Answer:
Minimum uncertainty in velocity of a proton,
Explanation:
It is given that,
A proton is confined to a space 1 fm wide, 
We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,
Since, p = mv
So, the minimum uncertainty in its velocity is greater than 
. Hence, this is the required solution.
We know that the Delta E + W(Work done by non-conservative
forces) = 0 (change of energy)
In here, the non-conservative force is the friction force
where f = uN (u =kinetic friction coefficient)
W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2
umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)
This will then give us:
1/2Vi^2-ugd = 1/2V^2
V^2 = Vi^2 - 2ugd
So plugging in our values, will give us:
V= Sqrt (5.6^2 -2.3^2)
=sqrt (26.07)
= 5.11 m/s
There is too much information given, it's hard to understand exactly which variables are important in this problem.
Explanation:
Acceleration is change in velocity over change in time:
a = Δv / Δt
a = (10 m/s - 25 m/s) / (240 s - 0 s)
a = -0.0625 m/s²
So the car decelerates at 0.0625 m/s².
