Answer:
The maximum shear stress set up in the material of the bar is:
τmax = 450 Nm / (0.038 m * 0.025 m) = 59.2 MPa
The angle of twist is:
θ = 450 Nm * (0.038 m^3) / (3 * 80 GN/m^2 * 0.025 m^2) = 0.0024 rad
The percentage error when using the approximate equations would be approximately 8.7%.
Carbon steel (AISI 1010) shafts of 0. 1 - m diameter are heat treated in a gas - fired furnace whose gases are at 1200 K and provide a convection coefficient of 100 W/m2,K. If the shafts enter the furnace at 300 K, how long must they remain in the furnace to achieve a centerline temperature of 800 K?
800 K's center line temperature must be attained in t = 58 seconds.
Explain about the centerline temperature?The normalized centerline temperature is determined by where the horizontal line crisscrosses the y-axis. Now, we'll use the second chart to determine the temperature at any point within the solid.
Following is the fundamental data: Surface energy balance equation, first: H = Rn - G - L. E, where H denotes the sensible heat flow from surface to air, Rn denotes net radiation to the surface, G denotes the heat absorbed into the soil, L denotes the latent heat of evaporation, and E denotes the rate of evaporation.
The temperature gradient of a heated body in a steady condition is defined as the ratio of the temperature difference to the distance between two places.
Given data
D = 0.1 m
h = 100
Carbon steel's specific heat (C) is 502.4.
1200 K is the furnace temperature.
800 K is the final temperature.
300 K at initial temperature
From the equation now (1)
㏑ 0.44 = -(0.001415 ) t
-(0.001415 ) t = - 0.82
t = 58 sec
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3. Soil has been compacted in an embankment at a bulk mass density rho of 2. 15 Mg/m^3 and water content w of 12%. Gs is 2. 65. Calculate dry mass density, void ratio, degree of saturation and air content. Would it be possible to compact this soil at w = 13. 5% to a dry mass density rhod of 2. 00 Mg/m^3?
S=110.07%, which is larger than 100% or full saturation. As a result, compact is impossible.
What is the soil compact ?[tex]$\gamma_b=2.15 \mathrm{Mg} / \mathrm{m}^3$[/tex]
[tex]$w=12 \%$\\$G_s=2.65$[/tex]
any list you can condense to [tex]$\omega=13.5 \%$[/tex]
[tex]r_d=2 \mathrm{Mg} / \mathrm{m} 3$$[/tex]
[tex]$\begin{aligned} \gamma d & =\frac{\gamma_b}{1+\omega} \\ & =\frac{2.15}{1+0.12}\end{aligned}$[/tex]
[tex]$\gamma d=1.919 \mathrm{Mg} / \mathrm{ms}^3$[/tex]
[tex]$\gamma_b=\frac{{Gs} \gamma \omega(1+\omega)}{1+e}$[/tex]
[tex]$2.15=\frac{(2.65)(1+0.12)}{1+e}$[/tex]
[tex]$e=0.38$[/tex]
[tex]$\begin{aligned} S & =\frac{w {Gos}}{e} \\ & =\frac{(0.12)(2.65)}{0.38}\end{aligned}$[/tex]
[tex]$S=83.68 \%$[/tex]
[tex]$\gamma_d=\frac{\left(1-n_a\right) G_{s \gamma_w}}{\left(1-h_a+w G_s\right)}$[/tex]
[tex]$h_a=0.165$[/tex]
[tex]$h _a=16.5 \%$[/tex]
Is it feasible to compact the aforementioned soil at a W=13.5 and [tex]$\gamma_d=2.00 \mathrm{Mg} / \mathrm{m} 3$[/tex]
[tex]$\begin{aligned} S & =\frac{\omega G_S}{e} \\ & =\frac{(0.135)(2.65)}{0.325} \times 100 \\ S & =110.07 \%\end{aligned}$[/tex]
[tex]$\begin{aligned} \text { rd } \quad & =\frac{{Cos} \delta \omega}{1+e} \\ 2 & =\frac{(2.65)}{1+e} \\ e & =0.325\end{aligned}$[/tex]
S=110.07%, which is larger than 100% or full saturation. As a result, compact is impossible.
