Answer:
No, there won't be a collision.
Explanation:
We will use the constant acceleration formulas to calculate,
v = u + a*t
0 = 25 + (-0.1)*t
t = 250 seconds (the time taken for the passenger train to stop)
v^2 = u^2 + 2*a*s
0 = (25)^2 + 2*(-0.1)*s
s = 3125 m (distance traveled by passenger train to stop)
If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur
Speed*time = distance
Distance = (15)*(250)
Distance = 3750 m
As the distance is way more, there won’t be a collision
<span>` You can consider T to be in units of seconds/step. Frequency is the inverse of period, so
1/T = frequency and has units of steps per second. There will be 60 times as many steps in a minute.</span>
Answer:
the magnitude of acceleration will be 1.50m/s^2
Explanation:
To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma
if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.
The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).
you will find that Fnet=15.0N
With that, plug in the values you know to calculate the acceleration of the block:
Fnet=ma
(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:
1.50=a (and now make sure to label the units of your answer)
a=1.50m/s^2 (which is the typical unit for acceleration)