The Lorentz transformation equations for space and time intervals are:
Ξt = Ξ³ (Ξtβ² + vΞxβ² / cΒ²)
Ξx = Ξ³ (Ξxβ² + vΞtβ²)
where:
Ξ³ = 1 / β(1 β vΒ² / cΒ²)
Ξ³ = 1 / β(1 β vΒ² / cΒ²)
= 1 / β(1 β (0.6)Β²)
= 1 / β(1 β 0.36) = 1 / β0.64 = 1.25
Using the Lorentz transformation for Ξt:
Ξt = Ξ³ Β· (vΞxβ² / cΒ²)
Substituting the values:
Ξt = 1.25 Γ (0.6c Γ 5) / cΒ²
= 1.25 Γ (3c / cΒ²)
= 1.25 Γ (3 Γ 10βΈ Γ 5) / (9 Γ 10ΒΉβΆ)
Simplify:
Ξt = (1.25 Γ 15) / (9 Γ 10βΈ)
= (1.25 Γ 5) / (3 Γ 10βΈ)
= 6.25 / (3 Γ 10βΈ)
Ξt = (6.25 / 3) ns = 12.5 ns
Using the Lorentz transformation for Ξx:
Ξx = Ξ³Ξxβ² = 1.25 Γ 5 = 6.25 m
A particle of mass 1kg, initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\frac{π}{6}\)\(\frac{\pi}{6}\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2seconds is ______N-m. (Rounded off to nearest integer)
(Take g = 10m/s2)
The P-V diagram of an engine is shown in the figure below. The temperatures at points 1, 2, 3 and 4 are T1, T2, T3 and T4, respectively. 1β2 and 3β4 are adiabatic processes, and 2β3 and 4β1 are isochoric processes
Identify the correct statement(s).
[Ξ³ is the ratio of specific heats Cp (at constant P) and Cv (at constant V)]