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New answer posted

4 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

1 f = ( μ - 1 ) 1 R 1 - 1 R 2

R 1 =

R 2 = - 30 c m

1 f = ( 1.5 - 1 ) 1 - 1 - 30

1 f = 0.5 30 f = 60 c m

New answer posted

4 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

V = B ?

V pipe   V air   = B 2 ? B ? = 1 2

V pipe   = V air   2

f n = ( n + 1 ) V pipe   2 l ; f 1 - f 0 = V pipe   2 l = 300 2 2

= 105.75 H z  (If 2 = 1.41  )

= 106.05 H z  (If 2 = 1.414  )

New answer posted

4 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

For elastic collision   K E i = K E r

1 2 m * 25 + 1 2 * m * 9 = 1 2 m * 32 + 1 2 m v 2

34 = 32 + v 2

K E = 1 2 * 0.1 * 2 = 0.1 J = 1 10

x = 1

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

580

X 1 = - 3 t 2 + 8 t + 10

v ? 1 = ( - 6 t + 8 ) i ˆ = 2 i ˆ

Y 2 = 5 - 8 t 3

v ? 2 = - 24 t 2 j ˆ

v = v ? 2 - v ? 1 = | - 24 j ˆ - 2 i ˆ |

v = 24 2 + 2 2

v = 580

New answer posted

4 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

10 R 10 + R * 12 = 15 * 4 on solving

R = 10 Ω

New answer posted

4 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

3 = G m 2 2 2 2 = G m 1 1 2

3 2 = 1 4 m 2 m 1

m 1 m 2 = 1 6

New question posted

4 months ago

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New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

E n e t = 4 k q d 2 * 2 c o s ? 30 ? = q 3 π ε 0 d 2

New answer posted

4 months ago

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R
Raj Pandey

Contributor-Level 9

m g = A ( 80 ) ρ 0 ? C g m g = A ( 79 ) ρ 4 ? C g ρ 4 ? C ρ 0 ? C = 80 79 = 1.01

 

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