| ⇦ | 
| ⇨ |
An object moving with a speed of 6.25 m/s, is deaccelerated at a rate given by (dv/dt)= -2.5√v, where v is the instantaneous speed.The time taken by the object ,to come to rest would be
Options
(a) 2 s
(b) 4 s
(c) 8 s
(d) 1 s
Correct Answer:
2 s
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - For having large magnification power of a compound microscope
- A long wire carrying a steady current is bent into a circular loop of one turn.
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
- Which one of the following equations of motion represents simple harmonic motion?
- A tuning fork gives 4 beats with 50 cm length of a sonometer
Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- For having large magnification power of a compound microscope
- A long wire carrying a steady current is bent into a circular loop of one turn.
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
- Which one of the following equations of motion represents simple harmonic motion?
- A tuning fork gives 4 beats with 50 cm length of a sonometer
Topics: Motion in Straight Line (93)
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