Answer:
the minimum shaft diameter is 35.026 mm
the maximum shaft diameter is 35.042mm
Explanation:
Given data;
D-maximum = 35.020mm and d-minimum = 35.000mm
we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6
so From table, Selection of International Trade Grades metric series
the grade tolerance are;
ΔD = IT7(0.025 mm)
Δd = IT6(0.016 mm)
Also from Table "Fundamental Deviations for Shafts" metric series
Sf = 0.026
so
D-maximum
Dmax = d + Sf + Δd
we substitute
Dmax = 35 + 0.026 + 0.016
Dmax = 35.042 mm
therefore the maximum diameter of shaft is 35.042mm
d-minimum
Dmin = d + Sf
Dmin = 35 + 0.026
Dmin = 35.026 mm
therefore the minimum diameter of shaft is 35.026 mm
Answer:
Blank wall
Explanation:
A wall that cannot be moved because it is carrying the weight of the roof is considered a blank wall.
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Answer:
(a) 0.243 m3/day
(b) 96 mg/l
(c) 0.426 m3/min
Explanation:
The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240= 9.6 mg/L
Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day
To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3
Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day
Approximately, the volume of sludge is 0.243 m3/day.
(b)
Efficiency of 60 percent is equivalent to 0.6
Efficiency=(influent concentration- flow rate)/influent concentration
0.6=(240-flow rate)/240
Flow rate= 96 mg/l
(c)
Cycle time= 0.243/0.57=0.4263157894736 m3/min
Rounded off, cycle time is 0.426 m3/day
Answer:
kindly check the explanation section.
Explanation:
In order to be able to preserve resources that are common there is need for each countries in the world to agree and cooperate. Also, policies should be set and made the standard for all countries involved. The developed countries should come together with developing measures to make sure that there is mutual agreements between these countries to preserve common resources. For instance, the most common resource is the Ozone layer, the industrialization in developed countries causes the depletion of ozone.
The main problem that the kind of preservation entails is how to get money for countries that are not rich enough to follow the agreed laydown rules. Other problems are: differences in economies and each countries have their own ways of regulating and preserving resources.
