Multiplying the ideal gas law constant
do u know the answer to this question yet
Answer:
1) an observer in B 'sees the two simultaneous events
2)observer B sees that the events are not simultaneous
3) Δt = Δt₀ /√ (1 + v²/c²)
Explanation:
This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems
1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events
Consequently an observer in B 'sees the two simultaneous events
2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.
Consequently observer B sees that the events are not simultaneous
3) let's calculate the times for each event
Δt = Δt₀ /√ (1 + v²/c²)
where t₀ is the time in the system S' which is at rest for the events
Answer:
dR/dt = 10.2 ft / s
Explanation:
Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance
Balloon
y = y₀ +
t
Motorcycle
x = v₀ₓ t
Distance, let's use Pythagoras' theorem
R² = x² + y²
R² = (v₀ₓ t)² + (y₀ + t)²
v₀ₓ = 88 ft / s
= 8 ft / s
y₀ = 150 ft
R² = (8 t)² + (150 + 8 t )²
R² = 64 t² + (150 + 8t )²
This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)
2RdR / dt = 64 2 t + 2 (150 + 8t) 8
dR / dt = [64 t + (1200 + 64t )] / R
dR/dt = (1200 +128 t)/R
Let's calculate for the time of 10 s
dR / dt = (1200 + 128 10) / R = 2480 /R
R = √ [64 10² + (150 + 8 10)²
R = √ [6400 + 52900]
R = 243.5 ft
dR / dt = (2480) / 243.5
dR / dt = 10.2 ft / s
