Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes
Now after applying brakes the distance covered by the car before it stops is given by kinematics equation
here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have
So total distance after which car will stop is
So car will not stop before the intersection as it is at distance 20 m
Answer:
B) 350 kg m/s
Explanation:
momentum or p is given by the equation p= mxv
We have the mass and velocity so we can use the equation directly
p= 72kg x 4.9 m/s
p= 352.8 kg m/s
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e (
)
with this expression we can find the closest approach distance (r)
