A baker is weighing the dough that will be used to make pastries. The line plot shows the weight of the dough for each pastry.

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

[tex]E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%[/tex]

Answer:

organizational goals

Explanation:

Answer:

4.2019 mts , 4.598 mts

Explanation:

center of flotation = 2 m fwd.

MCTC = 120 tonnes

Determine the position of  200 t of cargo to be loaded

first step : determine the change trim = trimming moment / MCTC

                                                              = 134 * 1 / 120 = 0.11 mts

∴ бtf = 0.11 * ( 2/100 ) = 0.0022 mts ( given that the ship trims forward )

бtf = - 0.0022 mts

also ; бTa = 0.0019 mts (  + due to increase in draught )

determine the Final position

Ta = 4.2 ( draught at AP )  + бTa

     = 4.2 +  0.0019

     = 4.2019 mts

T[tex]_{f}[/tex] = 4.6 ( draught at Fp )  + бtf

                                = 4.6 + ( - 0.0022 )  = 4.598 mts

Answer:

Technical communication is a means to convey scientific, engineering, or other technical information.[1] Individuals in a variety of contexts and with varied professional credentials engage in technical communication. Some individuals are designated as technical communicators or technical writers. These individuals use a set of methods to research, document, and present technical processes or products. Technical communicators may put the information they capture into paper documents, web pages, computer-based training, digitally stored text, audio, video, and other media. The Society for Technical Communication defines the field as any form of communication that focuses on technical or specialized topics, communicates specifically by using technology or provides instructions on how to do something.[2][3] More succinctly, the Institute of Scientific and Technical Communicators defines technical communication as factual communication, usually about products and services.[4] The European Association for Technical Communication briefly defines technical communication as "the process of defining, creating and delivering information products for the safe, efficient and effective use of products (technical systems, software, services)".[5]

Whatever the definition of technical communication, the overarching goal of the practice is to create easily accessible information for a specific audience.[6]

Answer:

12.332 KW

The positive sign indicates work done by the system ( Turbine )

Explanation:

Stagnation pressure( P1 ) = 900 kPa

Stagnation temperature ( T1 ) = 658K

Expanded stagnation pressure ( P2 ) = 100 kPa

Expansion process is  Isentropic, also assume steady state condition

mass flow rate ( m ) = 0.04 kg/s

Calculate the Turbine power

Assuming a steady state condition

( p1 / p2 )^(r-1/r)  = ( T1 / T2 )

= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )

=  ( 9 )^0.285 = 658 / T2

∴ T2 = 351.22 K

Finally Turbine Power / power developed can be calculated as

Wt = mCp ( T1 - T2 )

    = 0.04 * 1.005 ( 658 - 351.22 )

    = 12.332 KW

The positive sign indicates work done by the system ( Turbine )

Answer:

a) | v2 | ,  β2   ( load, bus voltage at bus 2 )

  p1 ,  q1 ( slack, bus power at bus 1 )

b) q2 , β2  

  p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

a) Identify the variables in the solution vector assume Bus 2 is a load bus

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 |  and β1 been specified while p1 and q1 are not specified

Hence the variables in the solution

= | v2 | ,  β2   ( load, bus voltage at bus 2 )

  p1 ,  q1 ( slack, bus power at bus 1 )

b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 |  and β1

unspecified : p1 and q1

Hence the variables in the solution

= q2 , β2  

  p1 and q1 ( slack, bus power at bus 1 )

Answer:

M/A = 0.425 g/cm²

Explanation:

Given the following data;

Mass = 192 grams

Dimensions = 7 * 10 inches.

To find the mass per unit area;

First of all, we would determine the area of the lead sheet;

Area = 7 * 10

Area = 70 in²

Conversion:

1 square inch = 6.452 square centimeters

70  inches = 70 * 6.452 = 451.64 square centimeters

Next, we find the mass per unit area;

M/A = 192/451.64

M/A = 0.425 g/cm²

Answer:

a. True

Explanation:

The study of fluids in a state of rest or in motion and the forces involved in it is called fluid mechanics. Fluid mechanics has a wide range of applications in the field of mechanical engineering as well as civil engineering.

When we study the flow of fluid between any two flat plates that is indefinitely flat and is parallel, the flow of the fluid is known as plane Poiseulle flow. The profile of a plane Poiseulle flow is parabolic.

The velocity profile of a plane Poiseulle flow is :

[tex]$\frac{u(y)}{U_{max}}=1-\left(\frac{2y}{h}\right)^2$[/tex]

Thus the answer is TRUE.

