Calculate the intensity of the sound waves from an electric guitar’s amplifier at a
distance of 5.0 m when its power output is equal to each of the following values:
a. 0.25 W
b. 0.50 W
c. 2.0 W

a. 0.8 mW/m² b. 1.6 mW/m² c. 6.4 mW/m²

Explanation:

Intensity,I = P/A were P = power and A = area = 4πr² were r = distance from electric guitar = 5.0 m.

a. When P = 0.25 W, I = 0.25 W/4π5² = 0.25 W/100π = 0.00079 W/m² ≅ 0.0008 W/m² =  0.8 mW/m²

b. When P = 0.50 W, I = 0.50 W/4π5² = 0.50 W/100π = 0.00156 W/m² ≅ 0.0016 W/m² =  1.6 mW/m²

c. When P = 2.0 W, I = 2.0 W/4π5² = 2.0 W/100π = 0.00636 W/m² ≅ 0.0064 W/m² =  6.4 mW/m²

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j