Explanation:
q= n e
6 × 10^-11 = n (1.6 × 10^-19)
n = 6×10^-11 / 1.6 × 10^-19
n= 3.75 × 10⁸ electrons
When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.
Now, given that:
- the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
- after hitting the wall, the final velocity will be (v) = 0 km/h
Assumptions:
- Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
- That the car's acceleration is also constant.
∴
For a motion under constant acceleration, we can apply the kinematic equation:

where;
v = final velocity
u = initial velocity
a = acceleration
s = distance
From the above equation, making acceleration (a) the subject of the formula:


The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.
since 1 km/h = 0.2778 m/s
100 km/h = 27.78 m/s


a = - 385.86 m/s²
Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;
v = u + at
where;
v = 0
-u = at


t = 0.07 seconds
An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.
Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.
Learn more about the kinematic equation here:
brainly.com/question/11298125?referrer=searchResults
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
Answer:
9 Brainly hahaha ............huh