# an observer detects 2 explosions that occur at the same time ,one near and the other 100km away.another onserver finds that the 2 explosions occur 160km apart.what time interval separates the exploson to the second observer.

Jhankaar Nayyar
27 Points
2 years ago
For the first observer:

(Δs)² =Δn² - (cΔt)²

∵ Δn = -100 and Δt²= 0

∴ Δs² = (-100)² = 10,000

For second observer:

(Δs’)² = (Δn’)² - (cΔt’)²

And  (Δn’) = (160)² = 25,600

We know that;

(Δs)² = (Δs’)²

∴ 1,000 = 25,600 - (cΔt’)²

⇒  (cΔt’)² = 25,600 - 10,000

= 15,600

⇒    (Δt’)² = 15,600/c²

⇒        Δt’ = 1.73 × 10⁻¹³ sec (Ans.)

Jhankaar Nayyar
27 Points
2 years ago

For the first observer:

(Δs)² =Δn² - (cΔt)²

∵ Δn = -100 and Δt²= 0

∴ Δs² = (-100)² = 10,000

For second observer:

(Δs’)² = (Δn’)² - (cΔt’)²

And  (Δn’) = (160)² = 25,600

We know that;

(Δs)² = (Δs’)²

∴ 1,000 = 25,600 - (cΔt’)²

⇒  (cΔt’)² = 25,600 - 10,000

= 15,600

⇒    (Δt’)² = 15,600/c²

⇒        Δt’ = √(1.73 × 10⁻¹³)

= 4.163 × 10⁻⁷ sec. (Ans.)