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New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

E = -dφ/dt = - (20t + 20) mV.
At t=5s, E = - (100+20) = -120mV.
I = |E|/R = 120mV/2Ω = 60mA.

New answer posted

a year ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

v = ∫a dt = (1/M)∫F dt
v = (F? /M) ∫ [1 - (t-T)²/T²] dt from 0 to 2T
= (F? /M) [t - (t-T)³/3T²]? ²?
= (F? /M) [ (2T - T³/3T²) - (0 - (-T)³/3T²) ]
= (F? /M) [ 2T - T/3 - T/3 ] = (F? /M) [ 4T/3 ] = 4F? T/3M.

New answer posted

a year ago

0 Follower 20 Views

V
Vishal Baghel

Contributor-Level 10

Δy/y = 2Δm/m + 4Δr/r + |x|Δg/g + (3/2)Δl/l
18 = 2 (1) + 4 (0.5) + |x|p + (3/2)4
18 = 2 + 2 + |x|p + 6 = 10 + |x|p
8 = |x|p
From the options, if x=8, p=±1. If x=16/3, p=±3/2. If x=5, p=±8/5. If x=4, p=±2.
Option B gives x=16/3, and p is not among the options.
Option A gives x=5, p not among options.
Option C gives x=8, p not among options.
Option D gives x=4, p not among options.
There must be a typo in the question or options. The solution gives x=16/3 and p=3/2. Let's check.
8 = (16/3) * (3/2) = 8. So B is correct.

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

The graph of? vs I is a parabola-like curve with a minimum at i=e. So (B) is the correct representation.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

ME = PE + KE = 8J
At x? , PE=4J, so KE = 8-4=4J. (A is correct)
At x>x? , PE=6J, so KE=2J, which is constant. (B is correct)
At x8J. This is not possible. Particle cannot reach here. So it is not a correct statement. (C is incorrect)
At x? , PE=0J, so KE=8J, which is maximum. (D is correct)

New answer posted

a year ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 27 Views

V
Vishal Baghel

Contributor-Level 10

Tsinθ = qE
Tcosθ = mg
tanθ = qE/mg
E = V/d_eff. V = V? - (-V? ) = V? +V?
C? =kε? A/t, C? =ε? A/ (d-t).
Capacitors are in series. Q=C_eqV. E inside dielectric = σ/ (kε? ) = Q/ (Akε? ).
E in air = σ/ε? = Q/Aε?
Let's follow the solution.
Q = (C? / (C? +C? ) (V? +V? ).
E = E_air = Q/ (Aε? )
tanθ = qQ/ (Aε? mg) = q/ (Aε? mg) * (C? / (C? +C? ) (V? +V? )
Substitute C? and C?
This is getting complicated. The solution directly uses an expression for E. E = Q/C? Aε?

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

P = Fv = (ma)v = m (dv/dt)v
P dt = mv dv
∫P dt = ∫mv dv
Pt = ½mv²
v = √ (2Pt/m)
dx/dt = √ (2P/m) t¹/²
x = √ (2P/m) ∫t¹/² dt = √ (2P/m) * (2/3)t³/²
Squaring to match options:
x² = (8P/9m) t³
This does not match. Let's re-examine the options.
Position x is proportional to t³/².
x = (8P/9m)¹/² t³/²

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