. Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high?

4.988kW

Explanation:

According to the question, energy E extracted from the ocean breaker is directly proportional to the intensity I. It can be expressed mathematically as E ∝ I

E = kI where k is the constant of proportionality.

From the formula; k = E/I

This shows that increase in energy extracted will lead to increase in its intensity and vice versa.

If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high

E = 10kW and I = 1.20m

k = 10/1.20

k = 8.33kW/m

To know how much energy E that will be produced when they are 0.600 m high, we will use the same formula

k = E/I where;

k = 8.33kW/m

I = 0.600m

E = kI

E = 8.33 × 0.6

E = 4.998kW

The device will produce energy of 4.998kW when they are 0.600m high.

It will produce 2.5 Kw at 0.6m high

Explanation:

We are given;

Initial Power output of device; P_i = 10 Kw

Initial amplitude; A_i = 1.2m

Final Amplitude; A_f = 0.6m

We know that power is directly proportional to energy because

P = Energy(work done)/time taken

Thus; P ∝ E - - - - (eq1)

Now,from the question, we are told that Energy is proportional to the intensity. Thus;

E ∝ I - - - - (eq2)

where I is intensity

Now, from formula of Intensity, which is; I = (1/2)(ρ²•β²•ω²•A²)

We can see that I is directly proportional to square of Amplitude A²

Thus, I ∝ A² - - - - (eq3)

Combining eq 1,2 and 3,we can deduce that;

P ∝ E ∝ I ∝ A²

Thus, P ∝ A²

Now, let's set up the proportion as;

P_i/P_f = A_i²/A_f²

Since we are looking for final power, let us make P_f the subject.

So,

P_f = (P_i•A_f²)/A_i²

Plugging in the relevant values to obtain ;

P_f = (10 x 0.6²)/1.2² = 2.5 Kw

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .