The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How wide is the "sound beam" 100. M outside the cave opening? Use v sound= 340. M/s.

The value is          

Explanation:

From the question we are told that

     The length of the crack is  

     The  frequency is  

      The distance outside the cave that is being consider is  

      The speed of sound is

Generally the wavelength of the wave is mathematically represented as

=>    

=>    

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

=>    

=>    

Generally the width of the sound beam is mathematically represented as

=>      

=>              

he made 11 calls in one day

the total distance traveled by the car is

30km + 49km = 79 km .