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New answer posted

12 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

Uinitial =k (4q) (q) (d/2)+k (q) (-q) (d/2)

6kq2dUfinal =4 (4q) (q)3d2+k (q) (-q) (d/2)23kq2dΔU=23-6kq2d-163kq2d

New answer posted

12 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

New answer posted

12 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

V 2 = U 2 + 2 g S V 2 = 0 + 2 g ( h - y ) V 2 = 2 g h - 2 g y V = 2 g h - 2 g y

New answer posted

12 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

T = 0.5 sec

No. of oscillation = 100

Resolution = 1 sec

l = 1 0 c m ± . 1 c m

               

New answer posted

12 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

As,  1v1u=1f1v'+f1μ'f=1f

Also,  μ'v'=225

Using Newton's formula - μ'v'=f2

So, f2 = 225

f = 15 cm

 

New answer posted

12 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

u = 100 cm

f=R2=100cm

after 10 sec i=66=1A

u = 80 m

1v=1f1u=1100180=1801100=54400=1400

v=400cm

New answer posted

12 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

New question posted

12 months ago

0 Follower 2 Views

New answer posted

12 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 v1v2=R1R2=P2P1=60100=35

v1=38*220=3*27.5=82.5v

P1= (v1v)2P

= 14 w

New answer posted

12 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

m = 4kg

U = 4 (1 – cos 4x) J

F=υx=4*4sin4x

a=164sin4x=4sin4x=16x

a=16x

ω2=16ω=4=2πTT=2π4=π2sec

k = 2

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