Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:

where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:

, so the work done by the brakes to stop the bus is

And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
You are asked to give the answer in <span>g/cm3. So without knowing any single formulae you can just divide grams by cm3.
</span>

= 4.5 g/cm3
The highest trophic level has the least available energy in kilojoules.
Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.
Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.
At the highest trophic level, the the least available energy in kilojoules in this food web is found.
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