2 A SQUARE GIVEN
LENGTH OF DIAGONAL = 70 mm
​

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

[tex]\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}[/tex]

[tex]Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t[/tex]

Therefore, we have;

[tex]Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522[/tex]

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

Answer:

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

Answer:

Explanation:

[tex]\text{The curve of the plot}[/tex] [tex]\mathbf{E_A,E_R, \ and \ E_N}[/tex] [tex]\text{can be seen in the attached diagram below}[/tex]

[tex]\text{From the plot}[/tex], [tex]\mathbf{r_o = 0.2 4nm \ and \ E_o =-5.3 eV}[/tex]

[tex]\mathbf{We \ knew \ that: E_N = E_A + E_R}[/tex]

[tex]\mathtt{GIven \ E_A = \dfrac{-1.436}{r}\ \ \ , E_R = \dfrac{7.32 \times 10^{-6}}{r^n} \ \ and \ \ n=8 }[/tex]

[tex]\mathtt{Then; E_N = -\dfrac{-1.436}{r}+ \dfrac{7.32\times 10^{-6}}{r^8}}[/tex]

[tex]\mathtt{Also; r_o = \Big( \dfrac{A}{nB} \Big)^{\dfrac{1}{1-n}}} \\ \\ \mathtt{ r_o = \Big( \dfrac{1.986}{8 \times 7.32\times 10^{-6}} \Big)^{\dfrac{1}{1-8}}} \\ \\ \mathbf{r_o = 0.236 nm}[/tex]

[tex]E_o = \dfrac{-1.436}{\Big[\dfrac{1.436}{8(732\times 10^{-6})}\Big]^{\dfrac{1}{1-8}}} + \dfrac{7.32 \times 10^{-6}}{\Big[ \dfrac{1.436}{8\times7.32 \times 10^{-6} } \Big]^{\dfrac{8}{1-8}}}[/tex]

[tex]\mathbf{E_o = -5.32 \ eV}[/tex]

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:

540 W

Explanation:

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:

Acef

Explanation:

Edginuity 2021

Answer:

2,3,4,5

Explanation:

guy above me is wrong

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

Explanation:

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

Answer:

a scale model each size is a certain amount smaller

Explanation:

Answer:

Hello your question has some missing part below is the missing part

Estimate The matching time for the Finish cut

answer:  79.588 seconds

Explanation:

Calculate the matching time for the Finish cut

Diameter of workpiece before cutting = 100 mm

hence

Diameter of workpiece before Finishing cut ( same as after rough cut )

D2 = D1 - 2 * d  = 100 - (2 * 12) = 76mm

step 1 : determine spindle speed

V = [tex]\frac{\pi D_{2}N_{s} }{60}[/tex]  -------- ( 1 ).  where : D2 = 76 mm , Ns = ? , V = 1.5  ( input values into equation 1 )

therefore Ns = 376.94 rpm

Finally : Determine the matching time

T = [tex]\frac{L + A}{Ft * Ns}[/tex] ---- ( 2 ) . where : Ft = 0.1  mm , Ns = 376.94 rpm, L = 50 mm , A = 0 ( input values into equation 2 )

note : A = allowance length ( not given ) , ft = feed rate

T = 50  / ( 0.1 * 376.94 )

   = 1.3265 minutes ≈ 79.588 seconds

Answer:

hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

attached below is a detailed solution to the given problem

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

[tex]I_{load}[/tex] = 0.75 < 25.84°

attached below is the remaining part of the solution

B) Find the input current on the primary side in real units

load current in primary = 31.38 < 25.84 A

C) find the input power factor

power factor = 0.9323 leading

attached below is the detailed solution

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:

The force needed to lift a 500 newton-load is 125 newtons.

The mechanical advantage of the lever is 4.

