Answer:
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The projectile has a height <em>h</em> at time <em>t</em> given by
<em>h</em> = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for <em>t</em> when <em>h</em> = 0 :
0 = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²
0 = 1/2 <em>t</em> (28.0 m/s - <em>g t</em> )
1/2 <em>t</em> = 0 <u>or</u> 28.0 m/s - <em>g</em> <em>t</em> = 0
The first equation says <em>t</em> = 0, which refers to the moment the gun is first fired, so we ignore that solution. We're left with
28.0 m/s - <em>g t</em> = 0
<em>t</em> = (28.0 m/s) / <em>g</em>
<em>t</em> = (28.0 m/s) / (9.80 m/s²)
<em>t</em> ≈ 2.86 s
20.9m = 1s
286.33m = 13.7s
To answer this you would multiply both sides by the amount of seconds she ran. The answer however is that she ran as far as 286.33m.
It’s either B or A, I hope this helps! I tried!
Answer:
Explanation:
the cross-sectional area of the gold nucleus / the cross-sectional area of the atom = 2.6 x 10⁻⁷
value of radius of gold atom = 1.4 x 10⁻¹¹
cross sectional area = π x (1.4 x 10⁻¹¹ )²
= 6.15 x 10⁻²² Putting this value in the ratio above
the cross-sectional area of the gold nucleus / 6.15 x 10⁻²² = 2.6 x 10⁻⁷
the cross-sectional area of the gold nucleus = 16 x 10⁻²⁹
radius of nucleus R
π R² = 16 x 10⁻²⁹
R² = 5.1 X 10⁻²⁹
R = 7.14 x 10⁻¹⁵ .