Answer:
option B.
Explanation:
The correct answer is option B.
Principal stress is the maximum normal stress a body can have. In principal stress, there is purely normal stress. On principal plane shear stress is zero.
In-plane shear stress are the shear stress which is acting on the plane.
The statement which is correct regarding principle plane and shear stress is that The shear stress over principal stress planes is always zero.
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
Answer:
Functional Relationship
Explanation:
A functional relationship is only achieved through control and involves a specific change in one event (dependent variable) that can reliably be produced by specific manipulations of another event (independent variable), and the change in the dependent variable is unlikely to be the result of other extraneous factors (confounding variables
Answer:
74 Ω
Explanation:
Data provided in the question:
Maximum value of the current that can be provided = 500 mA
= 500 × 10⁻³ A
Applied voltage set for the system = 37 V
Now,
The smallest resistance that can be measured
= [ Applied voltage ] ÷ [ Maximum current that can be applied ]
= 37 ÷ ( 500 × 10⁻³ )
= 74 Ω
