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New answer posted
10 months agoContributor-Level 10
Power dissipated is same
P? = P?
P_R = P_RLC
(V²/R) = (V²/Z)cos (φ) = (V²/Z) (R/Z) = V²R/Z²
R = R²/Z² => Z² = R²
R² + (ωL - 1/ωC)² = R²
ωL = 1/ωC
ω = 1/√ (LC) = 1/√ (0.1 * 40 * 10? ) = 1/√ (4 * 10? ) = 1/ (2 * 10? ³) = 500 rad/s
New answer posted
10 months agoContributor-Level 10
C = 200µF, R
Q/A = constant [Given]
A = A? e? & Q = Q? e? /?
Q/A = (Q? /A? ) e^ (-t/RC + λt) = constant
For this to be constant, the exponent must be zero.
-t/RC + λt = 0
1/RC = λ
R = 1/ (λC) =
New answer posted
10 months agoContributor-Level 10
r = 0.5Å = 0.5*10? ¹? m
v = 2.2*10? m/s
I =?
t = 2πr/v
I = e/t = ev/2πr = (1.6*10? ¹? * 2.2*10? )/ (2 * 22/7 * 0.5*10? ¹? )
I ≈ 1.12*10? ³ A = 1.12 mA = 112 * 10? ² mA
New answer posted
10 months agoContributor-Level 10
We have to find the point where the gravitational field must be zero.
EG = 0
GM/x² = G (9M)/ (8R-x)²
1/x² = 9/ (8R-x)²
8R - x = 3x => x = 2R
Potential at A (surface of first planet), VA = -GM/R - G (9M)/7R = -16GM/7R
Potential at point x, Vx = -GM/x - G (9M)/ (8R-x) = -GM/2R - G (9M)/6R = -2GM/R
ΔV = Vx - VA = -2GM/R - (-16GM/7R) = (-14+16)GM/7R = 2GM/7R
Using conservation of mechanical energy
ΔKE = ΔU = mΔV
½mv² = m (2GM/7R)
v² = 4GM/7R
v = √ (4GM/7R) => a = 4
New answer posted
10 months agoContributor-Level 10
M = 9.85*10? ² A/m²
I = 5*10? kg-m²
T = 5s/10osc = 0.5 sec
B =?
T = 2π√ (I/MB)
B = 4π²I/ (MT²) = (4*9.85*5*10? )/ (9.85*10? ² * 0.5²)
B = (20*10? )/ (10? ² * 0.25) = 800*10? = 0.8*10? ³ T = 8mT
New answer posted
10 months agoContributor-Level 10
Amplitude of sidebands = µAc/2
Here µ = Am/Ac = 5/15 = 1/3
Amplitude = (1/3) (15)/2 = 2.5 V
a/10 = 2.5 => a = 25
b/10 = 2.5 => b = 25
a/b = 1
New answer posted
10 months agoContributor-Level 10
For photon, λ? = h/p = 6.6*10? ³? /10? ²? = 6.6*10? m
For a particle, λ? = h/mv = 6.6*10? ³? / (9.1*10? ³¹ * 10? ) ≈ 7.25*10? ¹? m
λ? /λ? = 6.6*10? / 7.25*10? ¹? ≈ 910
New answer posted
10 months agoContributor-Level 10
Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 * (0.5*10? * 10? / 0.1) * (0.04)² = 20*10? ³ v²
0.5 * (5*10²) * 1.6*10? ³ = 20*10? ³ v²
0.4 = 20*10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s
New answer posted
10 months agoContributor-Level 10
When T? & T? are connected, at steady state current, I becomes
I = 6V/6Ω = 1A
Now when T? & T? are connected, the current through the inductor, just after connecting, remains same. So
V? Ω = 1A * 3Ω = 3volt.
New answer posted
10 months agoContributor-Level 10
Light rays travel undeviated if the refractive index of the medium does not change for any value of angle of incidence.
i.e. n? = n?
1.2 + 10.8*10? ¹? /λ² = 1.45 + 1.8*10? ¹? /λ²
9.0*10? ¹? /λ² = 0.25
λ² = 36*10? ¹?
λ = 6*10? m = 600 nm
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