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A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it for 4 sec. What will be the distance of the block from the point where the force started acting?
Options
(a) 6 m
(b) 8 m
(c) 10 m
(d) 2 m
Correct Answer:
10 m
Explanation:
Horizontal distance,x=ut ⇒1.5 x 4 = 6m
Vertical distance,y=(1/2)at² = (1/2)(F/m)t²
Two motions are in mutually perpendicular directions. Net displacement = √(6²+8²) = √(36+64) =√100 = 10m.
Related Questions: - Two thin dielectric slabs of dielectric constants K₁ and K₂, (K₁ < K₂) are inserted
- A body is thrown with a velocity of 9.8 m/s making angle of 30° with the horizontal
- In the given figure, a diode D is connected to an external resistance R=100 Ω
- A particle with total energy E is moving in a potential energy region U(x)
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two thin dielectric slabs of dielectric constants K₁ and K₂, (K₁ < K₂) are inserted
- A body is thrown with a velocity of 9.8 m/s making angle of 30° with the horizontal
- In the given figure, a diode D is connected to an external resistance R=100 Ω
- A particle with total energy E is moving in a potential energy region U(x)
- The kinetic energy of particle moving along a circle of radius R depends
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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