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New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

(n) λ=5

(n, m) : Integers

2m+12λ=323/25=2m+12n3n=10m+5

N, m are integers.

So,  m=1, n=5, λ=1

m=4n=15λ=13m=7n=25λ=15

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

ρvg-mg=ma

ρvgm=g+a

m=ρvgg+a

10343π*10-6 (9.8)9.898

4.15gm

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

(n) λ=5

(n, m) : Integers

2m+12λ=323/25=2m+12n3n=10m+5

N, m are integers.

So,  m=1, n=5, λ=1

m=4n=15λ=13m=7n=25λ=15

New answer posted

6 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

E= (I) (t) (A)cos2? θ
(3.3)2π31.43*10-4*12

0.99*10-4

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

m (L)=m1S1 (ΔT)

m3.4*105= (200) (4200) (25)

m=61.7

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

BA=μ0Iθ4πR

BABB=IAθARBIBθBRA

23π2 (4)35π3 [2]

65 .

New answer posted

6 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

V?1=2Ucm?-U1?2m/2Vom2+m3-Vo

65Vo-VoV05

λo=hcm2Voλf=hcM2Vo5

Δλ=8hcmVo

 

New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

K=QrΔxΔAT

ML2T-2 (L)L2 (θ) (T)M1L1-T-3θ-1

New question posted

6 months ago

0 Follower 8 Views

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