Physics

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New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

y = ( n λ ) D d

n 1 λ 1 = n 2 λ 2

(8) ( 600 n m ) = n 2 ( 400 n m )

n 2 = 12

New answer posted

a year ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

⇒ T = W + f

= 20000 + 3000

= 23000 N

⇒ Power = T v

= 23000 * 1.5

= 34500 watts 

New answer posted

a year ago

0 Follower 11 Views

R
Raj Pandey

Contributor-Level 9

(a) Radio wave (ii) ≈ 102 m   (ii)
(b) Microwave ≈ (iii) 10.2 m   (iii)

(c) Infrared radiations ≈ (iv) 10.4 m   (iv)

(d) X - ray (i) ≈ ? = 10.10 m   (i)

(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

X C M = 20 * 10 20 + 10 = 20 3 m

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

In half wave rectification ⇒ f output   = 60 H z

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

E = P * t = 100 * 10 3 * 3600

= 36 * 10 7 J

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

P = P 0 + 4 T R

on expansion R increase, as a result, P  decreases.

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

M L T - 2 A - 2 = Magnetic permeability

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

Electric field is always perpendicular to equipotential surface.

New answer posted

a year ago

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J
Jaya Sharma

Contributor-Level 10

You can identify whether a medium has higher or lower refractive index through the three ways. The first step is to observe the direction of bending. In this case, if the light is bending towards the normal, the second medium has higher refractive index. If the light bends away from normal, first medium has higher index of refraction. 

In the second method, you can use Snell's law. If the second angle is smaller than the first one, second medium has higher refractive index. In case the first angle is smaller than second one, first medium has higher index of refraction. The third method is critical angle method where, if the light u

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