A spaceship flies past earth with a speed of 0.990 c (about 2.97×108m/s) relativ
ID: 1782210 • Letter: A
Question
A spaceship flies past earth with a speed of 0.990c (about 2.97×108m/s) relative to earth. A crew member on the spaceship measures its length, obtaining the value 400 m. What is the length measured by observers on earth?
Suppose your car is 5.00 m long when it is sitting at rest in your driveway. How fast would it have to move for its length (as seen by someone at rest in the driveway) to be 3.60 m ?
Express your answer in meters per second to three significant figures.
MasteringPhysics: HW#5-Google Chrome Secure l https://session.masteringphysics.com/myct/itemView?assignmentProblemID-8741 5550&offsetsnext; HPractice Problem 273 e previous | 5of 27 Ined a Practice Problem 27.3 SOLUTION SET UP The lan ofthe spaceship nta tame n which SOLVE From the equatom Aspacerip figg past aanh with a speed at0000e (about 2.97 × 10° m/s) nalativa to ath. A crew momber on the spacechip maaaues ita largh, octanng tha value 400 m. ha is he length measured by obeervers on earth? at ast 400 m sapro erlangth in s hama comas ondng tol. in e 1 , Navant to tnd the en h 1 m asured by observa sonaathho seeneship mo n (400 m) Vi-(0.564 m REFLECT To messure I. we need tuo obeervers because we have to okeerve the positions of the two ends of the spaceship simulteneously in the esth's neference fame We could use two okeervers with synchronized clocks Ginr1 Thwe two otcarvations ara mutaneous in eam's rararonoe trama, but thayag nor simutanoous as seen by an oorver in the spaceship. Suppoe your car i 5.00m long when it is sttintrest in your driveway How fast wouild it have to move for is h (as seen by scmeone at rest in the driveway) to be 3.80 m? Express your answer in meters per seoond to three signficant figures. 09 o, o ohtnin the correct lengthl- Type here to search 6:05 PM 11/14/2017Explanation / Answer
1.
Given that,
v^2 / c^2 = 0.990
length measured by observers on earth,
L = Lo * sqrt (1 - (v^2 / c^2)) (relativity of length)
L = 400 * sqrt (1 - (0.990)^2)
L = 56.4 m
2.
Given that,
L = 3.6 m
Lo = 5 m
From the formula of relativity of length
L = Lo * sqrt (1 - (v^2 / c^2))
3.6 = 5 * sqrt (1 - v^2 / (3*10^8)^2)
v = 2.08*10^8 m/s
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