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
Here we know that



now from kinematics we have

now from above all values we have



so final angular speed is -12.6 rad/s
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
Answer:
A switch opens or closes the electrical circuit, turning the flow of electricity on or off.
Explanation: