Question Number 154853 by peter frank last updated on 22/Sep/21
Answered by peter frank last updated on 23/Sep/21
Commented by peter frank last updated on 23/Sep/21
![from Bernoullis principle p+(1/2)ρv_1 ^2 +ρgh=p_a +(1/2)ρv_2 ^2 +ρgh v_1 ≈0 (A_1 >>>>A_2 ) p+(1/2)ρv_1 ^2 =p_a +(1/2)ρv_2 ^2 p−p_a =(1/2)ρv_2 ^2 v_2 ^2 =((2(p−p_a ))/ρ) v_2 =(√((2(p−p_a ))/ρ)) [p−p_a ]=8×10^5 p_a ρ_(H_2 0) =1000kgm^(−3) v_2 =40m/s consider projectile motion of water y=(1/2)gt^2 [u=0] x=v_2 cos θt [θ=0] x=v_2 t y=(1/2)g((x/v_2 ))^2 y=2 v_2 =40m/s x=(√((1000×4)/(9.8))) x=25.6m (a) d=(√(x^2 +y^2 )) [pythagorous] v=(√((25.6)^2 +2^2 )) v=25.678m (b)vertical force=AρV_y V_y =? A=1cm^(2 ) V_y =v_y_o +gt [v_y_o ]=0 t=(x/v)=((25.6)/(40))= V_y =gt=6.27m/s vertical force=AρV_y =3.93N (c)Horizontal force exerted on the tank=ρAV^2 =1000×(1×10^(−4) )×(40)^2 =160N](https://www.tinkutara.com/question/Q154943.png)
$$\mathrm{from}\:\mathrm{Bernoullis}\:\mathrm{principle} \\ $$$$\mathrm{p}+\frac{\mathrm{1}}{\mathrm{2}}\rho\mathrm{v}_{\mathrm{1}} ^{\mathrm{2}} +\rho\mathrm{gh}=\mathrm{p}_{\mathrm{a}} +\frac{\mathrm{1}}{\mathrm{2}}\rho\mathrm{v}_{\mathrm{2}} ^{\mathrm{2}} +\rho\mathrm{gh} \\ $$$$\mathrm{v}_{\mathrm{1}} \approx\mathrm{0}\:\:\:\left(\mathrm{A}_{\mathrm{1}} >>>>\mathrm{A}_{\mathrm{2}} \right) \\ $$$$\mathrm{p}+\frac{\mathrm{1}}{\mathrm{2}}\rho\mathrm{v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{p}_{\mathrm{a}} +\frac{\mathrm{1}}{\mathrm{2}}\rho\mathrm{v}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{p}−\mathrm{p}_{\mathrm{a}} =\frac{\mathrm{1}}{\mathrm{2}}\rho\mathrm{v}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{v}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{p}−\mathrm{p}_{\mathrm{a}} \right)}{\rho} \\ $$$$\mathrm{v}_{\mathrm{2}} =\sqrt{\frac{\mathrm{2}\left(\mathrm{p}−\mathrm{p}_{\mathrm{a}} \right)}{\rho}}\:\:\:\: \\ $$$$\:\:\left[\mathrm{p}−\mathrm{p}_{\mathrm{a}} \right]=\mathrm{8}×\mathrm{10}^{\mathrm{5}} \mathrm{p}_{\mathrm{a}} \:\:\:\rho_{\mathrm{H}_{\mathrm{2}} \mathrm{0}} =\mathrm{1000kgm}^{−\mathrm{3}} \\ $$$$\mathrm{v}_{\mathrm{2}} =\mathrm{40m}/\mathrm{s} \\ $$$$\mathrm{consider}\:\mathrm{projectile}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{water} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\left[\mathrm{u}=\mathrm{0}\right] \\ $$$$\mathrm{x}=\mathrm{v}_{\mathrm{2}} \mathrm{cos}\:\theta\mathrm{t}\:\:\:\:\left[\theta=\mathrm{0}\right] \\ $$$$\mathrm{x}=\mathrm{v}_{\mathrm{2}} \mathrm{t} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\frac{\mathrm{x}}{\mathrm{v}_{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{y}=\mathrm{2}\:\:\:\mathrm{v}_{\mathrm{2}} =\mathrm{40m}/\mathrm{s} \\ $$$$\mathrm{x}=\sqrt{\frac{\mathrm{1000}×\mathrm{4}}{\mathrm{9}.\mathrm{8}}} \\ $$$$\mathrm{x}=\mathrm{25}.\mathrm{6m} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{d}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\left[\mathrm{pythagorous}\right] \\ $$$$\:\:\:\:\:\mathrm{v}=\sqrt{\left(\mathrm{25}.\mathrm{6}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\mathrm{v}=\mathrm{25}.\mathrm{678m} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{b}\right)\mathrm{vertical}\:\mathrm{force}=\mathrm{A}\rho\mathrm{V}_{\mathrm{y}} \\ $$$$\mathrm{V}_{\mathrm{y}} =?\:\:\:\mathrm{A}=\mathrm{1cm}^{\mathrm{2}\:\:\:} \\ $$$$\mathrm{V}_{\mathrm{y}} =\mathrm{v}_{\mathrm{y}_{\mathrm{o}} } +\mathrm{gt}\:\:\:\:\left[\mathrm{v}_{\mathrm{y}_{\mathrm{o}} } \right]=\mathrm{0} \\ $$$$\mathrm{t}=\frac{\mathrm{x}}{\mathrm{v}}=\frac{\mathrm{25}.\mathrm{6}}{\mathrm{40}}= \\ $$$$\mathrm{V}_{\mathrm{y}} =\mathrm{gt}=\mathrm{6}.\mathrm{27m}/\mathrm{s} \\ $$$$\mathrm{vertical}\:\mathrm{force}=\mathrm{A}\rho\mathrm{V}_{\mathrm{y}} =\mathrm{3}.\mathrm{93N} \\ $$$$\left(\mathrm{c}\right)\mathrm{Horizontal}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{tank}=\rho\mathrm{AV}^{\mathrm{2}} \\ $$$$=\mathrm{1000}×\left(\mathrm{1}×\mathrm{10}^{−\mathrm{4}} \right)×\left(\mathrm{40}\right)^{\mathrm{2}} \\ $$$$=\mathrm{160N} \\ $$$$ \\ $$$$ \\ $$