**Answer:**

A

**Explanation:**

The answer is towels because towels after a little bit of just sitting around have a chemical reaction that cam cause them to spontaneously combust

**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 l**n(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

**Answer:**

30.25 in Hg will be equal to 14.855 psi

**Explanation:**

We have given 30.25 Hg pressure

We have to convert the pressure of 30.25 Hg into psi

We know that 1 inch of Hg = 0.4911 psi

So to convert 30.25 inch Hg in psi we have to multiply with 0.4911

We have to convert 30.25 in Hg

So

So 30.25 in Hg will be equal to 14.855 psi

**Answer:**

R = 4.75 lb (↑)

**Explanation:**

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.