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Aconveyor belt ismoving at a constant speed of 2m/s.Abox is gently dropped on it. The coefficient of frictionbetween them is µ= 0.5. The distance that the box will move relative to belt before coming to rest on it takingg = 10 ms-2, is :a)1.2 mb)0.6 mc)zerod)0.4 mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared
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the NEET exam syllabus. Information about Aconveyor belt ismoving at a constant speed of 2m/s.Abox is gently dropped on it. The coefficient of frictionbetween them is µ= 0.5. The distance that the box will move relative to belt before coming to rest on it takingg = 10 ms-2, is :a)1.2 mb)0.6 mc)zerod)0.4 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam.
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Aconveyor belt ismoving at a constant speed of 2m/s.Abox is gently dropped on it. The coefficient of frictionbetween them is µ= 0.5. The distance that the box will move relative to belt before coming to rest on it takingg = 10 ms-2, is :a)1.2 mb)0.6 mc)zerod)0.4 mCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Aconveyor belt ismoving at a constant speed of 2m/s.Abox is gently dropped on it. The coefficient of frictionbetween them is µ= 0.5. The distance that the box will move relative to belt before coming to rest on it takingg = 10 ms-2, is :a)1.2 mb)0.6 mc)zerod)0.4 mCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Aconveyor belt ismoving at a constant speed of 2m/s.Abox is gently dropped on it. The coefficient of frictionbetween them is µ= 0.5. The distance that the box will move relative to belt before coming to rest on it takingg = 10 ms-2, is :a)1.2 mb)0.6 mc)zerod)0.4 mCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Aconveyor belt ismoving at a constant speed of 2m/s.Abox is gently dropped on it. The coefficient of frictionbetween them is µ= 0.5. The distance that the box will move relative to belt before coming to rest on it takingg = 10 ms-2, is :a)1.2 mb)0.6 mc)zerod)0.4 mCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice NEET tests.