a. If The Enterprise istraveling at 0.900 c , does Picard have enough time torec

Question

a.       If The Enterprise istraveling at 0.900 c, does Picard have enough time torecharge The Enterprise’s phasors before losing the theEarth’s shadow as a shield?

b.      How long does the Romulan Warshiphave to activate their cloaking device before The Enterpriseemerges from Earth’s shadow?

c.       How far in the Earth’sshadow will Captain Picard measure that The Enterprise has traveledby the time their phasors are recharged?

d.      How far will the Romulan shipmeasure that The Enterprise has traveled in the Earth’sshadow at the instant their phasors are recharged?

Explanation / Answer

a) First we will determine the length of time it takes for anobserver on earth to witness the distance traveled. Sincethis does not involve relativity since the speed was measuredrelative to earth and so was the distance we can use t= displacement/velocity = 1.5*10^7/0.9c =1.5*10^7/(2.998*10^8*.9) = 0.055593 s or 55.593 ms ________________________________________________________________________________ Now for the enterprise's frame of reference thissame time elapsed will be t_o= t*(1-u^2/c^2) =0.055593*(1-.9^2)=0.024232 s or 24.3232 ms So the enterprise will recharge their phasors with 4.3232 msto spare b) Since the romulun ship is at rest relative to the earth thetime it takes will be the same as the t measured from theearth and as previously found t=55.59 ms b) Since the romulun ship is at rest relative to the earth thetime it takes will be the same as the t measured from theearth and as previously found t=55.59 ms c) For the enterprise it takes 20 ms in their frame of referenceto recharge their sheilds So l_o=t_o*u = .020*(3.00*10^8*0.9) l_o= 5.4*10^6 m c) For the enterprise it takes 20 ms in their frame of referenceto recharge their sheilds So l_o=t_o*u = .020*(3.00*10^8*0.9) l_o= 5.4*10^6 m d) We can solve this 2 different ways to get the same answer Iwill do both just prove you can First we can use the distance found in part c and useequation l_o = l/(1-u^2/c^2) l = 5.4*10^6/(1-.9^2) = 1.24*10^7 m ______________________________________________________________________________________ Or we can figure out the time it takes for the phasors torecharge from the reference frame of the romulun ship and then usethe simple equation displacement= velocity*time t=t_o/(1-u^2/c^2) t = 0.020s/(1-.9^2) t= 0.045883 s or 45.883 ms So l=v*t l=(0.9*3.0*10^8)*0.045883 l=1.24*10^7 m Notice this is the same answer as before so we can be veryconfident that it is correct Hope this helped! d) We can solve this 2 different ways to get the same answer Iwill do both just prove you can First we can use the distance found in part c and useequation l_o = l/(1-u^2/c^2) l = 5.4*10^6/(1-.9^2) = 1.24*10^7 m ______________________________________________________________________________________ Or we can figure out the time it takes for the phasors torecharge from the reference frame of the romulun ship and then usethe simple equation displacement= velocity*time t=t_o/(1-u^2/c^2) t = 0.020s/(1-.9^2) t= 0.045883 s or 45.883 ms So l=v*t l=(0.9*3.0*10^8)*0.045883 l=1.24*10^7 m Notice this is the same answer as before so we can be veryconfident that it is correct Hope this helped!

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