Answer:176.73 N
Explanation:
Given data
diameter[tex]\left ( d\right )[/tex]=5cm
Velocity of jet[tex]\left ( v\right )[/tex]=40m/s
Velocity of plate[tex]\left ( v_0\right )[/tex]=10m/s
Cross-sectional area of jet=19.36 [tex]cm^2[/tex]
Mass flow rate which strikes the plate=[tex]\rho \times A\times \left ( v-v_0\right )[/tex]
Thus force(f) applied is =[tex]\rho \times A\times \left ( v-v_0\right )^{2}[/tex]
F=[tex]10^3\times 19.63\times 10^{-4}\times [/tex][tex]\left ( 40-10\right )^2[/tex]
F=176.73N