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In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1?.

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Here you can find the meaning of In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1? defined & explained in the simplest way possible. Besides giving the explanation of In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1?, a detailed solution for In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1? has been provided alongside types of In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1? theory, EduRev gives you an ample number of questions to practice In a two dimensional moyion of a particle ,the particle moves from point A, position vector r1to point B position vector r2 .If the magnitude of these vector are respectively r1 =3 and r2=4and the angle they make with the x-axis are $(theta)1=75 degree ,$2=15degree respectively ,then magnitude of the dispacement vector is -(A) √3 (B) √13 (C) √5 (D) √1? tests, examples and also practice NEET tests.