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New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

No. of electric field line per unit area = electric field

E = ρ r 3 0 f o r r = R

E = 4 5 * 1 0 1 0 N / C

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

v S = 0 , v O b = 5 m / s

f d i r e c t = ( 3 2 0 5 3 2 0 ) 6 4 0 = 6 3 0 H z

f b e a t = ( 6 5 0 6 3 0 ) = 2 0 H z

New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

l B ( 1 + α B Δ T ) l i ( 1 + α i Δ T ) = l B l i

α B l B = l i α i

l i = 6 0 c m

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Δ L 1 = F L A Y = F L π r 2 Y = 5 c m

Δ L 2 = 4 F 4 L π 1 6 r 2 y = F L π r 2 Y = 5 c m

New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

F 2 = 1 c o s 4 5 ° + 2 c o s 4 5 ° = 3 c o s 4 5 ° = 3 2 N

F 1 + 1 c o s 4 5 ° = 2 s i n 4 5 ° F 1 F 2 = 1 : 3

x = 3

New answer posted

5 months ago

0 Follower 17 Views

V
Vishal Baghel

Contributor-Level 10

Stopping distance = v 2 2 a = d

If speed is made 1 3 r d

d 1 = d 9 , d 1 = 2 7 9 = 3 m

Braking acceleration Remains same.

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

C = 500 μF,  V = 100 v, L = 50 mH

In this LC – oscillation

q = q0 cos ωt

i=dqdt=q0ωsinωtω=12c=150*103*5*104

10005=200

So, imaxq0ω=500*106*100*200

= 10A

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

λA=25λ,  λB=16λ

At t = 0 NA = NB = N0

after t = 1aλ:NBNA=N0e16λtN0e25λt=e (25λ16λ)t

e = e (9λ1aλ)

9a=1a=9

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

 i (zenermax)=25mA

20 – imax R – 8 = 0

imax R = 12

At minimum zener current  (μA):

20iminRiminRL=0

RRL=128=32

lminR=12

iminRL=8

At maxm zener current –

20imaxR8=0

iL=O {asizmaxm=25mA}

imaxR = 12v

25 * 103 R = 12

R=12*10325=12*40=480Ω

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

geff = g – (ρwρb)g

T=2πlg

T'=2πlgeff

T'=54T

54*10

55sec

So, x = 5

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