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
I believe it is acceleration
The third (left hand corner) since the x and y are both negative.
Answer:
The time constant is 1.049.
Explanation:
Given that,
Charge 
We need to calculate the time constant
Using expression for charging in a RC circuit
![q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]](https://tex.z-dn.net/?f=q%28t%29%3Dq_%7B0%7D%5B1-e%5E%7B-%28%5Cdfrac%7Bt%7D%7BRC%7D%29%7D%5D)
Where, 
= time constant
Put the value into the formula
![0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]](https://tex.z-dn.net/?f=0.65q_%7B0%7D%3Dq_%7B0%7D%5B1-e%5E%7B-%28%5Cdfrac%7Bt%7D%7BRC%7D%29%7D%5D)
Hence, The time constant is 1.049.
Answer:
the force exerted by the seat on the pilot is 10766.7 N
Explanation:
The computation of the force exerted by the seat on the pilot is as follows:
Hence, the force exerted by the seat on the pilot is 10766.7 N
