Answer:10.961 N
Step-by-step explanation:
Given
mass flow rate in bucket is
![\dot{m}=75.5 g/s](https://img.qammunity.org/2020/formulas/physics/high-school/n8aruiwccr46vivwn7t65b9uje0m542dva.png)
mass of bucket
![m=0.540 kg](https://img.qammunity.org/2020/formulas/physics/high-school/xwdfh7zg8ve64gh9x0vfpuhk0waiciy767.png)
velocity of sand
![v=3.70 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/q5rejjgf5b1eo6adnyjjwpqtllejqgn5fb.png)
Force acting on bucket due to incoming of sand is given by
![F=\frac{\mathrm{d} m}{\mathrm{d} t}v](https://img.qammunity.org/2020/formulas/physics/high-school/3xfh83cl7tj87du6v6jq0qv9eimmsj8ayo.png)
![F=75.5* 10^(-3)* 3.7=279.35* 10^(-3) N](https://img.qammunity.org/2020/formulas/physics/high-school/48su3lub4keg6y4t6zdvxd55m10f933l0z.png)
Now weight of sand and bucket when 0.55 kg sand is Present in bucket
![W_1=(0.54+0.55)* 9.8](https://img.qammunity.org/2020/formulas/physics/high-school/gl87x2vb42mvqqm9l5066fcwa6w8bmggjt.png)
![W_1=1.09* 9.8=10.68 N](https://img.qammunity.org/2020/formulas/physics/high-school/ha5ozfzsl3k6wjf89vr6wfh91zvqhr5wf1.png)
Total reading on scale will be addition of weight and force acting by incoming sand
![Reading =10.68+0.279=10.961 N](https://img.qammunity.org/2020/formulas/physics/high-school/rb4p7qrfk01f0l5xwg3oxkna2y50fxekyp.png)