Answer:
4.18
Explanation:
Givens
The car's initial velocity 
= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.
Knowns
We know that the maximum static friction force is given by:
μ_s*n (1)
Where μ_s is the coefficient of static friction and n is the normal force.
Calculations
(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:
Δx=
a=41 m/s
This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:
Substituting (3) into (1), we get:
μ_s*m*g
Equating this equation with (4), we get:
ma= μ_s*m*g
μ_s=a/g
=4.18
Answer:
the car will slow down due to the change in speed
Answer:
Explanation:
The equation that relates electric field strength and force that a charge experiments by it is F=qE.
Newton's 2nd Law states F=ma, which for our case will mean:
We know from accelerated motion that 
In our case the proton is released from rest, so 
and we get 
Substituting, we get our final velocity equation:
For the <em>proton </em>we know that 
and 
. Writing in S.I. d=0.0025m, we obtain:
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Answer:
(a) x=157 m
(b) No
Explanation:
Given Data
Mass of proton m=1.67×10⁻²⁷kg
Charge of proton e=1.6×10⁻¹⁹C
Electric field E=3.00×10⁶ N/C
Speed of light c=3×10⁸ m/s
For part (a) distance would proton travel
Apply the third equation of motion
In this case vi=0 m/s and vf=c
so
From the electric force on proton
put this a(acceleration) in Equation (i)
So
For part (b)
No the proton would collide with air molecule
