Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
When driver see the child standing on road his speed is 20 m/s
So here at that instant his reaction time is 0.80 s
He will cover a total distance given by product of speed and time
now after this he will apply brakes with acceleration a = 7 m/s^2
so the distance covered before it stop is given by
so the total distance covered by it
<em>so it will cover a total distance of 44.6 m</em>
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
It can never be shorter than a component - magnitude of avector is the square root of the sum of the components squared, and a square function never produces a negative number. However, it can be the same size as its component, if that component is the only one
Answer:
The final velocity of the vehicle is 10.39 m/s.
Explanation:
Given;
acceleration of the vehicle, a = 2.7 m/s²
distance moved by the vehicle, d = 20 m
The final velocity of the vehicle is calculated using the following kinematic equation;
v² = u² + 2ah
v² = 0 + 2 x 2.7 x 20
v² = 108
v = √108
v = 10.39 m/s
Therefore, the final velocity of the vehicle is 10.39 m/s.
