Class 11th

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New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

  L = m u c o s θ u 2 s i n 2 θ 2 g

= m u 2 1 4 2 g x = 8

New answer posted

3 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

  k e q = 2 k k 3 k + k = 5 k 3

Angular frequency of oscillation ( ω ) = k e q m  

ω = 5 k 3 m            

Period of oscillation (t) = 2 π ω = 2 π 3 m 5 k  

= π 1 2 m 5 k                                                   

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

p H = p K w + p K a ? p K b 2

pKa = pKb

p H = p K w 2 = 7

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Adenine base pairs with thymine with 2 hydrogen bonds and cytosine base pairs with guanine with 3 hydrogen bonds.

 

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

T = 2 π I g

g = 4 π 2 I T 2

Δ g g = Δ I I + 2 Δ T T

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

= 6%

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

The magnitude of resultant vector (R)

= a 2 + b 2 + 2 a b c o s θ            

here a = b = A

then R = A 2 + A 2 + 2 A 2 c o s θ  

= A 2 1 + c o s θ

= 2 A 2 c o s 2 θ 2

= 2 A c o s θ 2      

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

q = 0 as adiabatic process is given

            W = 0 as paxt = 0

            q + W = DU

            q = 0

            W = 0

            Þ DU = 0

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

PF5, PCl5, PBr5, Fe (CO)5 Þ Trigonal bipyramidal

BrF5 Þ Square pyramidal

[PtCl4]2– Þ Square planar

SF6 Þ Octahedral

New answer posted

3 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

a = ( m 1 m 2 ) g ( m 1 + m 2 ) = g 8

8m1 – 8m2 = m1 + m2

7m1 = 9m2

m 1 m 2 = 9 7          

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

U = nCvT

-> U = n 1 C v 1 T + n 2 C v 2 T  

-> 8 * 5 R 2 * T + 4 * 5 R 2 * T

= 30RT

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