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New answer posted

6 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

For angle of incidence 'i' :-

cos I = 548+27+25=5100

i = 60°

Using snell's law :-

μ1sini=μ2sinr

sin r = 23sin60°=23*32=12

r=45°

So, difference, I – r = 60° - 452 = 15°

New answer posted

6 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

l=2πr

314cm=2*3.14r

r=12m=0.5m

μ=iA

=14*πr2

=142*227*14=11

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

IA=9l+4l+29l*4lcos0°

=13l+12l=25l

IB=9l+4l+29l*4lcosπ

=13l12l=l

IAIB=24l

New answer posted

6 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 Bcentre=Nμ0i2r

100*4π*107*i2*5*102=37.68*104

i=37.684*3.14=3A

New answer posted

6 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

i = v R N e t = 6 3 = 2 A

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 Cv=α2R4J/molk

As, Cv (mix) = 1*32R+3*52R4=9R4=α2R4

α=3

New answer posted

6 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

For adiabatic process – PVY = const

T1V1Y1=T2V2Y1

T2T1= (v1v2)Y1= (d2d1)Y1= (32) (751)= (32)2/5

= (2)2=4

New question posted

6 months ago

0 Follower

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 g (ath)=g (atdepthαh) h << R

g (12hR)=g (1αhR)

12hR=1αhR2hR=αhRα=2

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 R=u2sin (2*45°)g=u2g

R2=u22g=u2sin20g

sin2θ=12

2θ=30°θ=15°

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