Explanation:
<u>Formula:</u>
![velocity = (d \div t)](https://tex.z-dn.net/?f=velocity%20%3D%20%28d%20%5Cdiv%20t%29)
<u>d = distance given</u>
<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>
<u>Substitute the given values into the formula for velocity</u><u>:</u>
![v = 8 \div 4](https://tex.z-dn.net/?f=v%20%3D%208%20%5Cdiv%204)
velocity is shortened for v.
8 (distance) divided by 4 (time) equals the velocity.
<u>Solve:</u>
![2 = 8 \div 4](https://tex.z-dn.net/?f=2%20%3D%208%20%5Cdiv%204)
The velocity of the toy car equals: B. 2 m/s.
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The motion of a falling whirligig is different to that of a falling paper ball due to spinning.
<h3>Type of motion performed by whirligig and falling paper ball </h3>
The motion of a falling whirligig is different from the motion of a falling paper ball because the paper ball falls on the ground without spinning while on the other hand, the whirligig falls on the ground along with spinning.
The falling whirligig performs two motion i.e. one is falling on the ground and the other is spinning during motion whereas paper ball performs one motion i.e. motion in the air towards the ground so we can conclude that the motion of a falling whirligig is different than of a falling paper ball.
Learn more about motion here: brainly.com/question/453639
Answer:
Along path BC of the Otto cycle, heat transfer Qh into the gas occurs at constant volume, causing a further increase in pressure and temperature. This process corresponds to burning fuel in an internal combustion engine, and takes place so rapidly that the volume is nearly constant.