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for his position as a lead mechanic, jeff needs to know when and how to use a 12 millimeter screwdriver and a 14 millimeter screwdriver for a specific task. this requires
Technical Knowledge is the understanding of the specific tools, techniques, and processes required to complete a task. In this case, Jeff needs to have a thorough understanding of when and how to use a 12 millimeter screwdriver and a 14 millimeter screwdriver for a specific task.
This requires him to have a deep understanding of the tools and techniques require completing the task, as well as the processes involved in using the tools correctly. Jeff must also be able to identify any potential risks or issues that may arise during the task and take steps to mitigate them. Additionally, Jeff must be able to troubleshoot any problems that may arise during the task and take the necessary steps to resolve them. Jeff must also be able to identify any potential safety hazards associated with the task and take steps to ensure that they are addressed. Jeff must be able to provide regular updates to his supervisor on the progress of the task and ensure that any changes or modifications are communicated in a timely manner. Finally, Jeff must be able to provide feedback and advice to his supervisor in order to ensure that the task is completed successfully.
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Combining data from two or more relational database tables is an example of ________.A. collationB. responseC. reviewD. detection
Combining data from two or more relational database tables is an example of collation. Hence option A is correct.
What is the data Collation about?Joining is a process of combining data from two or more relational database tables based on a related column between them
Collation is a term used in the context of relational databases to refer to the process of combining data from two or more tables. This can be done through joining the tables on a common column, creating a new table that contains all the columns from the joined tables and the matching rows from each table.
Therefore, Collation is a way to combine data from multiple tables into a single result set.
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given a string s of characters, a subsequence of s, defined by the indices i 1, i 2, ..., i m , is said to be a repeated subsequence if you can find another sequence of indices j 1, j 2, ..., j m, such that s[i k]
Answer:
= s[j k] for all 1 <= k <= m.
In other words, a repeated subsequence is a sequence of characters within the original string that appears in the same order at least twice. The indices i1, i2, ..., im refer to the positions of the characters in the original string that make up the repeated subsequence, and the indices j1, j2, ..., jm refer to the positions of the same characters in the second occurrence of the repeated subsequence.
this bed load is what causes streams to abrade their streambed sand, and pebbles that are light enough to be picked up by the water and bounced along the bottom of the stream are moved in a process called .
The process is called traction or saltation.
What is traction or saltation?Traction or saltation is the process of sediment transport in which particles are moved by rolling or bouncing along the surface of the Earth. This process occurs when the wind, water, or ice apply enough force to the sediment particles to move them along the Earth's surface. The particles may roll, bounce, or slide along the surface.
This type of sediment transport is often seen in arid and semi-arid regions, where the wind is strong enough to move the particles, although it can also occur in areas with strong currents such as rivers and oceans.
Traction or saltation is one of three major processes of soil erosion, along with suspension and sheet erosion. It is most effective in carrying coarse-grained sediment, such as sand and gravel, while finer-grained sediment is usually carried away by suspension.
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The rolling or dragging of large grains along a river bed or shore, aided by the push of smaller grains, is known as traction.
What is traction?Traction is a river transportation method for large stones or boulders. The water rolls the stones along the river bottom because they are too large to transport in the water.This type of sediment transport is common in arid and semi-arid regions where the wind is strong enough to move the particles, but it can also happen in areas with strong currents like rivers and oceans.Traction, along with suspension and sheet erosion, is one of the three major processes of soil erosion. It is most effective at transporting coarse-grained sediment, such as sand and gravel, whereas finer-grained sediment is usually transported by suspension.To learn more about traction refer to :
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Draw shear Force and bending moment diagram for the beam given below.
The primary purpose of beams as structural components is to support vertical loads.
What is Force and bending moment?The highest shear and maximum moment locations, as well as the corresponding magnitudes, should be noted while constructing a beam because there is where the structure is most likely to break.
We must measure the shear force and bending moment at every location along the whole length of the beam in order to identify these important places. The method to determine the shear and bending moment at a single location was described in the previous part.
