Respuesta :
Answer:
observer see the leading edges of the two ships pass each other at time 0.136 s
Explanation:
given data
spaceship length = 100 m
speed of 0.700 Co = [tex]0.700\times 3 \times 10^8[/tex] m/s
distance d = 58,000 km = 58000 × 10³ m
solution
as here distance will be half because both spaceship travel with same velocity
so they meet at half of distance
Distance Da = [tex]\frac{d}{2}[/tex] ............1
Distance Da = [tex]\frac{58000\times 10^3}{2}[/tex]
Distance Da = 29 ×[tex]10^{6}[/tex] m
and
time at which observer see leading edge of 2 spaceship pass
Δ time = [tex]\frac{Da}{v}[/tex] ..........2
Δ time = [tex]\frac{29 \times 10^6}{0.700\times 3 \times 10^8}[/tex]
Δ time = 0.136 s
Answer:
The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.
Explanation:
Given that,
Length = 100 m
Speed = 0.700 c
Distance = 58000 km
The distance should be halved because the spaceships both travel the same speed.
So they will meet at the middle of the distance
We need to calculate the distance
Using formula for distance
[tex]d'=\dfrac{d}{2}[/tex]
Put the value into the formula
[tex]d'=\dfrac{58000}{2}[/tex]
[tex]d'=29000\ km[/tex]
We need to calculate the time at which the observer see the leading edges of the two ships pass each other
Using formula of time
[tex]\Delta t=\dfrac{d}{V}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{29\times10^{6}}{0.700\times3\times10^{8}}[/tex]
[tex]\Delta t=0.138\ sec[/tex]
Hence, The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.
