The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Weak winds that blow for short periods of time with a short fetch.
The total work done is 5980 Joules and the power expended is 57 Watts.
<h3>What is work done?</h3>
The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;
Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules
Height of the center of mass of chain = 25.9 / 2 = 12.95 m
Work done by the chain Wc;
Wc = 12.95 * 19.3 * 9.8 = 2450 Joules
Total work = 3530 + 2450 = 5980 Joules
Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts
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Mr. Hitch taught us about sedimentary, metamorphic, and igneous rocks. He described how they were formed, what they contain, and showed us samples of each. He is a good geologist.
The missing word and answer is: geologist.
