Answer:
The application of regulations in locations are very important.
Explanation:
The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.
A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 22,000 N (4950 lbf), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0
Answer:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
attached below is a detailed solution
A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?
Answer:
10.007
Explanation:
Assuming we have to find out the compression ratio of the engine
Given information
Cubic capacity of the engine, V = 245 cc
Clearance volume, V_c = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]
plugging values we get
245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]
Solving we get D =7 cm
therefore, L= 7/1.1 =6.36 cm
Now,
the compression ratio is given as:
r =(V+V_c)/V_c
on substituting the values, we get
r = (245+27.2)/27.2 =10.007
Hence, Compression ratio = 10.007
Which one of the following answer options are your employers responsibility
Where are your answer options?
Answer:
Implement a hazard communication program
Explanation: i took the quiz
Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temperature, although lightly-crosslinked NR can partially crystallize at room temperature when stretched. Apparently, Tm is elevated upon stretching which allows for crystallization above the Tm of the unstretched polymer. Explain.
Answer:
Explanation:
Crystalline melting temperature (Tm) is termed as the temperature required for a crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).
Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.
For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.
The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.
The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:
1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.
2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.
3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.
Construct the plane-stress yield envelopes in a principle stress space for both the Tresca and the von Mises yield theories using your calculated value of the yield strength to scale the envelopes. Indicate the two equivalent load paths corresponding to pure shear on the yield envelopes. Calculate the shear yield strength of Al 6061-T6 aluminum predicted by the above theories.
Answer:
Explanation:
The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.
From the given information:
The elastic modulus is:
[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]
Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.
The shear yield strength by using von Mises criteria is estimated as;
[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]
The shear yield strength by using Tresca criteria is:
[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]
In Databrawl, what team is most powerful?
A) Virus
B) Malware
C) Firewall security
D) Programs
Answer:
I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.
Explanation:
Where do greywater pipes generally feed into?
-Vent stack
-Water heater
-Waste stack
-Main supply
Answer:
c Waste stack
Explanation:
what's nested piezometer?
Explanation:
Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube
What two units of measurement are used to classify engine sizes?
QUESTION 6
Which of the following is NOT a resume format?
01. Chronological
O2. Portfolio
3. Functional
04. Combination
The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter of the rod be not to deform
Answer:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
Forces always act in equal and opposite pairs
A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.
Answer: hello your question lacks the required diagram attached below is the diagram
answer : 29528.1 N/m^2
Explanation:
Given data :
dimensions of tank :
Length = 5-m
Width = 4-m
Depth = 2.5-m
acceleration of tank = 2m/s^2
Determine the maximum gage pressure in the tank
Pa ( pressure at point A ) = s*g*h1
= 10^3 * 9.81 * 3.01
= 29528.1 N/m^2
attached below is the remaining part of the solution
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree
Answer:
it is indeed C
Explanation:
Answer:
c
Explanation:
4. What element is missing from construction drawings?
A. Physical arrangement of specific electrical equipment
B. Electrical layout
C. Electrical connections
D. Side elevation views
Answer:
C Electrical Connections
Explanation:
In reading says . However, electrical
connections aren’t shown in construction drawings.
Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.
How does distribution add value to goods and services being sold,
including intellectual property?
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Explanation:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.
Explanation:
hope it helps <33