Answer:
b)1.08 N
Step-by-step explanation:
Given that
velocity of air V= 45 m/s
Diameter of pipe = 2 cm
Force exerted by fluid F
![F=\rho AV^2](https://img.qammunity.org/2020/formulas/engineering/college/oe5ziuizcckim4jmat3eelqfa3zh1vuag4.png)
So force exerted in x-direction
![F_x=\rho AV^2](https://img.qammunity.org/2020/formulas/engineering/college/y7sa9knyjvxdrb8obevn3i4j6in0tcb72d.png)
![F_x=1.2* (\pi)/(4)* 0.02^2* 45^2](https://img.qammunity.org/2020/formulas/engineering/college/qrwhj2tb8t7j7nqg2e9lp2qd1bvja7r3mm.png)
F=0.763 N
So force exerted in y-direction
![F_y=\rho AV^2](https://img.qammunity.org/2020/formulas/engineering/college/hwsojqje3b27hn04accacc4j77gzhiehnk.png)
![F_y=1.2* (\pi)/(4)* 0.02^2* 45^2](https://img.qammunity.org/2020/formulas/engineering/college/pr87dl1np5ovr42yxxzw023g1wkoz2qm85.png)
F=0.763 N
So the resultant force R
![R=√(F_x^2+F_y^2)](https://img.qammunity.org/2020/formulas/engineering/college/t2floi5qdpixx7uh5s6i6p8m0c34usphvz.png)
![R=√(0.763^2+0.763^2)](https://img.qammunity.org/2020/formulas/engineering/college/u72zdmjlf9474cp2ue98e8s26oqcah9w1u.png)
R=1.079
So the force required to hold the pipe is 1.08 N.