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New answer posted

a year ago

0 Follower 5 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

a year ago

0 Follower 6 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

a year 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

a year ago

0 Follower 13 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

a year ago

0 Follower 32 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

a year ago

0 Follower 4 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=q0ωsinωt  ω=12c=150*10−3*5*10−4

= 10005=200

So, imax = q0ω=500*106*100*200

= 10A

New answer posted

a year 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=N0e−16λtN0e−25λt=e (25λ−16λ)t

e = e (9λ1aλ)

⇒9a=1⇒a=9

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

 i (zener−max)=25mA

20 – imax R – 8 = 0

imax R = 12

At minimum zener current  (μA):−

20−iminR−iminRL=0

RRL=128=32

lminR=12

iminRL=8

At maxm zener current –

20−imaxR−8=0

iL=O {as  iz  maxm=25mA}

imaxR = 12v

25 * 103 R = 12

R=12*10325=12*40=480Ω

New answer posted

a year 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

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

m = 10g

l=50cm

A = 2mm2

Y = 1.2 * 1011N/m2

Δx=x*10−5m

As,  TA=YΔxl

Δx=TlAY

Tl=V2m

= V2mAY

=3600*10*10−32*10−6*1.2*1011=1800*10−3*1061.2

=1812*10−4=32*10−4=15*10−5m

So, x = 15

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