Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6

- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6
*(15s)²
Solving:
x=½*6
*(15s)²
x=½*6
*225 s²
x= 675 m
<u><em>
The object will travel 675 m during that time.</em></u>
Volume = mass / density = 45.6/10.5 = .... L