Physics

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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Active region of the CE transistor is linear region and is best suited for its use as an amplifier.

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

μ 0 = B 0 H = N A − 2

[ μ 0 ] = [ M 1 L 1 T − 2 A − 2 ]

It is not dimensionless

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

For x < a,

B1 = μ 0 i 0 x 2 π a 2  

For a < x < b,

B 2 = μ 0 i 0 2 π x   

B 1 B 2 = x 2 a 2           

 

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

l z = l x + l y

l z = m l 2 3 + m l 2 3 = 2 3 m l 2

New answer posted

a year ago

0 Follower 33 Views

V
Vishal Baghel

Contributor-Level 10

R = l θ

⇒ T i m e = 4 * 2 π R V = 4 * 2 π x V ( l θ )

Time = 4 * 2 π * 4 . 4 * 9 . 6 4 * 1 0 1 5 8 * 1 . 5 * 1 0 1 1 * 4 3 6 0 0 * π 1 8 0

⇒ T i m e = 4 . 5 * 1 0 1 0 s o c s

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

R = 7 5 * 1 0 2 ± 5 %

⇒ R = 7 5 0 0 ± 3 7 5 Ω

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Viscous force = Weight

⇒ F V = ρ * 4 3 π r 3 * g = 3 . 9 * 1 0 N − 1 0

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

λ > h ⇒ λ > 4 0 0 m

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

l 1 = 2 5 5 + R

l 2 = 5 R + 1 5

? l 1 = l 2 ⇒ 4 R = 4 ⇒ R = 1 Ω

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Q 1 = C 1 V = 2 V μ C

Q 2 = Q 3 = C 2 C 3 C 2 + C 3 V = 6 * 1 2 6 + 1 2 V = 4 V μ C

Q 1 : Q 2 : Q 3 = 1 : 2 : 2

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