A) 12.5 m/s
B) 5 m/s
C) 3.7 m/s
D) 2.5 m/s
Correct Answer: D
Solution :
Here: Mass of ship \[m=2\,\times {{10}^{7}}\,kg,\] Force \[F=25\,\times {{10}^{5}}\,N\] Displacement s = 25 m According to the Newtons second law of motion \[F=ma\] or \[a=\frac{F}{m}=\frac{25\times {{10}^{5}}}{2\times {{10}^{7}}}\] \[=12.5\times {{10}^{-2}}\,m/{{s}^{2}}\] The relation for final velocity is \[{{\upsilon }^{2}}=\upsilon +2as\] or \[{{\upsilon }^{2}}=0+2\times (12.5\times {{10}^{-2}})\times 25\] or \[\upsilon =\sqrt{6.25}=2.5\,m/s\]You need to login to perform this action.
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