Answer:
Code is as follows
country_pop={"United States": 318463000, "Indonesia" : 25216480, "India" : 1247220000, "China": 1365830000}
for key,value in country_pop.items(): print(key +" has "+ str(value) +" people.")
Answer:
Flow-rate = 0.0025 m^3/s
Explanation:
We need to assume that the flow-rate of pure water entering the pond is the same as the flow-rate of brine leaving the pond, in other words, the volume of liquid in the pond stays constant at 20,000 m^3. Using the previous assumption we can calculate the flow rate entering or leaving the tank (they are the same) building a separable differential equation dQ/dt, where Q is the milligrams (mg) of salt in a given time t, to find a solution to our problem we build a differential equation as follow:
dQ/dt = -(Q/20,000)*r where r is the flow rate in m^3/s
what we pose with this equation is that the variable rate at which the salt leaves the pond (salt leaving over time) is equal to the concentration (amount of salt per unit of volume of liquid at a given time) times the constant rate at which the liquid leaves the tank, the minus sign in the equation is because this is the rate at which salt leaves the pond.
Rearranging the equation we get dQ/Q = -(r/20000) dt then integrating in both sides ∫dQ/Q = -∫(r/20000) dt and solving ln(Q) = -(r/20000)*t + C where C is a constant (initial value) result of solving the integrals. Please note that the integral of dQ/Q is ln(Q) and r/20000 is a constant, therefore, the integral of dt is t.
To find the initial value (C) we evaluate the integrated equation for t = 0, therefore, ln(Q) = C, because at time zero we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q is equal to the concentration of salt (mg/m^3) by the amount of liquid (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. Now the equation is ln(Q) = -(r/20000)*t + 26.9378, the only thing missing is to find the constant flow rate (r) required to reduce the salt concentration in the pond to 500 mg/L = 500000 mg/m^3 within one year (equivalent to 31536000 seconds), to do so we need to find the Q we want in one year, that is Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, therefore, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s that is approximately 0.0025 m^3/s.
Note:
- ln() refer to natural logarithm
- The amount of liquid in the tank never changes because the flow-rate-in is the same as the flow-rate-out
- When solving the differential equation we calculated the flow-rate-out and we were asked for the flow-rate-in but because they are the same we could solve the problem
- During the solving process, we always converted units to m^3 and seconds because we were asked to give the answer in m^3/seg
Explanation:
Engineering is an important and learned profession. As members of this profession, engineers are expected to exhibit the highest standards of honesty and integrity. Engineering has a direct and vital impact on the quality of life for all people. Accordingly, the services provided by engineers require honesty, impartiality, fairness, and equity, and must be dedicated to the protection of the public health, safety, and welfare. Engineers must perform under a standard of professional behavior that requires adherence to the highest principles of ethical conduct.
Answer:
It means that a feature is dimensioned more than once.
For example, suppose you have a drawing of a rectangle, and the width is dimensioned twice. That's an example of over-dimensioning.
Of course, that's a rather obvious example. Sometimes it's more subtle. Suppose the rectangle has the width and height dimensioned, but also the diagonal. This is also an example of over-dimensioning. If the width and height are dimensioned, then you don't need to dimension the diagonal; that's already fixed by Pythagorean theorem.
Answer:
W = 166.35 KJ
Explanation:
P₁ = 600 KPa
V₁ = 0.4 m³
V₂ = 0.2 m³
T = 300 K
W = ?
We can find the solution from the standard relation for work in an isothermal process
W = -n*R*T*Ln (Vf / Vi)
We know that
n*R*T = P*V ⇒ P₁*V₁ = P₂*V₂ = n*R*T = 600 *0.4 = 240
Now, we use the equation
W = -n*R*T*Ln (Vf / Vi) = - P₁*V₁*Ln (Vf / Vi)
⇒ W = -240*Ln (0.2 / 0.4) = 166.35 KJ