C. Semiconductors.

They are made up of what is called a porous silicon material that is very light and extremely heat resistant.

Answer:

composites

Explanation:

Answer:

For the Top Side

- Strain ε  = 0.00021739

- Elongation is 0.00260868 cm

For The Right side

- Strain ε  = 0.00021739

-Elongation is 0.00347826 cm

Explanation:

Given the data in the question;

Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m

Thickness = 5 mm = 0.005 m

Force to the Top F[tex]_t[/tex] = 15 kN = 15000 Newton

Force to the right F[tex]_r[/tex] = 20 kN = 20000 Newton

elastic modulus, E = 115 GPa = 115 × 10⁹ pascal

Now, For the Top Side;

- Strain = σ/E = F[tex]_t[/tex]  / AE

we substitute

= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 15000 / 69000000

Strain ε  = 0.00021739

- Elongation

Δl = ε × l

we substitute

Δl = 0.00021739 ×  12 cm

Δl = 0.00260868 cm

Hence, Elongation is 0.00260868 cm

For The Right side

- Strain = σ/E = F[tex]_r[/tex]  / AE

we substitute

Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 20000 / 69000000

Strain ε = 0.000289855

- Elongation

Δl = ε × l

we substitute

Δl = 0.000289855×  12 cm

Δl = 0.00347826 cm

Hence, Elongation is 0.00347826 cm

Answer:

- the undrained friction angles for the soil is 25.02°

- the drained friction angles for the soil is 18.3°

Explanation:

Given the data in the question;

First we determine the major principle stress using the express;

σ₁ = σ₃ + (Δσ[tex]_d[/tex] )[tex]_f[/tex]

where σ₃ is the total minor principle stress at failure ( 15 lb/in² )

(Δσ[tex]_d[/tex] )[tex]_f[/tex] is the deviator stress ( -9 lb/in² )

so

σ₁ = 15 lb/in² + 22 lb/in²

σ₁ = 37 lb/in²

Now, we calculate the consolidated-undrained friction angle as follows;

∅ = sin⁻¹[ (σ₁ - σ₃ ) / ( σ₁ + σ₃ ) ]

∅ = sin⁻¹[ (37 - 15 ) / ( 37 + 15 ) ]

∅ = sin⁻¹[ 22 / 52  ]

∅ = sin⁻¹[ 0.423 ]

∅ = 25.02°

Therefore, the undrained friction angles for the soil is 25.02°

-  The drained friction angles for the soil;

∅ = sin⁻¹[ (σ₁ - σ₃ ) / ( σ₁ + σ₃ - 2(Δσ[tex]_d[/tex] )[tex]_f[/tex] ) ]

so we substitute

∅ = sin⁻¹[ (37 - 15 ) / ( 37 + 15 - 2( -9 ) ]

∅ = sin⁻¹[ 22 / ( 37 + 15 + 18 ) ]  

∅ = sin⁻¹[ 22 / 70 ]

∅ = sin⁻¹[ 0.314 ]

∅ = 18.3°

Therefore, drained friction angles for the soil is 18.3°

Answer:

ICC

Explanation:

The International Building Code (IBC) is a model building code developed by the International Code Council (ICC). It has been adopted for use as a base code standard by most jurisdictions in the United States.

Answer:

$Monitor

Explanation:

The command that would be used when running a test bench to monitor variables or signals ( i.e. changes in the values of specific variables and signa)

is the $Monitor command

This command is also used to monitor the varying values of signals during simulation.

Answer:

The receiving end Voltage ( VR )  will reduce significantly

Explanation:

Determine what happens to the receiving end voltage

Given that ; Load on transmission line is increased and Load has a lagging power factor

Increase in load = Increase in reactive/real power drawn from the transmission line and this will cause an increase in the current drawn as well.

Vs = VR + j XL I

Vs = sending end voltage

VR = receiving end voltage

jXL I = voltage drop across reactance

The rotating and shifting of Vs to a new power angle ( lagging power factor )  without a change in magnitude will cause the value of the VR receiving end voltage to drop significantly

Answer:

The Mechanical Properties include Elasticity, Plasticity, Ductility, Malleability, Hardness, Toughness, Brittleness, Tenacity, Fatigue, Fatigue resistance, Impact Resistance property, Machineability, Strength, Strain Energy, Resilience, Proof Resilience, Modulus of Resilience, Creep, Rupture, and Modulus of Toughness.

