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New answer posted
4 months agoContributor-Level 10
x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]
New answer posted
4 months agoContributor-Level 10
Binding energy = (50M? + 70M? – Msn)C²
= (50.3915 + 70.6069 – 119.902199)U²
= (1.0962U)C²
= 931 * 1.0962MeV
Binding energy per neucleon
= (931 x 1.0962)/120 MeV
= 8.5MeV
New answer posted
4 months agoContributor-Level 10
I = 0.8 kgM², |μ? | = 20Am²
U? = -μ? ⋅ B? = 0
Uf = -μBcos (30°) = -20 x 4 x √3/2
U? – Uf = 40√3 = ½ I ω² = 0.4ω²
ω² = 100√3; ω = 10 (3¹/? )
New answer posted
4 months agoContributor-Level 10
By conservation of energy
(K.E. + P.E.)? = (K.E. + P.E.)?
0 + (1/ (4πε? a³) * 2P * P = 2 * (1/2)mv² + 0
v = √ (p² / (2πε? ma³) = (P/a) * √ (1 / (2πε? ma)
New answer posted
4 months agoContributor-Level 10
Sol. eVslop = hc/λ - φ
Slope of curve
tan θ = hc/e = constant
as intensity of incident radiation is increased, there will be no effect on graph
New answer posted
4 months agoContributor-Level 10
In steady state rate of flow of heat in all three rods are same.
dQ/dt = (k? A (100 – 70)/? = (k? A (70 – 20)/? = (k? A (20 – 0)/?
30k? = 50k? = 20k?
∴ k? : k? = 2:3 & k? : k? = 2:5
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