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New answer posted

a year ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

n v 0 . 6 = 4 0 0 & ( n + 1 ) v 0 . 6 = 4 5 0

⇒ [ 0 . 6 * 4 5 0 v + 1 ] v 0 . 6 = 4 5 0

v = 30

⇒ T μ = 3 0

⇒ 2 7 0 0 μ = 9 0 0

μ = 3

New answer posted

a year ago

0 Follower 30 Views

V
Vishal Baghel

Contributor-Level 10

C e f f = [ ε 0 ( 7 * 4 ) 4 / 1 0 + 5 ε 0 ( 1 * 4 ) 4 / 1 0 ] * 1 0 − 2

C e f f = 1 . 2 ε 0

Energy = 1 2 C e f f V 2

1 2 ( 1 . 2 ε 0 ) ( 2 0 ) ( 2 0 ) = 2 4 0 ε 0

New answer posted

a year ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Modulating signal 2sin (6.28 * 106)t

Carrier signal  4 sin (12.56 * 109)t

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

q = C V 1 0 0 Ω

= ( 1 . 1 * 1 0 − 6 ) ( 1 0 R + r R )

= 1 . 1 * 1 0 − 6 ( 1 0 1 1 0 * 1 0 0 )

= 1 0 μ C

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

? = B → . A →

= ( 3 t 3 j ^ + 3 t 2 k ^ ) . ( π ( 1 ) 2 k ^ )

? = 3 t 2 π

E i n d = | d ? d t | = 6 t π

at t = 2, Eind = 12

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

δ = A ( μ y − 1 ) − A 1 ( μ y 1 − 1 )

= 6 ( 1 . 5 − 1 ) − 5 ( 1 . 5 5 − 1 )

= 1 4

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

m = 0.3g density,   ρ ( b a l l ) = 8 g / c c ∴ V = m ρ = 0 . 3 8 c c                                        

ρ g l y c e r e n e = 1 . 3 g / c c = 1 . 3 * 1 0 3 k g / m 3      

Fv + FB = mg.

=Fv = mg – FB = .3 * 10-3 * 10 – 1.3 *   0 . 3 8 * 1 0 − 6 * 1 0 * 1 0 3


= 3 * 1 0 − 3 − . 3 9 8 * 1 0 − 2 = 3 * 1 0 − 3 − . 5 * 1 0 − 3    

= 2.5 * 10-3N                       

= 25 * 10-4N

 x = 25

 

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Independent of are a is case of uniform wire

New answer posted

a year ago

0 Follower 21 Views

V
Vishal Baghel

Contributor-Level 10

Slope = Δ L / W L = Δ L / L W = 1 Y A

Y = 1 ( S l o p e ) A

Y = 1 ( 2 * 1 0 − 6 ) ( 0 . 2 5 * 1 0 − 5 )

Y = 2 * 1 0 1 1 N / m 2

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  l = 0 . 5 m

A  = 1 0 − 4 m 2

Breaking stress, F A = 5 * 1 0 8 N / m 2

F = 5 * 1 0 8 * 1 0 − 4 N

F = 5 * 104N

T = m V 2 l ⇒ V ( m a x ) 2 = = T l m

V m 2 = 2 . 5 * 1 0 3

 

Vmax = 50m/s

 

 

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