Answer:
196.07 m, 2.85 m and 0.15 m
Step-by-step explanation:
Frequency of AM radio signal, f₁ = 1530 kHz
Its wavelength can be given by :
![\lambda_1=(c)/(f_1)\\\\=(3* 10^8)/(1530* 10^3)\\\\=196.07\ m](https://img.qammunity.org/2021/formulas/physics/college/1jc4ty5t7qvys15eg8ex4kkd5cyt4476go.png)
Frequency of FM radio signal, f₂ = 105.1 MHz
Its wavelength can be given by :
![\lambda_2=(c)/(f_2)\\\\=(3* 10^8)/(105.1* 10^6)\\\\=2.85\ m](https://img.qammunity.org/2021/formulas/physics/college/2wakx65s1gxn0omne2qoudvaftrjmgtqob.png)
Frequency of cell phone signal, f₃ = 1.9 GHz
Its wavelength can be given by :
![\lambda_2=(c)/(f_2)\\\\=(3* 10^8)/(1.9* 10^9)\\\\=0.15\ m](https://img.qammunity.org/2021/formulas/physics/college/86xy1dgeh4lkedss1l5hzqqefs4yvcnqqg.png)
Hence, the required wavelengths are 196.07 m, 2.85 m and 0.15 m respectively.