The easiest way to answer this question is by realizing there are relating the velocities of the two cars. To tackle this problem, you have to understand the picture. Car 1 travels at 35m/s and Car 2 travels at 25m/s. Based on relative velocities, we can understand that Car 1 travels 10m/s faster than Car 2 every second. So we can interpret Car 1's relative velocity to Car 2 as 10m/s. Car 1 needs to travel 10m/s till a point of catching up to Car 2 which is 462m away.
v = 10m/s
d = 462m
v = d/t
(10) = (462)/t
t = 46.2s
So it takes 46.2 seconds for Car 1 to catch up to Car 2, but the question is asking how far does Car 1 travel to catch up. So we have to use Car 1's velocity and not the relative velocity:
v = 35m/s
v = d/t
(35) = d/(46.2)
d = 1617m
Car 1 traveled a total distance of 1617m.
<span>W=98*1.49/9.8 N
</span>soo <span>14.9 Newtons</span>
Hello there.
<span>In the context of depth perception, which of the following is a monocular cue?
</span><span>(C) Convergence
</span>
Answer:
160000000 kg.
Explanation:
p=mv
p=1.6x10^9
v=10m/s
rearrange and substitute:
(1.6x10^9)=m(10)
m=(1.6x10^9)/10
m= 1.6x10^8 kg.
Answer:
v = √[ GM / (R + 160934.4 meters) ]
Explanation:
The speed of an object in a circular orbit (in meters per second) is
v = √(GM/r)
where
G = 6.6743e-11 m³ kg⁻¹ sec⁻²
M = the sum of the masses of the planet and the satellite in kilograms
r = the radius of the circular orbit in meters
If the planet's radius, R, is 100 miles, then
v = √[ GM / (R + 160934.4 meters) ]
