Answer:
The minimum thickness t of the wall is 0.00446 mm
Explanation:
Solution
Given that
Pressure =670kPa = 0.670
σ allowable normal stress = 150 MPa
Inner diameter = 2mm
Steel = A516 grade 60
Now,
Since the hoop stress is twice the longitudinal stress, the cylindrical tank is more likely to fail from the hoop stress.
Thus
σ allowable = σₙ = pμ/t
=p (d/2)
150 MPa =0.670MPa * 2/2/t
=0.67/t
t=0.67/150
t =0.00446 mm
Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline
Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?
Answer: Technician B is correct
Explanation: Two types of engines exist , the two stroke (example, used in chainsaws) is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the four stroke engines(eg lawnmowers) which uses four strokes, 2-strokes during compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.
While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before forcing the mixture into the cylinder and do not require a pressurized system. The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and directed to the main moving parts of crankshaft through its channels.
We can therefore say that Technician A is wrong while Technician B is correct
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!