Answer:
The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.
Explanation:
In this ideal engine, none of these events would happen due to the nature of the efficiency.
We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.
n=W, total/(E, available).
However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.
Acceleration = (change in speed) / (time for the change)
= (49 m/s) / (5 seconds)
= (49 / 5) m/s / s
= 9.8 m/s²
The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.
Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal
The force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.
<h3>How to calculate force?</h3>
The force needed to push an object can be calculated by multiplying the mass of the object by its acceleration as follows:
Force = mass × acceleration
According to this question, a car of mass 800 kg has an acceleration of 2.0 ms−². The force is calculated as follows:
Force = 800kg × 2m/s²
Force = 1600N
Therefore, the force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.
Learn more about force at: brainly.com/question/13191643
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Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
