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A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer?.

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Here you can find the meaning of A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A block of mass 0.38 kg is kept at rest on a frictionless surface and attached to a wall with a spring of negligible mass. A bullet weighing 0.02 kg moving with a speed of 200 m/s hits the block at time t = 0 and gets stuck to it. The displacement of the block (in metre) with respect to the equilibrium position is given by(Spring constant = 640 N/m.)a)2 sin5t.b)cos10t.c)0.4 cos25t.d)0.25 sin40t.Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice IIT JAM tests.