| ⇦ | 
| ⇨ |
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: - The moderator in a reactor
- The two ends of a metal rod are maintained at temperatures 100°C and 110°C.
- The force of repulsion between two electrons at a certain distance is F.
- Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial.
- The phase difference between the instantaneous velocity and acceleration of a particle
Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moderator in a reactor
- The two ends of a metal rod are maintained at temperatures 100°C and 110°C.
- The force of repulsion between two electrons at a certain distance is F.
- Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial.
- The phase difference between the instantaneous velocity and acceleration of a particle
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