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New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Since speed of light is constant for all colour so red colour and blue colour have different frequencies and different wavelengths.

New answer posted

10 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

N? /N? = e?
For 33% decay, N? /N? = 0.67 ≈ 2/3.
2/3 = e? ⇒ t? = (1/λ)ln (3/2)
For 67% decay, N? /N? = 0.33 ≈ 1/3.
1/3 = e? ⇒ t? = (1/λ)ln (3)
Δt = t? - t? = (1/λ) [ln (3) - ln (3/2)] = (1/λ)ln (2) = T? /? = 20 min

New answer posted

10 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

λ = kT / (√2πd²P)
= (1.38*10? ²³ * 300) / (√2 * 3.14 * (0.3*10? )² * 1.01*10? )
≈ 102 nm

New answer posted

10 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

Since B,  v and length are perpendicular
ε = Bvl
emf will induce only in wire CD
ε = B (d)v? (d) = B? (d/a)v? d = B? v? d²/a

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

using hook's law:
σ = Yε ⇒ f/A = Y (x/l) ⇒ Y = fl/ (xA) = fl/ (xπr²)
Using error analysis formula:
ΔY/Y = Δf/f + Δl/l + Δx/x + 2Δr/r
%error in Y = [ (Δm/m) + (Δl/l) + (Δx/x) + 2 (Δr/r) ] * 100
= [ (1/1000) + (1/1000) + (0.001/0.5) + 2 (0.001/0.2) ] * 100
= [ 0.001 + 0.001 + 0.002 + 0.01 ] * 100 = 1.4%

New answer posted

10 months ago

0 Follower 37 Views

A
alok kumar singh

Contributor-Level 10

v = 0.5t² i + 3t j + 9 k m/s ⇒ a = dv/dt = (ti + 3j) m/s²
At t=2sec,  v = 2i + 6j + 9k m/s and a = (2i + 3j) m/s²
The direction of acceleration of mosquito after 2s is given by the angle θ with the y-axis, where tanθ = a? /a? = 2/3.
So, the direction is tan? ¹ (2/3) from the y-axis.

New answer posted

10 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

10 months ago

0 Follower 25 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

10 months ago

0 Follower 11 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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