(1100 ft/sec) x (1 mile / 5280 ft) x (3600 sec / 1 hr)
= (1100 x 3600 / 5280) mile/hour
= 750 miles per hour .
Answer:
![R_f=\dfrac{4}{5}R](https://tex.z-dn.net/?f=R_f%3D%5Cdfrac%7B4%7D%7B5%7DR)
Explanation:
Here, the kinetic and potential energy is conserved
![K_f-K_i=U_f-U_i\\\Rightarrow \dfrac{1}{2}mv_i^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{r}\\\Rightarrow \dfrac{1}{2}m(\dfrac{1}{2}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R}](https://tex.z-dn.net/?f=K_f-K_i%3DU_f-U_i%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7Dmv_i%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7Br%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7Dm%28%5Cdfrac%7B1%7D%7B2%7Dv_f%29%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR%7D)
For the second case
![K_f-K_i=U_f-U_i\\\Rightarrow \dfrac{1}{2}mv_i^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{r}\\\Rightarrow \dfrac{1}{2}m(\dfrac{1}{4}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R_f}](https://tex.z-dn.net/?f=K_f-K_i%3DU_f-U_i%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7Dmv_i%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7Br%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7Dm%28%5Cdfrac%7B1%7D%7B4%7Dv_f%29%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR_f%7D)
Subtract the equations we get
![\dfrac{1}{2}m(\dfrac{1}{2}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R}\\ \dfrac{1}{2}m(\dfrac{1}{4}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R_f}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dm%28%5Cdfrac%7B1%7D%7B2%7Dv_f%29%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR%7D%5C%5C%20%5Cdfrac%7B1%7D%7B2%7Dm%28%5Cdfrac%7B1%7D%7B4%7Dv_f%29%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR_f%7D)
![\\\Rightarrow \dfrac{1}{4}mv_f^2-mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R}\\ \dfrac{1}{16}mv_f^2-mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R_f}\\\Rightarrow -\dfrac{3}{4}mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R}\\ -\dfrac{15}{16}}mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R_f}](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B4%7Dmv_f%5E2-mv_f%5E2%3D%5Cdfrac%7B1%7D%7B2%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR%7D%5C%5C%20%5Cdfrac%7B1%7D%7B16%7Dmv_f%5E2-mv_f%5E2%3D%5Cdfrac%7B1%7D%7B2%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR_f%7D%5C%5C%5CRightarrow%20-%5Cdfrac%7B3%7D%7B4%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B2%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR%7D%5C%5C%20-%5Cdfrac%7B15%7D%7B16%7D%7Dmv_f%5E2%3D%5Cdfrac%7B1%7D%7B2%5Cpi%20%5Cepsilon%7D%5Cdfrac%7Bq_1q_2%7D%7BR_f%7D)
Divide the equations
![\dfrac{\dfrac{3}{4}}{\dfrac{15}{16}}=\dfrac{R_f}{R}\\\Rightarrow R_f=\dfrac{4}{5}R](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B3%7D%7B4%7D%7D%7B%5Cdfrac%7B15%7D%7B16%7D%7D%3D%5Cdfrac%7BR_f%7D%7BR%7D%5C%5C%5CRightarrow%20R_f%3D%5Cdfrac%7B4%7D%7B5%7DR)
![R_f=\dfrac{4}{5}R](https://tex.z-dn.net/?f=R_f%3D%5Cdfrac%7B4%7D%7B5%7DR)
Yes because it helps with my work
I think that the girl has greater tangential acceleration because she is closer to the center and the acceleration is greater there.
It also get triple in magnitude as compared to it's initial value, as they are directly proportional.