Observer A, who is at rest in the laboratory, is studing a particle that is moving through the laboratory at a speed of 0.624c and determines its lifetime to be 159ns a) Obsever A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory? b) Obsever B, who observes the particle to be at rest and measures its lifetime to be 124ns. According to B, how far apart are the two markers in the laboratory?

a) the markers are 29.76 m far apart in the laboratory

b) According to B, the markers are 23.21 m far apart in the laboratory

Explanation:

Given that;

speed of particle = 0.624c = (0.624 × 3×10⁸m/s)

{ where c is the speed of light}    

lifetime of the particle t = 159ns  = 1.59 × 10⁻⁷

a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory

lets apply the formula distance = velocity × time

lets substitute

distance = (0.624×3×10⁸ m/s)  × (1.59 × 10⁻⁷ s)

distance = 29.76 m

Therefore, the markers are 29.76 m far apart in the laboratory

b) Observer B, who observes the particle to be at rest and measures its lifetime to be 124ns. According to B, how far apart are the two markers in the laboratory?

lifetime of the particle t = 124ns  = 1.24 × 10⁻⁷

distance = (0.624×3×10⁸ m/s)  × (1.24 × 10⁻⁷ s)

distance = 23.21 m

Therefore,  According to B, the markers are 23.21 m far apart in the laboratory

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid

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