Answer:
It will produce 2.5 Kw at 0.6m high
Step-by-step 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