A) \[2.16\times {{10}^{-3}}kg-m/s\]
B) \[1.52\times {{10}^{-5}}kg-m/s\]
C) \[8.31\times {{10}^{-8}}kg-m/s\]
D) \[18.2\times {{10}^{-6}}kg-m/s\]
Correct Answer: A
Solution :
Total energy falling \[U=(18W/c{{m}^{2}})(30\times 60s)(20c{{m}^{2}})\] \[=6.48\times {{10}^{5}}J\] Total, momentum delivered is \[P=\frac{U}{c}=\frac{6.48\times {{10}^{5}}}{3\times {{10}^{8}}}=2.16\times {{10}^{-3}}kg-m/s\]You need to login to perform this action.
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