We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
(2.00 hours) x (3,600 seconds/hour) = 7,200 seconds
(9.00 minutes) x (60 seconds/minute) = 540 seconds
The record time = (7,200 + 540 + 21) = 7,761 seconds
Distance = (speed) x (time)
= (5.436 m/s) x (7,761 sec) =<span> 42,188.8 meters
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The official length of the marathon run is 42,195 meters.
If we divide that by the record time in the question, we get
5.4368... m/s .
Rounded to the nearest thousandth, that's 5.437 m/s.
If the question had given the speed as 5.437 instead of 5.436 ,
then we would have calculated the distance to be
(5.437 m/s) x (7,761 sec) =<span> 42,196.6 meters,
4.6 meters closer to the official distance than the answer we did get.
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Answer:
A.) 1430 metres
B.) 80 seconds
Explanation:
Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:
U = acceleration × time
U = 1.1 × 20 = 22 m/s
It then proceeds at constant speed for 1100 m
Then, time t will be
Time = distance/ velocity
Time = 1100/22
Time = 50 s
before slowing down at 2.2m/s^2 until it stops at the station.
Deceleration = velocity/time
2.2 = 22/t
t = 22/2.2
t = 10s
Using area under the graph, the distance between the two stations will be :
(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)
220 + 1100 + 110
1430 m
The time taken between the two stations will be
20 + 50 + 10 = 80 seconds
Answer:
The magnitud of the velocity is

and the direccion:
degrees from the horizontal.
Explanation:
Fist we define our variables:

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.
velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.
Using conservation of momentum:

Clearing for the velocity of the stone after the crash:

Substituting known values:

The magnitud of the velocity is :

and the direction:

this is -28.3 degrees from the +i direction or the horizontal direcction.
Note: i and j can also be seen as x and y axis.