MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass
        a=15m/s^2 represent total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5m at a given instant of time the speed of particle is
2 years ago

Answers : (2)

Piyush Kumar Behera
394 Points
							
see this acceleration is the centripetal acceleration as the particle is moving in the constant velocity .
so a=v2/r
v=\sqrt{ar}
so speed of the particle is \sqrt{2.5*15}
=6.1237243569579452454932101867647m/s
hope it helps please approve!!
2 years ago
Shefali
11 Points
							The acceleration given in question is the resultant of tangential and centripetal acceleration.So first we will calculate the centripetal acceleration. Here,Centripetal acceleration =a cos 30°                                   =a √3/2=15*√3/2We know,                Centripetal acceleration=v*v/r                           15*√3/2=v*v/2.5                            15*√3/2*2.5=v*v                             31.875= v*v                             √31.875=v                            5.7=v
						
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details