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
<h2>Answer: protons and neutrons.
</h2>
The atomic nuclei of almost all elements consist of protons and neutrons.
The nucleus of an atom has very small dimensions. However, it <u>occupies its central part and concentrates more than 99% of its total mass.
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It is in the nucleus that the protons (positive charge) and neutrons (neutral charge) are found.
Sound waves....................................
