Answer: car B has travelled 4times as far as Car A
d=vi*t+1/2at^2
No initial velocity so equation becomes;
d=1/2at^2 and the acceleration is the same between both only time is different;
Car A d=1/2a(1)^2
Car B d=1/2a(2)^2
Car A d= 1^2=1
Car B d= 2^2=4
Car B d=4*Car A
So car B has travelled 4 times as far as car A
The simple answer would be; closeness to the equator combined with general climate.
Answer:
1170 m
Explanation:
Given:
a = 3.30 m/s²
v₀ = 0 m/s
v = 88.0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)
x = 1173.33 m
Rounded to 3 sig-figs, the runway must be at least 1170 meters long.
15.0 I’m pretty sure that’s the answer to your question
Do not approach within 100-yards, and slow to minimum speed within 500-yards of any U.S. naval vessel.
