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:

Explanation:
<u>Dimensional Analysis</u>
It's given the relation between quantities A, B, and C as follows:

and the dimensions of each variable is:



Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

Operating:


Equating the exponents:


Adding both equations:

Solving:


Answer:

It depends on the indirect taxes and the share market values
Answer:
The second system must be set in motion
seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion
seconds later