Answer:
V = 0.714m/s
Explanation:
Full solution calculation can be found in the attachment below.
From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.
Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.
See the attachment below for the solution calculation.
Answer:
0.2687 approximately 0.27
Explanation:
Diameter = 0.320
Speed = 40.0 rev/min
We are required to find coefficient of static friction between friction and button
The radius can be calculated as
0.320/2
= 0.160m
Then we have the rotational speed w = 40rev/min x 2pi/60
= 4.19 rad/s
umg = mrw²
u = mrw²/mg
u = rw²/g -------(1)
g = 9.8
When we put values into equation 1
0.150m x 4.19² / 9.8
= 0.150m x 17.5561 /9.8
= 0.2689
This is approximately 0.27
Neither set of choices is correct.
If the distance is tripled, then the forces decrease to
1/9 Fg. and. 1/9 Fe.
Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.
