Laws of Motion

Get insights from 108 questions on Laws of Motion, answered by students, alumni, and experts. You may also ask and answer any question you like about Laws of Motion

Follow Ask Question
108

Questions

0

Discussions

4

Active Users

0

Followers

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Let the spring is in extension state and   a B > a A , a A B = a A i ^ ( a B i ^ ) = ( a A + a B ) i ^  

Hence we can say that block moves away from block B in the frame of B

F – kx = MaB           . (1)

kx = Ma                  . (2)

a B = F M a

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10


? v

mv=16mv1

V1=V16

Δk loss =12mv2-12 (16m)V162

12mv21516

% loss =1516*100=93.75%

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Mass = m

Length = L

mL=?

at l=Lx, achain=g2

a=FPullingmass= (L? l)? g? ? lg? L= (L? 2l)? g? L=g2

l=L4? x=4

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Using newton's law:

2mg – T = 4ma  - (1)

T – mg = ma       - (2)

From (1) & (2) :

m g = 5 m a a = g 5 = 2 m / s 2                

T = m g + m g 5 = 6 m g 5                     

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

a = k2rt2

v 2 r = k 2 r t 2                         

             

v = k r t

a t = d v d t = k r                

⇒ tangential force, Ft = mat = mkr

P o w e r d e l i v e r e d , P = F t v = ( m k r ) ( k r t ) = m k 2 r 2 t                

Note Power delivered by centripetal force will be zero.

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

goff = g cos

T=2πLgcosα

New answer posted

3 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

N = m v 2 R

 

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Sol. Sol.  Electric field due to infinite sheet is uniform

 

New answer posted

3 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

F.B.D. of point 'P'

T s i n θ = 3 0            - (1)

T c o s θ = 1 0 0          - (2)

( 1 ) ÷ ( 2 )

t a n θ = 3 1 0 θ = t a n 1 ( 3 * 1 0 1 )    

New answer posted

3 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

F.B.D, with respect to Non-Inertial frame

k x = m ω 2 ( l 0 + x )

x | k m ω 2 ( l 0 + x )

x = m ω 2 l 0 k m ω 2

        

               

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.