While this method is helpful, you will need a more potent strategy to determine the shear and moment at every point in the item. A shear and bending moment diagram can be made to do this.
Therefore, The primary purpose of beams as structural components is to support vertical loads.
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a 200-ft-long section of a steam pipe whose outer diameter is 4 inches passes through an open space at 50 f. the average temperature of the outer surface of the pipe is measured to be 280 f, and the average heat transfer coecient on that surface is determined to be 6 btu/(h ft2 f). determine: (a) the rate of heat loss from the steam pipe (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an eciency of 86 percent,
Therefore, the annual cost of this energy loss can be calculated as: Cost = 1,531,200 * 0.86 * (Cost of natural gas per BTU) * 8760 = $XXXX.
The energy cost of the steam pipethe cost of natural gas is $10/mmbtua) The rate of heat loss from the steam pipe is:Q = hA(Tsurf - Tamb)
Q = (6 BTU/h ft2 F)(π*D*L)(280-50)
Q = (6 BTU/h ft2 F)(π*4 in * 200 ft)(230)
Q = 4,711,200 BTU/h
b) The annual cost of this energy loss is:Cost = (Q * 8760 hr/yr * $10/MMBtu)/(1,000,000 BTU/MMBtu)
Cost = (4,711,200 BTU/h * 8760 hr/yr * $10/MMBtu)/(1,000,000 BTU/MMBtu)
Cost = $39,690/yr
The rate of heat loss from the steam pipe can be calculated using the equation: Q = hAΔT, where Q is the rate of heat loss, h is the heat transfer coefficient, A is the surface area of the pipe, and ΔT is the difference between the average surface temperature (280F) and the ambient temperature (50F).The surface area of the pipe is A = π * (D/2)^2 * L, where D is the diameter of the pipe and L is the length of the pipe. Therefore, the rate of heat loss can be calculated as: Q = hAΔT = 6 (BTU/(h ft2 F)) * π * (4/2)^2 * 200 * (280-50) = 1,531,200 BTU/h.The annual cost of this energy loss can be calculated using the equation: Cost = Q * E * C * 8760, where Q is the rate of heat loss, E is the furnace efficiency, C is the cost of natural gas per BTU, and 8760 is the number of hours in a year.This is called the energy cost of the steam pipe.To learn more about the energy cost of the steam pipe refer to:
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As a result, the yearly cost of this energy loss may be computed as follows: Cost = 1,531,200 * 0.86 * (natural gas cost per BTU) * 8760 = $XXXX.
hieHow to get the solutionThe steam pipe’s energy cost
Natural gas costs $10 per mmbtu.
A)The heat loss rate from the steam pipe is:
Q = hA(Tsurf – Tamb)
Q = (6 BTU/h ft2 F)(π*D*L)(280-50)
Q = (6 BTU/h ft2 F)(π*4 in * 200 ft)(230)
Q = 4,711,200 BTU/h
b) The yearly cost of this energy loss is as follows:
Cost = (Q * 8760 hours per year * $10/MMBtu)/(1,000,000 BTU/MMBtu)
(4,711,200 BTU/h * 8760 hours per year * $10/MMBtu)/(1,000,000 BTU/MMBtu)
$39,690 per year
The following equation may be used to calculate the rate of heat loss from the steam pipe: Q = hAT, where Q is the heat loss rate, h is the heat transfer coefficient, A is the pipe surface area, and T is the difference between the average surface temperature (280F) and the ambient temperature (50F).
The surface area of the pipe is A = * (D/2)2 * L, where D is the pipe’s diameter and L is its length. As a result, the heat loss rate may be estimated as follows: Q = hAt = 6 (BTU/(h ft2 F)) * * (4/2)2 * 200 * (280-50) = 1,531,200 BTU/h.
The yearly cost of this energy loss may be computed as
Cost = Q * E * C * 8760,
where Q is the rate of heat loss,
E is furnace efficiency,
C is natural gas per BTU cost, and 8760 is the number of hours in a year.
This is referred to as the energy cost of the steam pipe.
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