The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.


Answer 1


The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L


Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

Answer 2

Final answer:

The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.


To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.

Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.

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Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?


The force between objects that are any distance apart is expressed as P'=(100P)/(r^2)

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,


M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

P=(GMm)/(10^2)\nP=(GMm)/(100)\nGMm = 100P

To find the force between objects that are any distance apart, we will use the same formula above to have;


Substitute the result above into the expression to have:


Hence the force between objects that are any distance apart is expressed as P'=(100P)/(r^2)

Learn more on gravitational law here:


F' = 100 F/r²


The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²


F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m


F = Gm₁m₂/10²

Gm₁m₂ = 100F   --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

1. Towards the end of a 400m race, Faisal and Edward are leading and are both running at 6m/s. While Faisal is 72m from the finish line Edward is 100m from the finish line. Realising this and to beat Faisal, Edward decides to accelerate uniformly at 0.2 m/s2 until the end of the race while Faisal keeps on the same constant speed. Does Edward succeed in beating Faisal?





Faisal will finish the race in ...

  (72 m)/(6 m/s) = 12 s

In order to beat Faisal, Edward's average speed in those 12 seconds must exceed ...

  (100 m)/(12 s) = 8 1/3 m/s

To achieve that average speed, Edward's acceleration must be ...

  (8 1/3 m/s -6 m/s)/(12 s/2) = 7/18 m/s² ≈ 0.3889 m/s²

Accelerating at only 0.2 m/s², Edward will not beat Faisal.


Additional comment

When acceleration is uniform, the average speed is reached halfway through the period of acceleration.

Much of our knowledge of the interior of the Earth comes from the study of planetary vibrations, which is the science of





  • Seismology is the beach of physical science that deals with the study of vibrations that comes out from the interior of the earth onto the surface and these vibrations are in the form of seismic waves that are primary, secondary, and surface waves.
  • The science of seismology tells about the magnitude and intensity of these waves that lead to planetary vibrations. These waves trigger earthquakes, floods, and even landslides.

A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is at rest just after the separation. How much work was done by the internal forces that caused the separation



W = ½ m v²


In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_(f) = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_(f)

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body


initial energy

         K₀ = ½ m v²

final energy

        K_(f) = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²


the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

Final answer:

The principle of conservation of momentum implies that no work is performed by the internal forces during the separation of the space vehicle. This is granted that external forces are ignored and the total momentum and kinetic energy of the closed system remain constant.


The subject you're asking about centers around the principle of conservation of momentum. In the case of this space vehicle, before separation, the momentum of the whole system is given by the product of the mass and velocity, mv. After separation, one piece is at rest, leaving the other piece with momentum mv. As there is no external force, the total momentum does not change, so no work is performed by the internal forces causing the separation.

In more detail, the principle of conservation of momentum states that the total linear momentum of a closed system remains constant, regardless of any interactions happening within the system. The system is 'closed' meaning that no external forces are acting upon it. In this case, the space vehicle and the two smaller pieces it separates into form a closed system. This is consistent with your question's stipulation to ignore external forces, such as gravitational forces.

This can also be understood from the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. If we consider the vehicle before and after the separation, the kinetic energy of the system remains the same: initially all the energy is concentrated in the moving vehicle, and after the separation, all the kinetic energy is transferred to the moving piece while the at-rest piece has none. Therefore, the work done by the internal forces - which would change the kinetic energy - must be zero.

Learn more about Conservation of momentum here:


A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.39 nm . Find the electric force on the proton.Express your answer using two significant figures. Enter your answers numerically separated by a comma.



The electric force on the proton is 8.2x10^-10 N


We use the formula to calculate the distance between two points, as follows:

r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:

r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m

We will use the following expression to calculate the electrostatic force:

F = (q1*q2)/(4*pi*eo*r^2)

Here we have:

q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2

Replacing values:

F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N

Some amount of ideal gas with internal energy U and initial temperature 1000C was compressed to half of volume meanwhile absolute pressure inside of a container increased twice. We can say that internal energy of this gas after compression in terms of U is (20.2, 20.1, 19.4, 19.5) Group of answer choices



U. With no variation.


Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.