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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: - A spring when stretched by 2mm containing energy 4 J. If it is stretched
- In an ac circuit an alternating voltage e = 200 √2 sin 100 t volts is connected
- A body of mass m is thrown upwards at an angle θ with the horizontal with velocity
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A spring when stretched by 2mm containing energy 4 J. If it is stretched
- In an ac circuit an alternating voltage e = 200 √2 sin 100 t volts is connected
- A body of mass m is thrown upwards at an angle θ with the horizontal with velocity
- Potentiometer measures the potential difference more accurately than a voltmeter
- If we study the vibration of a pipe open at both ends
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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