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英文物理問題 四條
發問:
1. what will be the final velocity of a 5.0 g bullet starting from rest , if a net force 45 n is applied over a distance of 0.8 m?2A. what is the net force nedded to lift a full grocery sack (weighing 210N)2B.what is the net force needed to accelerate the grocery sack upward at 1.5 m/s^23.A car has a... 顯示更多 1. what will be the final velocity of a 5.0 g bullet starting from rest , if a net force 45 n is applied over a distance of 0.8 m? 2A. what is the net force nedded to lift a full grocery sack (weighing 210N) 2B.what is the net force needed to accelerate the grocery sack upward at 1.5 m/s^2 3.A car has a mass of 820 kg. It starts from rest and travels 41 m in 3.0 s. What is the net force applied to the car? 最好比埋解釋
1. what will be the final velocity of a 5.0 g bullet starting from rest , if a net force 45 n is applied over a distance of 0.8 m? Work done = FS = (45)(0.8) = 36 J This energy changes to kinetic energy of bullet 36 = (1/2) mV2 = (1/2)(5/1000)V2 V = 120 m/s 2A. what is the net force nedded to lift a full grocery sack (weighing 210N) 210 N is needed 2B. what is the net force needed to accelerate the grocery sack upward at 1.5 m/s^2 since F=mg ; 210 = m(10) ; (這裏，我用了 g=10 ，如果有需要改回 9.8 ，自己改啦) m = 21 kg The force that needed : F=ma = (21)(1.5) = 31.5 N 3. A car has a mass of 820 kg. It starts from rest and travels 41 m in 3.0 s. What is the net force applied to the car? S = ut + (1/2)at2 41 = 0 + (1/2)a(3.0)2 a = 9.111 m/s2 F = ma = (820)(82/9) = 7471 N
其他解答:
英文物理問題 四條
發問:
1. what will be the final velocity of a 5.0 g bullet starting from rest , if a net force 45 n is applied over a distance of 0.8 m?2A. what is the net force nedded to lift a full grocery sack (weighing 210N)2B.what is the net force needed to accelerate the grocery sack upward at 1.5 m/s^23.A car has a... 顯示更多 1. what will be the final velocity of a 5.0 g bullet starting from rest , if a net force 45 n is applied over a distance of 0.8 m? 2A. what is the net force nedded to lift a full grocery sack (weighing 210N) 2B.what is the net force needed to accelerate the grocery sack upward at 1.5 m/s^2 3.A car has a mass of 820 kg. It starts from rest and travels 41 m in 3.0 s. What is the net force applied to the car? 最好比埋解釋
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最佳解答:1. what will be the final velocity of a 5.0 g bullet starting from rest , if a net force 45 n is applied over a distance of 0.8 m? Work done = FS = (45)(0.8) = 36 J This energy changes to kinetic energy of bullet 36 = (1/2) mV2 = (1/2)(5/1000)V2 V = 120 m/s 2A. what is the net force nedded to lift a full grocery sack (weighing 210N) 210 N is needed 2B. what is the net force needed to accelerate the grocery sack upward at 1.5 m/s^2 since F=mg ; 210 = m(10) ; (這裏，我用了 g=10 ，如果有需要改回 9.8 ，自己改啦) m = 21 kg The force that needed : F=ma = (21)(1.5) = 31.5 N 3. A car has a mass of 820 kg. It starts from rest and travels 41 m in 3.0 s. What is the net force applied to the car? S = ut + (1/2)at2 41 = 0 + (1/2)a(3.0)2 a = 9.111 m/s2 F = ma = (820)(82/9) = 7471 N
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