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A space station sounds an alert signal at time intervals of 1.00 h . Spaceships A and B pass the station, both moving at 0.400c0 relative to the station but in opposite directions.A.How long is the time interval between signals according to an observer on A?B.How long is the time interval between signals according to an observer on B?C.At what speed must A move relative to the station in order to measure a time interval of 2.00 hbetween signals?

Respuesta :

Answer:

a)  Time interval between signals according to an observer on spaceship A  = 1.09h

b)  Time interval between signals according to an observer on spaceship B = 1.09h

c)The speed of A relative to the station to measure a time interval of 2.00h = 2.6 * 10⁸ m/s

Explanation:

a) Time interval between signals according to an observer on spaceship A

Initial time intervals for the alert signal, Δt₀ = 1.00h

The speed of spaceship A, v = 0.400c

The formula for the time interval is given by

Δt = ∆t₀ /√(1-(v/c)² )

Δt = 1 / √(1-(0.4c/c)² )

Δt = 1 / √(1-(0.4)²

Δt = 1.09 h

b) The intervals between signals according to an observer on spaceship A and B are equal because both spaceships are moving at the same speed.

Therefore, the time interval between signals according to an observer on B is Δt = 1.09 h

c) The speed of A relative to the station to measure a time interval of 2.00h

Δt = ∆t₀ /√(1-(v/c)² )

2 = 1 /√(1-(v/c)² )

√(1-(v/c)² ) = 0.5

1-(v/c)² = 0.25

0.75 = (v/c)²

v/c = 0.866

v = 0.866c

v = 0.866 * 3 * 10⁸

v = 2.6 * 10⁸ m/s

onyebuchinnaji

(a) The time interval between signals according to observer on A is 1.091 h.

(b) the time interval between signals according to observer on B is 1.091 h.

(c) The speed of the spaceship A at the given 2.0 h time interval is [tex]2.6 \times 10^8 \ m/s[/tex]

The given parameters;

  • time interval of alert signal, Δt₀ = 1.0 h
  • speed of A = 0.4c

The time interval equation is calculated as follows;

[tex]\Delta t = \frac{\Delta t_0}{\sqrt{1 -(\frac{v}{c} })^2 } \\\\\Delta t = \frac{1}{\sqrt{1 - (\frac{0.4c}{c} })^2 } \\\\\Delta t = \frac{1}{\sqrt{0.84} } \\\\\Delta t = 1.091 \ h[/tex]

Thus, the time interval between signals according to observer on A is 1.091 h.

(b)

the time interval between signals according to an observer on B is 1.09 h, since the both spaceships are moving at the same speed.

(c) The speed of the spaceship A at the given 2.0 h time interval is calculated as follows;

[tex]\Delta t = \frac{\Delta t_0}{\sqrt{1 -(\frac{v}{c} })^2 } \\\\2 = \frac{1}{\sqrt{1 -(\frac{v}{c} })^2} \\\\2\sqrt{1 -(\frac{v}{c} })^2 = 1\\\\\sqrt{1 -(\frac{v}{c} })^2 = 0.5\\\\1 - (\frac{v}{c} )^2 = (0.5)^2\\\\(\frac{v}{c} )^2 = 1 - (0.5)^2\\\\(\frac{v}{c} )^2 =0.75\\\\\frac{v}{c} = \sqrt{0.75} \\\\\frac{v}{c} = 0.866\\\\v = 0.866c\\\\v = 0.866 \times 3\times 10^8\\\\v = 2.6 \times 10^8 \ m/s[/tex]

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