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, [tex]$h_0=2.2$[/tex] inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

[tex]$\Delta h = H_0-h_f=\mu^2R$[/tex]

     [tex]$=0.2^2 \times 14 = 0.56$[/tex] inches

[tex]$h_f = 2.2 - 0.56$[/tex]

     = 1.64 inches

Roll strip contact length (L) = [tex]$\sqrt{R(h_0-h_f)}$[/tex]

                                             [tex]$=\sqrt{14 \times 0.56}$[/tex]

                                             = 2.8 inches

Absolute value of true strain, [tex]$\epsilon_T$[/tex]

[tex]$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$[/tex]

Average true stress, [tex]$\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$[/tex] Psi

Roll force, [tex]$L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$[/tex]

                                 = 788,900 lb

For SI units,

Power = [tex]$\frac{2 \pi FLN}{60}$[/tex]  

           [tex]$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$[/tex]

           = 10399.81168 W

Horse power = 13.9357

Answer:

Explanation:

Pie charts generally should have no more than eight segments.

Answer:

rqbynqnyeuenqrqununrqruqrun

Answer:

A

Explanation:

2/10 = 1/5 = 0.2

Answer:

Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.

hope this helps

have a good day :)

Explanation:

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

h = 389.55213360 W/m²

h ≅ 389.6 W/m²

Answer:

I think D is the correct answer

Explanation:

hope it is

The container that traps the most heat is a shoebox covered in aluminum foil.

An aluminum foil is used for food storage, and to wrap items, such as meats, to minimize moisture loss while cooking.

Aluminum foil is also an excellent insulator. This is due to the fact that it reduces heat emission by reflecting it back.

Therefore, in a shoebox container, the aluminum foil helps to trap most heat better than a plastic wrap or wax paper.

Learn more about aluminum foil here:

https://brainly.com/question/277354

#SPJ2

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = [tex]$7.2 \ g/cm^3$[/tex]

                                                  = [tex]$7200 \ kg/m^3$[/tex]

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

[tex]$F=\rho g A h$[/tex]

Area, A [tex]$=2 \pi r^2$[/tex]

            [tex]$=2 \pi (0.23)^2$[/tex]

            [tex]$=0.3324 \ m^2$[/tex]

[tex]$F=\rho g A h$[/tex]

   [tex]$=7200 \times 9.81 \times 0.3324 \times 0.3$[/tex]

     = 7043.42 N

Answer:

W18 * 106

Explanation:

Section modulus of wide flange = 200 m^3

Determine the value of the lightest beam possible

The lightest beam possible that will satisfy the given condition will  have a section modulus ≥ 200m^3 ( note: it will also be the nearest to 200 in^3 )

From Beam Table ; The Lightest beam with its section modulus( 204 in^3) > 200in^3  is   W18 * 106

Answer:

A)

  D = 158.42 kmol/h

  B =  191.578 kmol/h

B) Rmin = 1.3095

Explanation:

a) Determine the distillate and bottoms flow rates ( D and B )

F = D + B ----- ( 1 )

Given data :

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B  ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D )  ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

B) Determine the minimum reflux ratio Rmin

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x   +  ( 0.97 / Rm + 1 )

q - line ;  y =  ( 9 / 9-1 ) x -  xf/9-1

therefore ; x = 0.45

Finally Rmin

=  (( 0.97 / (Rm + 1 ))  = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

Answer:

The composition of an alloy is 0.75%wt

Explanation:

Let alloy is a hypoeutectoid alloy.

So, we can apply lever rule which is shown below.

[tex]W_{a}=[/tex][tex]\frac{C_{0} -C_{a} }{C_{b}-C_{a} }[/tex]

We know that [tex]C_{a}=0.022[/tex] and [tex]C_{b}=6.7[/tex]

Given that [tex]W_{a}=0.109[/tex], we have to find [tex]C_{0}[/tex]

Thus,

[tex]0.109=\frac{C_{0}-0.022 }{6.7-0.022}[/tex]

Hence

[tex]C_{0}=0.75[/tex]%wt

Answer:

106 kips

Explanation:

Determine the service live load for a first floor interior column

following ASCE-7  guidelines

Area = 20 * 25 = 500 ft^2 which is > 400 ft^2

Nominal live load on every floor = 80 psf

Hence the reduced service live load ( L )

L = Lo [ 0.25 + 15 / √AI ]  ------ ( 1 )

Lo = unreduced load

AT = Tributary area = 500 * number of floors = 2000 ft^2

AI ( influence area ) = AT * number of floors = 2000 * 4 = 8000 ft^2

Lo = 1.6 * 80 psf * 2000 ft^2 = 256 kips

Input values into equation 1 above

L = 106 kips