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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: - When one of the slits of Young’s experiment is covered with a transport sheet
- If the unit of force and length are doubled, then unit of energy will be
- Which component of electromagnetic spectrum have maximum wavelength?
- There is a ring of radius r having linear charge densityλ and rotating with a uniform
- An experiment takes 10 minutes to raise temperature of water from 0⁰C to 100⁰C
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When one of the slits of Young’s experiment is covered with a transport sheet
- If the unit of force and length are doubled, then unit of energy will be
- Which component of electromagnetic spectrum have maximum wavelength?
- There is a ring of radius r having linear charge densityλ and rotating with a uniform
- An experiment takes 10 minutes to raise temperature of water from 0⁰C to 100⁰C
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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