Answer:
In Crystalline metals or materials, plasticity is examined from the perspective of the motion of linear defects or dislocations within the polymer chains.
Explanation:
When a temperature range below and near the glass transition temperature is reached, there is warping or contortion of structureless or malformed polymers. This warping happens as the polymer chains move over one another.
Unlike elasticity when requires or enables an object to resume its original dimensions, ductility is the quality of an element or material to change form albeit permanently.
Cheers
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top 
= 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π
/4) ×
= √(4Q/π)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though
Answer:
Explanation:
(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄
Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14
1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54
= 153.12 + 21.28 + 148.16 + 7.56
= 330.12 g/mol
which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable
The moles of methane produced will be given as
m = (4a + b -2c - 3d)/8
= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8
= (51.04 + 21.28 - 18.52 - 1.62)/8
= 52.18/8
= 6.5225
(b) Volume of methane V is given as
V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)
= 0.1461 m³ CH₄/kg lawn trimmings
(c) Energy will be given as
CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol
= 5805.025
≈ 5805 kJ/kg
