A) zero
B) \[\frac{1}{2}m{{s}^{-2}},N-W\]
C) \[\frac{1}{\sqrt{2}}m{{s}^{-2}},N-E\]
D) \[\frac{1}{\sqrt{2}}m{{s}^{-2}},S-W\]
Correct Answer: B
Solution :
\[\Delta u=2u\,\sin \left( \frac{\theta }{2} \right)=2\times 5\times \sin 45{}^\circ =\frac{10}{\sqrt{2}}\] \[\therefore \] \[a=\frac{\Delta u}{\Delta T}=\frac{10/\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m{{s}^{-2}}\]
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