<span>F=ma = 3000x2m/sec^2 =6000 newtons. </span>
Current will be
![I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz](https://tex.z-dn.net/?f=%20I%3D%5Cdfrac%7BV_%7Brms%7D%7D%7BZ%7D%3D%5Cdfrac%7BV_%7Brms%7D%7D%7B%5Csqrt%7B%20R%5E%7B2%7D%2B%28X_%7BC%7D-X_%7BL%7D%29%5E%7B2%7D%7D%7D%5C%5Cwhere~X_%7Bc%7D%3D%5Cdfrac%7B1%7D%7Bj.%5Comega%20.C%7D~and~X_%7BL%7D%3Dj.%5Comega.L~where~%5Comega%3D2.%5Cpi%20f~and~f%3D60Hz)
now just pluf in the values and Voila..
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Answer: F = 102141N
Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by
F = m.a
whose units are:
[F] = N
[m] = kg
[a] = m/s²
Jenny's car, at the moment of the break, had acceleration:
![a=\frac{\Delta v}{\Delta t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D)
![a=\frac{11}{0.14}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B11%7D%7B0.14%7D)
a = 78.57 m/s²
Then, Force is
F = 1300*78.57
F = 102141 N
<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>
distance traveled by a uniformly accelerated bike is given as
![d = \frac{v_f + v_i}{2} (t)](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7Bv_f%20%2B%20v_i%7D%7B2%7D%20%28t%29)
here we know that
![v_f = 25 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2025%20m%2Fs)
![v_i = 15 m/s](https://tex.z-dn.net/?f=v_i%20%3D%2015%20m%2Fs)
![t = 15 s](https://tex.z-dn.net/?f=t%20%3D%2015%20s)
now we will have from above equation
![d = \frac{15 + 25}{2} (15)](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B15%20%2B%2025%7D%7B2%7D%20%2815%29)
![d = 20 (15) = 300 m](https://tex.z-dn.net/?f=d%20%3D%2020%20%2815%29%20%3D%20300%20m)
so it will cover the total distance of 300 m