First we have to calculate the acceleration of the car,

Here, u is initial velocity of the car and its value is given 12.2 m/s and v is final velocity of the car and it comes to stop, so its value zero.
.
As during the braking the acceleration is constant, from the kinematic equation,

Here, s is the distance traveled by the car during braking and its value is given 36. 5 m.
Substituting all the values in kinematic equation, we get

Therefore, car will stop after 5.98 s.
Answer:
The initial acceleration is 0,221 m/s^2.
Explanation:

Answer:
Acceleration,
Explanation:
Given that,
Electric field, E = 680 N/C
Speed of the proton, v = 1.3 Mm/s
We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :

a and m are the acceleration and mass of the proton.



So, the acceleration of the proton is
. Hence, this is the required solution.
Statements A, B, and C are true.
Statement <em> D is false</em>.
Answer:
the length of stretched spring in cm is 22
Explanation:
given information:
spring length, x1 = 20 cm = 0.2 m
force, F = 100 N
the length of spring streches, x2 = 22 cm = 0.22 m
According to Hooke's law
F = - kΔx
k = F/*=(x2-x1)
= 100/(0.22 - 0.20)
= 5000 N/m
if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end
m = 10.2 kg
W = m g
= 10.2 x 9.8
= 99.96 N
F = - k Δx
Δx = F / k
= 99.96 / 5000
= 0.02
Δx = x2- x1
x2 = Δx + x1
= 0.20 + 0.02
= 0.22 m
= 22 cm