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New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

V A B = 5 5 * 4 = 4 V

⇒ Q = C e q V = 2 x = 8 μ c

New answer posted

a year ago

0 Follower 22 Views

V
Vishal Baghel

Contributor-Level 10

Emf = − d ? d t

When magnet passes through the centre rogion of loop, no emf is induced and the polarity of emf will be negative while entering the loop.

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

F. B. D of Beaker

By NLM2    

f = m a = m ω 2 R ⇒ m ω 2 R ≤ μ N R ≤ μ g / ω 2

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

PT3 = k (constant)

⇒ ( n R T V ) T 3 = k

⇒ T 4 V − 1 = k

⇒ T 4 = k V

⇒ 4 * Δ T T = Δ V V - (i) >

⇒ Δ V = V Y Δ T - (ii) >

Comparing I > & ii >

⇒ Y = 4 T

New answer posted

a year ago

0 Follower 20 Views

V
Vishal Baghel

Contributor-Level 10

dA = 2r dr

d q = σ * 2 π r     d r

⇒ E = ∫ 0 R k d q z ( z 2 + r 2 ) 3 2

⇒ E = σ 2 ε 0 [ 1 − z R 2 + z 2 ]

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

P = ρ R T

⇒ ρ 1 ρ 2 = P 1 P 2 * T 2 T 1 = 7 6 4 5 * 2 6 0 3 0 0

⇒ ρ 1 ρ 2 = M 1 M 2

⇒ M 2 = 4 5 * 3 0 0 * 1 8 5 7 6 * 2 6 6 = 1 2 3 . 5 4 k g

New answer posted

a year ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

X P ( t ) = α t + β t 2

V P ( t ) = α + 2 β t  - (i)

X Q ( t ) = f t − t 2

V Q ( t ) = f − 2 t - (ii)
(i) = (ii)

α + 2 β t = f − 2 t ⇒ t ( 2 β + 2 ) = f − α

t = f − α 2 ( β + 1 )

New answer posted

a year ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

First angle, θ1= θ, R = u 2 s i n 2 θ g      

Another angle (θ2 = 90 - θ) for which range will be Same as that of θ1 =θ

R 1 = u 2 s i n 2 ( 9 0 − θ ) g = u 2 g s i n 2 θ = R

at θ 1 = θ ,           h 1 = u 2 s i n 2 θ 2 g

&       θ 2 = 9 0 − θ , h 2 = u 2 s i n 2 θ 2 2 g = u 2 c o s 2 θ 2 g

⇒ h 1 h 2 = 1 4 2 [ u 2 s i n 2 θ g ] 2

h 1 h 2 = 1 4 2 R 2

R = 4 h 1 h 2

So, Both statement is true & Reason is correct explanation for statement 1.

New answer posted

a year ago

0 Follower 1 View

S
Syed Aquib Ur Rahman

Contributor-Level 10

A beat wave arises due to constructive and destructive interference between two sound waves. These two frequencies have nearly equal frequencies but never the same. The resulting amplitude envelope oscillates at the beat frequency.  

New answer posted

a year ago

0 Follower 2 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

Beats form when two harmonic waves of nearly equal frequencies and identical amplitudes travel in the same direction, whic allows the superposition of waves. 

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