Answer:

Answer:

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

Explanation:

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:
### Final answer:

### Explanation:

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.

Learn more about **downstream** here:

#SPJ3

) A circular coil of diameter 20. cm, with 16. turns is in a 0.13 Tesla field. (a) Find the total flux through the coil when the field is perpendicular to the coil plane. (b) If the coil is rotated in 10. ms so its plane is parallel to the field, find the average induced emf.

The table below shows the mass and velocity of four objects. Which object has the least inertia?A. YB. ZC. XD. W

As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Monochromatic coherent light shines through a pair of slits. If the distance between these slits is decreased, which of the following statements are true of the resulting interference pattern?A) The distance between the maxima stays the same.B) The distance between the maxima decreases.C) The distance between the minima stays the same.D) The distance between the minima increases.E) The distance between the maxima increases.

The table below shows the mass and velocity of four objects. Which object has the least inertia?A. YB. ZC. XD. W

As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Monochromatic coherent light shines through a pair of slits. If the distance between these slits is decreased, which of the following statements are true of the resulting interference pattern?A) The distance between the maxima stays the same.B) The distance between the maxima decreases.C) The distance between the minima stays the same.D) The distance between the minima increases.E) The distance between the maxima increases.

The **force** between objects that are any **distance apart** is expressed as

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;

To find the **force **between objects that are any **distance apar**t, 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

Learn more on **gravitational law **here: brainly.com/question/11760568

**Answer:**

F' = 100 F/r²

**Explanation:**

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**

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

**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.

**Answer:**

no

**Explanation:**

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.

**Answer:**

**Seismology.**

**Explanation:**

**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.**

**Answer:**

W = ½ m v²

**Explanation:**

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

= m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2

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

= 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

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.

#SPJ11

**Answer:**

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

**Explanation:**

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

**Answer:**

**U. With no variation.**

**Explanation:**

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.