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New answer posted
7 months agoContributor-Level 10
For electron,
λe = h/p ⇒ (KE)e = ½mv² = p²/ (2m) = (h²/λ²)/ (2m)
For photon,
λp = h/p ⇒ (K.E.)p = pc = hc/λ
KEe / KEp = (h²/ (2mλ²) / (hc/λ) = h/ (2mcλ)
But for electron p = mv = h/λ so h/λ = mv
KEe / KEp = mv / (2mc) = v/2c
New answer posted
7 months agoContributor-Level 9
Displacement in 4? second
= S? - S?
= (1/2)g [2n - 1]
= (1/2)g [2*4 - 1]
= (1/2) * 9.8 * 7
= 34.3m
As this distance matches with data given in question for position of next drop. So drops are falling at the rate of 1 drop/second.
New answer posted
7 months agoContributor-Level 10
Body is dropped from height 75m with initial velocity (upward) 10m/s
So, s = ut + ½ at²
-75 = 10t - ½ gt²
5t² - 10t - 75 = 0
By solving t = 5 sec.
In 5sec. balloon covered
h = 10 * 5 = 50m
Now height of balloon = 75 + 50 = 125m
New answer posted
7 months agoContributor-Level 10
ε? = 250 * Potential Gradient
ε? + ε? = 400 * Potential Gradient
(ε? )/ (ε? + ε? ) = 250/400 = 5/8
By solving above
ε? /ε? = 5/3
New answer posted
7 months agoContributor-Level 10
By shell's law
(sin θ)/ (sin θ') = 4/3 . (i)
For TIR on second surface
sin θ' > sin θc
(θ' + θ' = 90)
sin (90 - θ') > sin θc
cos θ' > sin θc
1 - sin²θ' > sin²θc (By equation (i) )
1 - (3/4 sin θ)² > sin²θc
[sin θc = 3/4]
1 - (9/16)sin²θ > (9/16)
1 - 9/16 > (9/16)sin²θ
7/16 > (9/16)sin²θ
sinθ < 7/3
New answer posted
7 months agoContributor-Level 9
λ = h/mv = h/√2mK
For same K:
λ ∝ 1/√m
λ? : λ? : λ? = 1/√m? : 1/√m? : 1/√4m?
As m? > m? ,
λ? > λ? > λ?
New answer posted
7 months agoContributor-Level 9
3N? /4 = N? e?
⇒ e? = 4/3
t? = (1/λ) ln (4/3)
t? = ln2/λ
t? - t? = (1/λ)ln2 - (1/λ)ln (4/3)
= (1/λ)ln (2/ (4/3) = (1/λ)ln (3/2)
New answer posted
7 months agoContributor-Level 9
β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)
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