Okay, so they want to basically Increase their grip, and they are taking advantage of the force of friction
Answer:
v = 57.2 m/s
Explanation:
The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:
v = s/t
where,
s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m
t = time taken to cover the distance = 2 h 14 min
Now, we convert it into minutes:
t = (2 h)(60 min/1 h) + 14 min
t = 120 min + 14 min = (134 min)(60 s/1 min)
t = 8040 s
Therefore, the value of velocity will be:
v = (4.6 x 10⁵ m)/8040 s
<u>v = 57.2 m/s</u>
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
solid state has <u>the </u><u>most</u> intermolecular force of attraction.
Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.
The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.
Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.
