Explanation:
These are probably the most used tool in any Plumber’s tool box. Pliers are not just another tool for a Plumber, they become an extension of their arms. Most people think that sounds odd, but pliers are more than just a tool to grab or turn things.
These are probably the most used tool in any Plumber’s tool box. Pliers are not just another tool for a Plumber, they become an extension of their arms. Most people think that sounds odd, but pliers are more than just a tool to grab or turn things.Sometimes a piece of copper pipe won’t quite go into a fitting. By using the handle end as a mallet you can gently force it in without damaging/denting the pipe or fittings. Or, when a brute force is needed the jaw end becomes a hammer. On an old pair of pliers I took a grinder to form one side of the handle into a flathead screwdriver/pry bar.
Answer:
A) 209.12 GPa
B) 105.41 GPa
Explanation:
We are given;
Modulus of elasticity of the metal; E_m = 67 GPa
Modulus of elasticity of the oxide; E_f = 390 GPa
Composition of oxide particles; V_f = 44% = 0.44
A) Formula for upper bound modulus of elasticity is given as;
E = E_m(1 - V_f) + (E_f × V_f)
Plugging in the relevant values gives;
E = (67(1 - 0.44)) + (390 × 0.44)
E = 209.12 GPa
B) Formula for upper bound modulus of elasticity is given as;
E = 1/[(V_f/E_f) + (1 - V_f)/E_m]
Plugging in the relevant values;
E = 1/((0.44/390) + ((1 - 0.44)/67))
E = 105.41 GPa
Answer:
See the attached file for the calculation
Explanation:
Find attached of the calculations
Answer:
a) t = 165 days
b) 9.9 kg/day
Explanation:
Given data:
final lelvel of BOD is 20 mg/l
Ks = 100 mg BOD/L,
kd = 0.10 day-1 ,
μm = 1.6 day-1 ,
Y = 0.60 mg SS/mg BOD
a) we know that
solving for t
t = 165 days
b) mass of microbes
The tensile stress at the yield point is 286 MPa, the ultimate tensile stress is 509 MPa and the average stress is 1,018 MPa.
<h3>Tensile stress</h3>
I. Tensile stress at the yield point
Tensile stress=Fy/A1=4×90×10^3÷πd1²
=360×10^3÷(2×10^-2)²π
=286 MPa
II. Ultimate tensile stress
Fmax/A1=4×160×10^3÷πd1²
=640×10^3÷(2×10^-2)²π
=509 MPa
III. Average stress at the breaking point
Fd/A2=4×80×10^3÷πd2²
=320×10^3÷(10^-2)²π
=1,018 MPa
Therefore the tensile stress at the yield point is 286 MPa, the ultimate tensile stress is 509 MPa and the average stress is 1,018 MPa.
Learn more about tensile stress here:brainly.com/question/12937199
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