Team communication is often more formal than other types of group communication.
True
False

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:

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:

(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:

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:

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:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

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

Explanation:

Answer:

a. The expected number of failures due to a problem with inside wiring is 8 failures

b. The probability that at least 10 failures are due to inside wiring is approximately 0.176

c. It will be not unusual

Explanation:

The probability that the problem of an out of order is the inside wiring, P(x) = 8%

The number of complaints in a month period, x = 100

a. The expected number of failures due to a problem with inside wiring, E(x) = x·P(x)

∴ E(x) = 100 × 8% = 8

The expected number of failures due to a problem with inside wiring, E(x) = 8 failures

b. The probability that at least 10, P₁₀, failures are due to inside wiring is given as follows;

The probability of success, P = 0.08, therefore, the probability of failure, q = 1 - 0.08 = 0.92

P = [tex]_nC_r[/tex]·[tex]P^r[/tex]·[tex]q^{n-r}[/tex]

P₀ = ₁₀₀C₀·(0.08)⁰·(0.92)¹⁰⁰ = 0.000239211874657

P₁ = ₁₀₀C₁·(0.08)¹·(0.92)⁹⁹ = 0.00208010325

P₂ = ₁₀₀C₂·(0.08)²·(0.92)⁹⁸ = 0.00895348793

P₃ = ₁₀₀C₃·(0.08)³·(0.92)⁹⁷ = 0.02543309616

P₄ = ₁₀₀C₄·(0.08)⁴·(0.92)⁹⁶ = 0.0536306593

P₅ = ₁₀₀C₅·(0.08)⁵·(0.92)⁹⁵ = 0.0895398833653

P₆ = ₁₀₀C₆·(0.08)⁶·(0.92)⁹⁴ = 0.123279549561

P₇ = ₁₀₀C₃·(0.08)⁷·(0.92)⁹³ = 0.143953759735

P₈ = ₁₀₀C₈·(0.08)⁸·(0.92)⁹² = 0.145518474516

P₉ = ₁₀₀C₉·(0.08)⁹·(0.92)⁹¹ = 0.129349755125

P₁₀ = ₁₀₀C₁₀·(0.08)¹⁰·(0.92)⁹⁰ = 0.10235502362

∴ P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀ = 0.82433004464

The probability of at least 10 failures are due problem with the inside wiring = 1 - (P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀) = 1 - 0.82433004464 = 0.175666995536

The probability of at least 10 failures are due problem with the inside wiring = 0.175666995536 ≈ 0.176

c. The probability of at most 5 failures are due problem with the inside wiring = P₀ + P₁ + P₂ + P₃ + P₄ + P₅ = 0.179876441908

Therefore, given that probability of at most 5 failures > The probability of 8 failures it will be not unusual since the cause of failure is more (92%) due to other causes which are more likely and therefore increase in the probability that there are fewer failures due inside wiring

Answer:

Your question has some missing information below is the missing information

Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )

answer : 300 kW , 95°c

Explanation:

Given data:

Fluid A ;

Temperature of Fluid ( Th1 )  = 420° C

mass flow rate (mh)  = 1 kg/s

Fluid B :

Temperature ( Tc1) = 20° C

mass flow rate ( mc ) = 1 kg/s

effectiveness of heat exchanger = 75% = 0.75

Determine the heat transfer rate and  exit temperature of fluid B

Cph = 1000 J/kgk

Cpc = 4000 J/Kgk

Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B

exit temp of fluid  B = 95°C

heat transfer = 300 kW

attached below is a the detailed solution

Answer:

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

Answer:

apple

Explanation:

is the answer toWhat is your favorite Electronic company? (E.g. Windows, Apple, Samsung...)

Explanation:

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

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:

length of tube = 4.12 m

Explanation:

Given data:

flow rate of engine oil = 1 kg/s

diameter of tube = 5-mm

inlet temperature of oil = 45°C

exit temperature of oil = 80°C

surface temperature of tube = 150°C

Determine the required length of the tube

attached below is a detailed solution to the given problem

length of tube = 4.12 m

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:

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:

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:

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:

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:

Acef

Explanation:

Edginuity 2021

Answer:

2,3,4,5

Explanation:

guy above me is wrong

Answer:

Hello your question is poorly written attached below is the complete question

answer :

1) 60 kHz

2)  Tmax  = ( 1 / 34000 ) secs

Explanation:

1) Determine the Nyquist sampling frequency, (in kHz), for the input signal.

F(s) = 2 * Fmax

Fmax = 30 kHz  ( since Xa(t) is band limited to 30 kHz )

∴ Nyquist sampling frequency ( F(s) ) = 2 * 30 = 60 kHz

2) Determine the largest sampling period (in s) .

Nyquist sampling period = 1 / Fs  = ( 1 / 60000 ) s

but there is some aliasing of the input signal ( minimum aliasing frequency > cutoff frequency of filter ) hence we will use the relationship below

=  2π - 2π * T * 30kHz  ≥  2π * T * 4kHz

∴ T ≤ [tex]\frac{1}{34kHz}[/tex]

largest sampling period ( Tmax ) = ( 1 / 34000 ) secs

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:

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