Question Number 47377 by rahul 19 last updated on 09/Nov/18
Commented by rahul 19 last updated on 09/Nov/18
$${Ans}:\:\mathrm{10}^{\mathrm{5}} {Pa}. \\ $$
Answered by ajfour last updated on 09/Nov/18
![P = (F/A) = ((△p/△t)/A) = ((△p)/(Al))×(l/(△t)) = ((v△p)/(Al)) = ((v[n(2m_0 vcos 30°)])/(Al)) & with (n/(Al)) = n_0 P = (√3)m_0 n_0 v^2 = (√3)(((28×10^(−3) )/(6.02×10^(23) )))(0.9×10^(25) )(16×10^4 ) = (((√3)×0.028×9×16)/6)×10^5 Pa = 1.4×0.48(√3)×10^5 Pa ≈ 1.164×10^5 Pa .](https://www.tinkutara.com/question/Q47406.png)
$${P}\:=\:\frac{{F}}{{A}}\:=\:\frac{\bigtriangleup{p}/\bigtriangleup{t}}{{A}}\:=\:\frac{\bigtriangleup{p}}{{Al}}×\frac{{l}}{\bigtriangleup{t}} \\ $$$$\:\:\:=\:\frac{{v}\bigtriangleup{p}}{{Al}}\:=\:\frac{{v}\left[{n}\left(\mathrm{2}{m}_{\mathrm{0}} {v}\mathrm{cos}\:\mathrm{30}°\right)\right]}{{Al}} \\ $$$$\:\:\&\:\:{with}\:\:\frac{{n}}{{Al}}\:=\:{n}_{\mathrm{0}} \\ $$$$\:{P}\:=\:\sqrt{\mathrm{3}}{m}_{\mathrm{0}} {n}_{\mathrm{0}} {v}^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{28}×\mathrm{10}^{−\mathrm{3}} }{\mathrm{6}.\mathrm{02}×\mathrm{10}^{\mathrm{23}} }\right)\left(\mathrm{0}.\mathrm{9}×\mathrm{10}^{\mathrm{25}} \right)\left(\mathrm{16}×\mathrm{10}^{\mathrm{4}} \right) \\ $$$$\:\:\:\:\:=\:\frac{\sqrt{\mathrm{3}}×\mathrm{0}.\mathrm{028}×\mathrm{9}×\mathrm{16}}{\mathrm{6}}×\mathrm{10}^{\mathrm{5}} \:{Pa} \\ $$$$\:\:\:\:\:=\:\mathrm{1}.\mathrm{4}×\mathrm{0}.\mathrm{48}\sqrt{\mathrm{3}}×\mathrm{10}^{\mathrm{5}} \:{Pa} \\ $$$$\:\:\:\:\:\approx\:\mathrm{1}.\mathrm{164}×\mathrm{10}^{\mathrm{5}} {Pa}\:. \\ $$