is a certain amount smaller

Explanation:

Answer:

fan

Explanation:

Answer:

Explained below

Explanation:

The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.

Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.

Answer:

$14253

Explanation:

Answer:

The answer is below

Explanation:

v = velocity = 30 m/min = 30 * 10³ mm/min, D =  diameter = 10 mm, f = feed = 0.25 mm/rev, point angle = 118, cutting time = Tm, d = depth = 60 mm

[tex]a)\\N=\frac{v}{\pi D}=\frac{30*10^3}{\pi * 10}=954.9\ rev/min\\\\f_r=Nf =954.9(0.25)=238\ mm/min\\\\A=0.5Dtan(90-\frac{point\ angle}{2} )=0.5*10*tan(90-\frac{118}{2} )=3\ mm\\\\T_m=\frac{(d+A)}{f_r} =\frac{60+3}{238}=0.265 \ s\\\\b)\\metal\ removal\ rate(R_{MR})=0.25\pi D^2f_r\\\\R_{MR}=0.25\pi (10)^2(238)=18692\ mm^3/min[/tex]

Answer:

try to 36v power and take 1a and intersect to 3

Answer:

Area of square = 2,450 mm²

Explanation:

Given:

Length of diagonal = 70 mm

Find:

Area of square

Computation:

Area of square = diagonal² / 2

Area of square = 70² / 2

Area of square = 4900 / 2

Area of square = 2,450 mm²

Answer:

A)

It should be Non- toxic

It should possess high Thermal conductivity

It should have the Required Thermal diffusivity

B)

C)   All the materials are suitable because they serve different purposes when making modern kitchen cookware

Explanation:

A) characteristics required of a ceramic material to be used as a kitchen cookware

B) comparison of three ceramic materials as to their relative properties

C) material most suitable for the cookware.

 All the materials are suitable because they serve different purposes when making modern kitchen cookware

Answer:

To drink unchlorinated groundwater with 10 ppb of TCE ( A )

Explanation:

The option that presents the greatest cancer  risk for ingesting water

contaminated with Trichloroethylene under the residential exposure parameters is to drink unchlorinated groundwater with 10 ppb of TCE

This is because suitable water for drinking  has chloroform concentration that ranges from 4 to 44 ppb but drinking under groundwater with ppb value above 4 ppb will have a more severe damage to the body

Answer:

preguntas a parte o no???????

Explanation:

the Ethics of emerging Technology can only make use of speculative data about future products,uses and impacts.

Answer:

Execution time is 13.65ms

CPI is 5.46

MIPS is 73.3\ MIPs

Explanation:

Given:

[tex]s = 40MHz[/tex] --- processor speed [Missing from the question]

[tex]\begin{array}{ccc}{Instruction\ Type} & {Instruction\ Count} & {Cycles\ per\ Instruction} & {Integer\ Arithmetic} & {45000} & {4} \ \\ {Data\ Transfer} & {32000} & {6} & {Floating\ point\ arithmetic} & {15000} & {10} & {Control\ Transfer} &{8000} & {3} \ \end{array}[/tex]

Solving (a): The program execution time

First, we solve for (b)

Solving (b): The effective CPI

This is calculated as:

[tex]CPI = \frac{\sum IC * CI}{\sum IC}[/tex]

Where: IC = Instruction Count and CI = Cycles per Instruction

So, the equation becomes:

[tex]CPI = \frac{45000 *4 + 32000 * 6 + 15000 * 10 + 8000 * 3}{45000+32000+15000+8000}[/tex]

[tex]CPI = \frac{546000}{100000}[/tex]

[tex]CPI = 5.46[/tex]

Solving (c): MIPS

This is calculated as:

[tex]MIPS = Speed * \frac{1}{CPI} * \frac{1}{\sum IC}[/tex]

[tex]MIPS = 40 * \frac{1}{5.46} * \frac{1}{100000}[/tex]

[tex]MIPS = \frac{40 * 1 * 1}{5.46*100000}[/tex]

[tex]MIPS = \frac{40}{546000}[/tex]

[tex]MIPS = 0.00007326007[/tex]

