Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
57 views
in Physics by (90.2k points)
closed by
A particle moves on a circle of radius r with centripetal accelration as function of time as `a_(c)=K^(2)rt^(2)` where k is a positive constant , find the resu ltant acceleration.
A. `kt^(2)`
B. `kr`
C. `krsqrt(k^(2)t^(4)+1)`
D. `krsqrt(k^(2)t^(2)-1)`

1 Answer

0 votes
by (90.9k points)
selected by
 
Best answer
Correct Answer - c
From given equation
`omega=kt`
`alpha=(domega)/(dt)=k, alpha_(t)=ralpha, alpha=sqrt(a_(c)^(2)+a_(r)^(2))`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...