Answer:
A. 171.24 Ibs
Explanation:
To find the amount of salt in the tank,
Let Q = Amount of salt in the mixture
And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.
Rate of gain - Rate of loss = dQ / dt
Concentration of salt = Q / (100+t)
For the linear differential equation,
dQ / dt = 3(2) - 2 [Q/ (100 + t)]
dQ /dt + Q [2 / (100 + t)] = 6
The general solution of the linear differential equation is:
Q (i.f) = ∫ A(t) (i.f) dt + C
Therefore,
i.f = e ^ ∫ P(t) dt
And P(t) = 2 / (100 + t)
i.f = e ^ ∫ 2 / (100 + t)
= e ^ 2㏑ (100 + t)
= e ^ ㏑ (100 + t) ^2 = (100 + t) ^2
Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C
Q(100 + t) ^2 = 2(100 + t) ^ 3 + C
When t = 0, Q = 50
Therefore,
50( 100) ^2 = 2(100) ^3 + C
C = -1.5 * 10 ^6
therefore, when t = 30,
Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6
Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6
Q = 171.24 Ibs
The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.
The given parameters:
Initial volume of the tank, i = 100 gallonsRate of gain of salt = 3 gpmRate of loss of salt = 2 gpmThe linear differential equation of the salt solution is calculated as follows;
[tex]\frac{dx}{dt} = Gain - loss[/tex]
where;
x is the salt concentrationThe salt concentration at time t, is calculated as follows;
[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]
Apply the general solution of linear differential equation as follows;
[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]
[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]
When t = 0 and X = 50
[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]
When t = 30 min, the concentration is calculated as;
[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]
Learn more about solution of Linear differential equation here: https://brainly.com/question/5508539
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
[tex] v = \frac{I}{nqA} [/tex]
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]
Now, we can find the drift speed:
[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
