Ultrasound of intensity 1.50 x 10^2 W/m^2 is produced by the rectangular head of a medical imaging device measuring 3 cm by 5 cm. What is it's power output?
To find the power output, we need to calculate the total power (P) produced by the rectangular head of the medical imaging device.
First, let's convert the dimensions of the rectangular head from centimeters to meters:
Length (L) = 3 cm = 3/100 m = 0.03 m
Width (W) = 5 cm = 5/100 m = 0.05 m
Now, the intensity of the ultrasound wave is given as 1.50 x 10^2 W/m^2, which represents the power per unit area (I = P/A), where P is the power output and A is the cross-sectional area of the rectangular head.
Since the rectangular head has a length of 0.03 m and a width of 0.05 m, the cross-sectional area is:
A = L x W = 0.03 m x 0.05 m = 0.0015 m^2
Now, we can rearrange the formula for intensity to solve for power:
I = P/A -> P = I x A
Substituting the given values:
P = (1.50 x 10^2 W/m^2) x (0.0015 m^2)
P = 0.225 W
Therefore, the power output of the ultrasound produced by the rectangular head of the medical imaging device is 0.225 watts.