A) \[\frac{\sqrt{2}}{3}\]
B) \[\frac{\sqrt{3}}{2}\]
C) 0.5
D) 1.0
Correct Answer: B
Solution :
By the law of parallelogram of forces \[F={{(1)}^{2}}+{{(1)}^{2}}+2\times 1\times 1\times \cos {{60}^{o}}\] \[=1+1+2\times \frac{1}{2}=3\] \[\Rightarrow \] \[\sqrt{F}=\sqrt{3}Newton\] Now \[F=ma\] \[\Rightarrow \] \[a=\frac{F}{m}=\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}\]
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