You have to reduce 2.00 an5.00 I order to use the×that=0.800
Answer:
From what I see, it's saying that every minute, the ant can move 30 meters. So how many meter would it move in 45 minutes?
30 meters = 1 min
x meters = 45 min
1 min x 45 = 45 min
30 meters x 45 = 1,350 meters
So, I believe the answer would be 1,350 meters.
hope this helps. :>
Because of Gravity, Basically a force so strong it constantly pulls us to the earth with 1 G (Maybe 100 pounds of force constantly pulling us to the earth)
2a for example the first one,2sec. You know that every second it moves 3metres further. So 2x3=6 but you start at 0.50m so 6+0.50=6.5
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s