Answer:
![\lambda=4000\ km](https://img.qammunity.org/2020/formulas/physics/high-school/f820jpuwqs3zbgi84q6kdr4ibjtit4xmws.png)
Step-by-step explanation:
It is given that,
Frequency for submarine communications, f = 76 Hz
We need to find the wavelength of those extremely low-frequency waves. the relation between the wavelength and the frequency is given by :
![c=f* \lambda](https://img.qammunity.org/2020/formulas/physics/middle-school/zyr5jhd8r4hsmcmkclmt2knxuok87as7cg.png)
![\lambda=(c)/(f)](https://img.qammunity.org/2020/formulas/physics/college/bvf2efxdcei63o2p5hc47z9087chl772m8.png)
![\lambda=(3* 10^8\ m/s)/(76\ Hz)](https://img.qammunity.org/2020/formulas/physics/high-school/jdksfddhf28lry9t176mbh1udyjbs7ivko.png)
![\lambda=3947368.42\ m](https://img.qammunity.org/2020/formulas/physics/high-school/af5dyd2d7wh0aruiw341vxcnusrjg2a8wp.png)
![\lambda=3947\ km](https://img.qammunity.org/2020/formulas/physics/high-school/9ti32eogpquza110fjh9ygvyxvx5etdxh0.png)
or
![\lambda=4000\ km](https://img.qammunity.org/2020/formulas/physics/high-school/f820jpuwqs3zbgi84q6kdr4ibjtit4xmws.png)
So, the wavelength of those extremely low-frequency waves is 4000 km. Hence, this is the required solution.