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New answer posted

10 months ago

0 Follower 19 Views

R
Raj Pandey

Contributor-Level 9

dC? = (ε? + kx)A / dx [For 0 < x < d/2]
1/C? = ∫ dx / (ε? + kx)A) from 0 to d/2
= (1/Ak) [ln (ε? + kx)] from 0 to d/2
= (1/kA) ln (1 + kd/ (2ε? )
C? = kA / ln (1 + kd/ (2ε? )

Similarly dC? = (ε? + k (d-x)A / dx [For d/2 ≤ x ≤ d]
C? = kA / ln (1 + kd/ (2ε? )
Clearly, C? = C? = C
For series combination:
C_eq = C? / (C? + C? ) = C/2 = kA / (2ln (2ε? + kd)/2ε? )

New answer posted

10 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

i = constant
v ∝ E & E ∝ 1/r² So, E increases with decrease in the radius.
also v ∝ E
So, drift speed increases.

New answer posted

10 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

I? / I? = (MR²/2) / (mr²/4)
[ M = σπR², m = σπr² ]
= 2 (M/m) (R²/r²)
= 2 (σπR²/σπr²) (R²/r²)
= 2 (R²/r²)² = 2R? /r?

New answer posted

10 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

(a) I = ML²/12
(b) I = (2M)L²/3
(c) I = M (2L)²/12 = ML²/3
(d) I = (2M) (2L)²/3 = 8ML²/3

New answer posted

10 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

Y? = (F/A) / (Δl? /l)
Y? = (F/A) / (Δl? /l)
Y = (F/A) / (Δl? + Δl? )/ (2l)
= (F/A) / ( (1/2) * (Δl? /l + Δl? /l) )
= (F/A) / ( (1/2) * (F/ (AY? ) + F/ (AY? ) )
= 2Y? Y? / (Y? + Y? )

 

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Bandwidth
= (ω? + ω? ) - (ω? - ω? )
= 2ω?
= 2 (2πf? )
= 4πf?
= 2π (2f? )
= 2π (2*10? ) rad/s
= 2 * 10? Hz
= 200 kHz

New answer posted

10 months ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

T sin θ = (1/4πε? ) * q²/ (2lsinθ)²
T cos θ = mg
∴ tan θ = q² / (4πε? mg * 4l²sin²θ)
[tan θ ≈ θ, for small angle]
So, θ³ = q² / (16πε? mgl²)
θ = ( q² / (16πε? mgl²) )¹/³
Also separation = 2l sin θ ≈ 2lθ
= 2l ( q² / (16πε? mgl²) )¹/³
= ( 8q²l³ / (16πε? mgl²) )¹/³
= ( q²l / (2πε? mg) )¹/³

New answer posted

10 months ago

0 Follower 92 Views

A
alok kumar singh

Contributor-Level 10

Each side of the square has a resistance of 16/4 = 4Ω.
The side AC has the 9V, 1Ω source.
The other three sides (AB, BD, DC) form a path with resistance 4+4+4 = 12Ω.
This is in parallel with the side AC (4Ω).
Total resistance of the loop part: (12*4)/ (12+4) = 48/16 = 3Ω.
Total resistance of the circuit: R_total = 3Ω + 1Ω (internal) = 4Ω.
Total current from source I = V/R_total = 9/4 A.
This current splits.
Current through path ABDC, I? = I * (R_AC / (R_ABDC + R_AC) = (9/4) * (4/16) = 9/16 A.
Potential at B: V_B = V_A - I? R_AB = 9 - (9/16)*4 = 9 - 9/4 = 27/4 V.
Potential at D: V_D = V_C + I? R_CD = 0 + (9/16)*4 = 9/4 V.
Potential drop a

...more

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

τ = Iα
I = ½MR² = ½ (10) (0.2)² = 0.2 kgm²
α = (ωf - ωi)/Δt = (0 - 600*2π/60)/10 = -2π rad/s²
τ = |Iα| = 0.2 * 2π = 0.4π = 4π * 10? ¹ Nm

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

W = ∫Fxdy
W = ∫ (5y + 20)dy from 0 to 10
W = [5y²/2 + 20y] from 0 to 10
W = (5 (10)²/2 + 20 (10) - 0 = 250 + 200 = 450 J

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