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JulsSmile [24]
3 years ago
10

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2

?
Physics
1 answer:
Shtirlitz [24]3 years ago
4 0
99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
 = (0²-27.5²)/(2x50.0)
 =-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
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Explanation:

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How would the model change as the atom forms bonds? The third shell would have eight electrons after the atom gains seven electr
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The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

Explanation:

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3 years ago
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Explanation:

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Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

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