Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
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Chacken Wangs
Answer:
In chemistry and atomic physics, an electron shell may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the "1 shell", followed by the "2 shell", then the "3 shell", and so on farther and farther from the nucleus.
Explanation:
