A) 1.5 m/s
B) 60 m/s
C) 0.1 m/s
D) 5 m/s
Correct Answer: C
Solution :
Given, \[m=3\times {{10}^{7}}kg\text{ }u=0\] \[F=5\times {{10}^{4}}N\text{ }S=3\text{ }v=?\] From the relation F = ma, we get We know, \[{{v}^{2}}{{u}^{2}}=2as\] ...................(1) Substituting the above values in equation (1) we get \[\Rightarrow \] \[{{v}^{2}}0=2\times \left( \frac{5}{3}\times {{10}^{-3}} \right)\times 3\] \[\Rightarrow \] \[{{v}^{2}}=10\times {{10}^{-3}}={{10}^{-2}}\] \[\Rightarrow \]\[v=0.1\text{ }m/s\]
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