Explanation:

Needed force is equal to the weight of the load. The Law of Lever, which is a particular case of the definition of Torque, states that force is inversely proportional to distance from fulcrum, that is:

[tex]F_{F} \cdot r_{F} = F_{L}\cdot r_{L}[/tex] (1)

Where:

[tex]F_{F}[/tex] - Needed force, in newtons.

[tex]r_{F}[/tex] - Force arm, in meters.

[tex]F_{L}[/tex] - Load force, in newtons.

[tex]r_{L}[/tex] - Load arm. in meters.

The mechanical advantage of the lever ([tex]n[/tex]), no unit, is determined by following formula:

[tex]n = \frac{F_{F}}{F_{L}}[/tex] (2)

If we know that [tex]F_{F} = 500\,N[/tex], [tex]r_{F} = 2\,m[/tex] and [tex]r_{L} = 8\,m[/tex], then the load force needed to lift is:

[tex]F_{L} = F_{F} \cdot \left(\frac{r_{F}}{r_{L}} \right)[/tex]

[tex]F_{L} = 125\,N[/tex]

The force needed to lift a 500 newton-load is 125 newtons.

And the mechanical advantage of the lever is:

[tex]n = 4[/tex]

The mechanical advantage of the lever is 4.

Answer:

The chip will operate below a maximum allowable temperature of 85°C

Explanation:

Given data:

8-mm-thick aluminum

0.02 mm-thick epoxy joint

chip and substrate = 10 mm on a side

temperature = 25°C

attached below is a detailed solution

Tc = 75.3 ° c   which is less than 85°c . hence the chip will operate below a maximum allowable temperature of 85°C

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.

Answer:

6156 kg /day

Explanation:

Determine the COD mass loading in the river upstream of the industrial source discharge

Given data:

Flow rate of river = 8.7 m^3/s

Average COD concentration in river = 32 mg/L

Industrial source continuous discharge ( Qw )= 18,000 m^3/d

Yw = 342 mg/l

since :

1 m^3 = 1000 liters

Qw = 18 * 10^6  liters = ( 18 million per day )

Hence the COD mass loading

= Yw * Qw

= 342 * 18 liters

= 6156 kg /day

Answer:

a) 0.488 days ≈ 11.72 hour ,  8542.38 m^3

b) 2.05 kg/m^3.day

c)  0.37 day^-1

d)

i) volume of wasted solid = 783 m^3/day

ii) mass of wasted solid = 7830 kg/day

e) 1.12

Explanation:

Given data :

Flow ( Q ) = 17500 m^3/day

BOD 5 ( influent ) = 1000 mg/L

BOD 5 ( efficient ) = 120 mg/L

mean cell residence time ( ∅ c ) = 6-day

MLSS concentration ( X ) = 5500 mg/L

a) Determine the hydraulic retention time and volume of the activated sludge

i) Hydraulic retention time ( OH ) we will use the relation below

X = [tex]\frac{OcY(sin-s)}{OH(1 + Kdθc)}[/tex]  

5500 = [tex]\frac{6*0.6(1000-120)}{OH(1 + 0.03*6)}[/tex] .  therefore  OH = 0.488 days ≈ 11.72 hour

ii) volume of the activated sludge  using the relation below

 OH = V / Q

V = OH * Q

  = 0.488 * 17500

  = 8542.38 m^3

b) Determine the  volumetric loading rate in kg BOD,/m- day to the reactor

loading rate = ( Q * Sin ) /  V

                     = ( 17500 * 1000 ) / 8542.38

                     = 2.05 kg/m^3.day

c) Determine The F/M ratio in the reactor

F/M ratio :  F/M = ( Q .sin ) / VX

                         = ( 17500 * 1000 ) / ( 8542.37 * 5500)

                         = 0.37 day^-1

d) Determine The mass and volume of solids wasted each day, at an underflow solids concentration

i) volume of wasted solid = 783 m^3/day

ii) mass of wasted solid = 7830 kg/day

attached below is the remaining part of the solution

e) Sludge recirculation ratio

= 1.12

attached below is the remaining part of the solution