Simplify -8.6 + 2.3 + 1 -5.3 -7.3 -9.9

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

3 years ago

Find all solutions of the equation in the interval [0, 2π). cotθ+ √3= 0

Ilia_Sergeevich [38]

Answer:

{5π/6, 11π/6}

Step-by-step explanation:

Since you have memorized the trig values of common angles, you know tan(π/6) = 1/√3, so cot(π/6) = √3.

The solution to this equation is ...

cot(θ) = -√3

so θ = -π/6 or, in the domain of interest, 11π/6. There is a corresponding quadrant II angle, 5π/6.

5 0

3 years ago

Sunlight is a clean energy source that does no reproduce

Alex777 [14]

Energy is the correct answer

4 0

2 years ago

Find the equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5).

neonofarm [45]

The equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5) is $y = \frac{-1}{3}x - 8$

Solution:

Given that line perpendicular to y = 3x + 6 and containing the point (-9, -5)

We have to find the equation of line

The slope intercept form is given as:

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Let us first find the slope of line

The given equation of line is y = 3x + 6

On comparing the given equation of line y = 3x + 6 with eqn 1, we get,

m = 3

Thus the slope of given equation of line is 3

We know that product of slopes of given line and slope of line perpendicular to given line is equal to -1

Slope of given line $\times$ slope of line perpendicular to given line = -1

$3 \times \text{ slope of line perpendicular to given line }= -1$

$\text{ slope of line perpendicular to given line } = \frac{-1}{3}$

Let us now find the equation of line with slope $m = \frac{-1}{3}$ and containing the point (-9, -5)

Substitute $m = \frac{-1}{3}$ and (x, y) = (-9, -5) in eqn 1

$-5 = \frac{-1}{3}(-9) + c\\\\-5 = 3 + c\\\\c = -8$

Thus the required equation of line is:

Substitute $m = \frac{-1}{3}$ and c = -8 in eqn 1

$y = \frac{-1}{3}x - 8$

Thus the required equation of line perpendicular to given line is found

6 0

3 years ago

Can an eulerian circuit also be an eulerian path?

kap26 [50]

Answer: i think yes

Step-by-step explanation:

5 0

3 years ago