| ⇦ |
| ⇨ |
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: - What will be the ratio of the distance moved by a freely falling body from rest in 4th
- Electrons used in an electron microscope are accelerated by a voltage of 25 kV
- The wavelength of first line of Balmer series is 6563Å. The wavelength of first line
- ₉₂U²³⁵ undergoes successive disintegrations with the end product of ₈₂P²⁰³
- Avalanche breakdown is due to
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- Electrons used in an electron microscope are accelerated by a voltage of 25 kV
- The wavelength of first line of Balmer series is 6563Å. The wavelength of first line
- ₉₂U²³⁵ undergoes successive disintegrations with the end product of ₈₂P²⁰³
- Avalanche breakdown is due to
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply