You are inside a room, 15 m from the west wall, which has two doors (similar to slits) and you are centered between the two open doors that are 3 m apart. Someone is blaring a 200 Hz tone outside the door so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. a. As you walk (north) along the direction of the west wall with the doors (but maintain a distance 15 m from the west wall), how far will you walk (in m) to hear a minimum in the sound intensityb. What is the spacing (in m) between maxima in sound intensity along that axis?

The distance of first minimum from central maxima is 4.25 m

The spacing between maxima in sound intensity is 8.5 m.

Explanation:

Given that,

Distance D= 15 m

Frequency = 200 Hz

distance between slits a=3 m

We know that,

The width of fringe,

(a). We need to calculate the distance of first minimum from central maxima

Using formula for distance

Put the value into the formula

(b). We need to calculate the spacing between maxima in sound intensity

Using formula for the distance of first maxima

Hence, The distance of first minimum from central maxima is 4.25 m

The spacing between maxima in sound intensity is 8.5 m.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2