Answer:
A(t) = 160 - 130 e^(-t/40)
Explanation:
At the start, the tank contains A(0) = 30 g of salt.
Salt flows in at a rate of
(1 g/L) * (4 L/min) = 5 g/min
and flows out at a rate of
(A(t)/160 g/L) * (4 L/min) = A(t)/40 g/min
so that the amount of salt in the tank at time t changes according to
A'(t) = 4 - A(t)/40
Solve the ODE for A(t):
A'(t) + A(t)/40 = 4
e^(t/40) A'(t) + e^(t/40)/40 A(t) = 4e^(t/40)
(e^(t/40) A(t))' = 4e^(t/40)
e^(t/40) A(t) = 160e^(t/40) + C
A(t) = 160 + Ce^(-t/40)
Given that A(0) = 30, we find
30 = 160 + C
C = -130
so that the amount of salt in the tank at time t is
A(t) = 160 - 130 e^(-t/40)
Answer:
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Explanation:
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Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress, , is given by the following formula;
= 1 cm = 0.01
h = 29 cm = 0.29 m
= 25 cm = 0.25 m
b = 15 cm = 0.15 m
= The centroidal moment of inertia
= 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;
= 1.2257083 × 10⁻⁴ m⁴
From which we have;
Which gives;
W = 11,416.6879 N
= 1500 N/cm² = 15,000,000 N/m²
= 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;
L ≈ 0.64417 m ≈ 64.417 cm.