Convert to MIPs

[tex]MIPS = 73.3\ MIPs[/tex]

Solving (a): Execution Time

This is calculated as:

[tex]Time = Instructions * CPI * \frac{1}{Speed}[/tex]

[tex]Time = 100000 * 5.46* \frac{1}{40M}[/tex]

[tex]Time = \frac{546000}{40M}[/tex]

[tex]Time = \frac{546000}{40*1000000}[/tex]

[tex]Time = \frac{546000}{40000000}[/tex]

[tex]Time = 0.01365s[/tex]

[tex]Time = 13.65ms[/tex]

Execution time is 13.65ms

Answer:Architect-> Construction Manager-> Plumber

Explanation:

Answer:

Rate of Heat Loss = 336 W

Explanation:

First, we will find the surface area of the cylinder that is modelled as the man:

[tex]Area = A = (2\pi r)(l)[/tex]

where,

r = radius of cylinder = 30 cm/2 = 15 cm = 0.15 m

l = length of cylinder = 170 cm = 1.7 m

Therefore,

[tex]A = 2\pi(0.15\ m)(1.7\ m)\\A = 1.6\ m^2[/tex]

Now, we will calculate the rate of heat loss:

[tex]Rate\ of\ Heat\ Loss = hA\Delta T[/tex]

where,

h = convective heat tranfer  coefficient = 15 W/m²K

ΔT = Temperature difference = 34°C - 20°C = 14°C

Therefore,

[tex]Rate\ of\ Heat\ Loss = (15\ W/m^2K)(1.6\ m^2)(14\ K)\\[/tex]

Rate of Heat Loss = 336 W

Answer:

#!/usr/bin/env python                                                          

def calculate(x, y):                                                          

   return {                                                                  

       "MOD": x % y,                                                          

       "DIV": x/y,  # you mean div instead of “VID”, right?                  

       "floating-point division": float(x)/y,                                

   }                                                                          

def calculateInBothSettings(x, y):                                            

   return {                                                                  

       "x/y": calculate(x, y),                                                

       "y/x": calculate(y, x),                                                

   }                                                                          

if __name__ == "__main__":                                                    

   x = int(input("x: "))                                                      

   y = int(input("y: "))                                                      

   print(calculateInBothSettings(x, y))

Explanation:

I wrote a python script. Example output:

x: 2

y: 3

{'x/y': {'MOD': 2, 'DIV': 0.6666666666666666, 'floating-point division': 0.6666666666666666}, 'y/x': {'MOD': 1, 'DIV': 1.5, 'floating-point division': 1.5}}

Answer:

(a) The distance up the slope the wagon moves before coming to rest is approximately 21.74 m

(b) The distance the wagon comes to rest from the starting point is approximately 12.06 m

(c) The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is approximately 3.214 m/s (the difference in value can come from calculating processes)

Explanation:

The wagon motion parameters are;

The mass of the wagon, m = 7,200 kg

The initial velocity with which the wagon is projected along the horizontal rail, v = U

The length of the horizontal portion of the rail = 100 m

The angle of inclination of the inclined portion of the rail, θ = sin⁻¹(0.01)

The exerted frictional resistance to motion of the rail, [tex]F_f[/tex] = 140 N

∴ θ = sin⁻¹(0.01)

The work done by the frictional force on the horizontal portion of the rail = 140 N × 100 m = 14,000 J

(a) If U = 3 m/s, we have;

Kinetic energy = 1/2·m·v²

The initial kinetic energy of the wagon, K.E. is given with the known parameters as follows;

K.E. = 1/2 × 7,200 kg × (3 m/s)² = 32,400 J

The energy, E, required to move a distance, 'd', up the slope is given as follows;

E = [tex]F_f[/tex] × d + m·g·h

Where;

[tex]F_f[/tex] = The friction force = 140 N

m = The mass of the wagon = 7,200 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height reached = d × sin(θ) = d × 0.01

Therefore;

E = 140 N × d₁ + 7,200 kg × 9.81 m/s² × d₁ × 0.01 = 846.32 N × d

The energy, [tex]E_{NET \ horizontal}[/tex], remaining from the horizontal portion of the rail is given as follows;

