Yes. B'coz its magnification is always negative.
Answer:
The final position is 36 feet.
Explanation:
initial position, d = 330 feet
speed, v = 3 feet per minute
time, t = 30 minute
now the time is 32 minute
time interval = 2 minute
So, the distance in 2 minutes is
d' = 2 x 3 = 6 feet
So, the final position is
D = 30 + 6 = 36 feet
Answer:
v_average = 500 m / min
Explanation:
Average speed is defined
v = (x_{f} -x₀) / Δt
let's look in each section
section 1
the variation of the distance is 800 in a time of 1.4 min
v₁ = 800 / 1.4
v₁ = 571.4 m / min
section 2
distance interval 500 in a 1.6 min time interval
v₂ = 500 / 1.6
v₂ = 312.5 m / min
section 3
distance interval 1200 m in a time 2 min
v₃ = 1200/2
v₃ = 600 m / min
taking the speed of each section we can calculate the average speed
the distance traveled
Δx = 800 + 500 + 1200
Δx = 2500 m
the time spent
Δt = 1.4 + 1.6+ 2
Δt = 5 min
v_average = Δx / Δt
v_average = 2500/5
v_average = 500 m / min
The transfer is perpendicular
Surprised
I knew this<span />
