A) 3.0 atm
B) 2.0 aim
C) 1.5 atm
D) 1.0 atm
Correct Answer: A
Solution :
\[P=\frac{1}{3}\rho {{e}^{2}}\] \[=\frac{1}{3}\times 8.99\times {{10}^{-2}}\times {{(3180)}^{2}}N\text{/}{{m}^{2}}\] \[=\frac{1}{3}\times \frac{8.99\times {{10}^{-2}}\times {{(3180)}^{2}}}{1.01\times {{10}^{5}}}\text{atm}=3\,\text{atm}\]You need to login to perform this action.
You will be redirected in
3 sec