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New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

In adiabatic process
PV? = constant
P (m/ρ)? = constant
As mass is constant
P ∝ ρ?
P_f/P_i = (ρ_f/ρ_i)? = (32)? /? = 2? = 128

New answer posted

a year ago

0 Follower 17 Views

V
Vishal Baghel

Contributor-Level 10

Now, using junction analysis
We can say, q? + q? + q? = 0
2 (x - 6) + 4 (x - 6) + 5 (x) = 0
x = 36/11, q? = 36 (5)/11 = 180/11
q? = 16.36µC

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

At T°C L = L? + L?
At T +? T Leq = L'? + L'?
where L'? = L? (1 + α? T)
L'? = L? (1 + α? T)
Leq = (L? + L? ) (1 + αavg? T)


⇒ (L? + L? ) (1 + αavg? T) = L? + L? + L? α? T + L? α? T
⇒ (L? + L? )αavg = L? α? + L? α?
⇒ αavg = (L? α? + L? α? )/ (L? + L? )

New answer posted

a year ago

0 Follower 27 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

a year ago

0 Follower 14 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 question posted

a year ago

0 Follower 6 Views

New answer posted

a year 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

a year ago

0 Follower 8 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

a year ago

0 Follower 9 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

a year ago

0 Follower 3 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

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