A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb a
1 answer:
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Answer:
1) the volume of solid (Vs) is 4733.33 ft³
2) the total cost to pay Pit A is $58220
3) the total cost to pay to Pit B is $53344.67
4) the total cost to pay to Pit C is $34269.32
5) the total cost to pay to Pit D is $49132.02
Explanation:
given the data in the question;
Vs = ?, V = 7100 ft³, e = 0.5
1) Calculate the volume of solid (Vs)
Vs = V / ( 1 + e )
we substitute
Vs = 7100 / ( 1 + 0.5 )
Vs = 7100 / 1.5
Vs = 4733.33 ft³
Therefore, the volume of solid (Vs) is 4733.33 ft³
2) Calculate the total cost to pay to Pit A
VpitA / Vcomp = ( 1 + e) / ( 1 + e
)
we substitute
VpitA / 7100 = ( 1 + 1.5) / ( 1 + 0.5)
VpitA / 7100 = 1.3666666
VpitA = 1.3666666 × 7100
VpitA = 9703.33 ft³
Total cost = volume × unit cost = 9703.33 ft³ × 6 = $58220
Therefore, the total cost to pay Pit A is $58220
3) Calculate the total cost to pay to Pit B = Borrow Pit C: void ratio = 0.81 and the unit cost = $ 4 /ft3
Vpitb / Vcomp = ( 1 + e) / ( 1 + e
)
we substitute
VpitB / 7100 = ( 1 + 0.61) / ( 1 + 0.5)
VpitB / 7100 = 1.0733333
VpitB = 1.0733333 × 7100
VpitB = 7620.666 ft³
Total cost = volume × unit cost = 7620.666 × 7 = $53344.67
Therefore, the total cost to pay to Pit B is $53344.67
4) Calculate the total cost to pay to Pit C = Borrow Pit D: void ratio = 0.73 and the unit cost = $6 /ft3
VpitC / Vcomp = ( 1 + e) / ( 1 + e
)
we substitute
VpitC / 7100 = ( 1 + 0.81) / ( 1 + 0.5)
VpitC / 7100 = 1.2066666
VpitC = 1.2066666× 7100
VpitC = 8567.33 ft³
Total cost = volume × unit cost = 8567.33 × 4 = $34269.32
Therefore, the total cost to pay to Pit C is $34269.32
5) Calculate the total cost to pay to Pit D
VpitD / Vcomp = ( 1 + e) / ( 1 + e
)
we substitute
VpitD / 7100 = ( 1 + 0.73) / ( 1 + 0.5)
VpitD / 7100 = 1.1533333
VpitD = 1.1533333 × 7100
VpitD = 8188.67 ft³
Total cost = volume × unit cost = 8188.67 × 6 = $49132.02
Therefore, the total cost to pay to Pit D is $49132.02
Answer:
c) 1.75 g/cm³
Explanation:
Given that
Radii of the A ion, r(c) = 0.137 nm
Radii of the X ion, r(a) = 0.241 nm
Atomic weight of the A ion, A(c) = 22.7 g/mol
Atomic weight of the X ion, A(a) = 91.4 g/mol
Avogadro's number, N = 6.02*10^23 per mol
Solution is attached below
