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New answer posted

7 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

7 months ago

0 Follower 4 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

7 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

7 months ago

0 Follower 11 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

7 months ago

0 Follower 66 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

7 months ago

0 Follower 2 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

7 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

New answer posted

7 months ago

0 Follower 20 Views

A
alok kumar singh

Contributor-Level 10

Binding Energy = (Δm)c²
= [Zmp + (A-Z)mn - MAl]c²
= [ (13*1.00726 + 14*1.00866) - 27.18846] u
= [ (13.09438 + 14.12124) - 27.18846] u
= [27.21562 - 27.18846] u
= 0.02716 u
= 0.02716 x = 27.16x * 10? ³

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

C? = dQ/ndT = (dU + pdV)/ndT
= C? + (pdV/ndT)
C? - C? = pdV/ndT = R - For ideal Gas [Box: PV = nRT, pdV = nRdT]
C? - C? = 1.1R - For Non – Idea gas (for Real gas)
And Real gas behaves as ideal gas at high temperature & low pressure.
∴ T_B > T_A

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

v = ω√ (A² - x²)
K.E. = ½mv² = ½mω² (A² - x²)
Total Energy T.E = ½mω²A²
At x = A/2
K.E. = ½mω² (A² - (A/2)²) = ½mω²A² (3/4)
K.E./T.E. = (½mω²A² (3/4) / (½mω²A²) = 3/4

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