PhysicsEasy145×since 2002Q6610If the total energy transferred to a surface in time t\mathrm{t}t is 6.48×105 J6.48 \times 10^5 \mathrm{~J}6.48×105 J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:A2.16×10−3 kg m/s2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}2.16×10−3 kg m/sB2.46×10−3 kg m/s2.46 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}2.46×10−3 kg m/sC1.58×10−3 kg m/s1.58 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}1.58×10−3 kg m/sD4.32×10−3 kg m/s4.32 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}4.32×10−3 kg m/sCheck answerSkip