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]
____ are intended to help provide the correct amount of ventilation and also help reduce cooling costs by allowing outside air to be introduced to the conditioned space when conditions are correct
HELP! It’s for an architecture class on PLATO
Select the correct answer.
Which association maintains the International Building Code?
A. NFPA
B. ICC
C. EPA
D. DOJ
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.
A small hot surface at temperature Ti-430K having an emissivity 0.8 dissipates heat by radiation into a surrounding area at T2-400K. If this radiation transfer process is characterized by a radiation heat transfer coefficient h, calculate the value of h (a) 14.4 W/m2.C (b) 114.4 W/m2C (c) 314.4 W/m2.C ( 514.4 W/m2.c
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
You will only have two attempts to answer this question correctly. Assuming you determine the required section modulus of a wide flange beam is 200 in3, determine the lightest beam possible that will satisfy this condition.
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
Pie charts should have no more than eight segments. True or False?
Answer:
Explanation:
Pie charts generally should have no more than eight segments.
A large tank, at 500 K and 200 kPa, supplies isentropic air flow to a nozzle. At section 1, the pressure is only 120 kPa. What is the temperature at section 1
Select the correct answer.
Which type of material functions as heat shields in space shuttles?
ОА. .
elastomers
OB.
metals
O c.
semiconductors
D.
plastics
E.
composites
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:
What is the mass per unit area of an 7 inch by 10 inch lead sheet that weighs 192 gm. ____________g / cm2
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
Explain in your own words how standard form can be used to better communicate information in certain scientific or engineering contexts.
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]
A clay sample was consolidated in a triaxial test under an all-around conning pressure of 15 lb/in.2. The sample was then loaded to failure in undrained condition by applying an additional axial stress of 22 lb/in.2. A pore water pressure sensor recorded an excess pore pressure of (Dud)f 5 29 lb/in.2 at failure. Determine the undrained and drained friction angles for the soil
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°
When you're running a test bench, you would like to include a printout to the screen to inform yourself on the status of the current simulation. Which command you could use to monitor specific variables or signals in a simulation every time one of the signals changes value?
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.
A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entirely in the upper part of the corresponding mold, with the maximum circle on the partition surface. If the density of the molten metal is 7.2 g / cm3 and the height of the pouring cavity above the partition surface is 300 mm, determine the metallostatic thrust that will be exerted on the upper mold at the end of casting.
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
IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The following properties of annealed copper are given: strength coefficient is 90,000 psi; true strain at the onset of non-uniform deformation is 0.45; and, engineering strain at yield is 0.11. The coefficient of friction between strip and roll is 0.2. The roll radius is 14inches and the rolls rotate at 150 rpm. Calculate the roll-strip contact length. Calculate the absolute value of thetrue strain that the strip undergoes in this operation. Determine the average true stress of the strip in theroll gap. Calculate the roll force. Calculate the horsepower required.
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
_____ is the set of processes used to get members of an organization to work together to further the interests of the organization.
Answer:
organizational goals
Explanation:
Which type of container traps the most heat a shoe box covered in aluminum foil plastic wrap or wax paper?
The container that traps the most heat is a shoebox covered in aluminum foil.
What is the use of 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:
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