. 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

at right angles to each other