Answer:
F= 0.6321 N
Step-by-step explanation:
Given
Area ,A= 14 cm²
Density ,ρ = 1.204 kg/m³
Velocity ,v= 50 m/s
Drag coefficient ,C=0.3
The drag force on the golf given as
![F=(1)/(2)\rho CAv^2](https://img.qammunity.org/2020/formulas/physics/high-school/f5ds2voamo4qdj9xtrqybhutpfljmk68wb.png)
Now by putting the values
![F=(1)/(2)\rho CAv^2](https://img.qammunity.org/2020/formulas/physics/high-school/f5ds2voamo4qdj9xtrqybhutpfljmk68wb.png)
![F=(1)/(2)* 0.3* 1.204 * (14* 10^(-4))* 50^2\ N](https://img.qammunity.org/2020/formulas/physics/high-school/whno9i3a5lhwgvj2ambqillpws7rpg05hk.png)
F= 0.6321 N
Therefore force due to drag on the golf is 0.6321 N