[tex]E_{NET \ horizontal}[/tex] = Initial kinetic energy of the wagon - Work done on frictional resistance on the horizontal portion of the rail

∴ [tex]E_{NET \ horizontal}[/tex] = 32,400 J - 14,000 J = 18,400 J

[tex]E_{NET \ horizontal}[/tex] = 18,400 J

Therefore, for the wagon with energy, [tex]E_{NET \ horizontal}[/tex] to move up the train, we get;

[tex]E_{NET \ horizontal}[/tex] = E

∴ 18,400 J = 846.32N × d

d₁ = 18,400 J/(846.36 N) ≈ 21.7401579 m

d₁ ≈ 21.74 m

The distance up the slope the wagon moves before coming to rest, d₁ ≈ 21.74 m

(b) Given that the initial velocity of the wagon, U = 3 m/s, the distance up the slope the wagon moves before coming to rest is given above as d₁ ≈ 21.74 m

The initial potential energy, PE, of the wagon while at the maximum height up the slope is given as follows;

P.E. = m·g·h = 7,200 kg × 9.81 m/s² × 21.74 × 0.01 m = 15,355.3968 J

The work done, 'W', on the frictional force on the return of the wagon is given as follows;

W = [tex]F_f[/tex] × d₂

Where d₂ = the distance moved by the wagon

By conservation of energy, we have;

P.E. = W

∴  15,355.3968 = 140 × d₂

d₂ = 15,355.4/140 = 109.681405714

Therefore;

The distance the wagon moves from the maximum height, d₂ ≈ 109.68 m

The distance the wagon comes to rest from the starting point, d₃, is given as follows;

d₃ = Horizontal distance + d₁ - d₂

d₃ = 100 m + 21.74 m - 109.68 m ≈ 12.06 m

The distance the wagon comes to rest from the starting point, d₃ ≈ 12.06 m

(c) For the wagon to come finally to rest at it starting point, we have;

The initial kinetic energy = The total work done

1/2·m·v² = 2 × [tex]F_f[/tex] × d

∴ 1/2 × 7,200 × U² = 2 × 140 × d₄

d₄ = 100 + (1/2·m·U² - 140×100)

(1/2·m·U² - 140×100)/(m·g) = h = d₁ × 0.01

∴ d₁ = (1/2·m·U² - 140×100)/(m·g×0.01)

d₄ = 100 + d₁

∴ d₄ = 100 + (1/2·m·U² - 140×100)/(m·g×0.01)

∴ 1/2 × 7,200 × U² = 2 × 140 × (100 + (1/2 × 7,200 × U² - 140×100)/(7,200 × 9.81 ×0.01))

3,600·U² = 280·(100 + (3,600·U² - 14,000)/706.32)

= 28000 + 280×3,600·U²/706.32 - 280 × 14,000/706.32

= 28000 - 280 × 14,000/706.32 + 1427.11518858·U²

3,600·U² - 1427.11518858·U² = 28000 - 280 × 14,000/706.32

U²·(3,600 - 1427.11518858) = (28000 - 280 × 14,000/706.32)

U² = (28000 - 280 × 14,000/706.32)/(3,600 - 1427.11518858) = 10.3319363649

U = √(10.3319363649) = 3.21433295801

The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is U ≈ 3.214 m/s

Percentage error = (3.214-3.115)/3.214 × 100 ≈ 3.1% < 5% (Acceptable)

The difference in value can come from difference in calculating methods

Answer:

Attenuators are electrical components designed to reduce the amplitude of a signal passing through the component, without significantly degrading the integrity of that signal. They are used in RF and optical applications

Explanation:

Answer:

hydraulic flute is energy transfer medium in all hydraulic system this simple function is only achieved by flute that does not easily trap gases trapped gas and forming problems should bring a higher level of compressibility to a fluid than its usual at a point to support a very steep pass trekking system

Answer:

A voltage divider is a simple series resistor circuit. It's output voltage is a fixed fraction of its input voltage. The divide-down ratio is determined by two resistors.