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if I start out happy and go down the hill at a 15 degree angle, how long will it take me to reach rock bottom?

Martinez Jessica Rodriguez Carol Thomas Helen

Soul Limbo, very nice.

NOTE: Problem #2 - I should have used sine instead of tangent. Answer should be 107.22

You did the mistake at 40:17 while using a calculator. Use the right parenthesis when applying cosine or sine at any angle. That was the issue you are suffering there.

Problem 2: Setting our variables q1=q2 m=0.005kg g=9.8 R=distance between balls=0.5m*sin(5°)*2 Fe=kq1q2/r^2= kq^2/r^2 Tx=Tsin(5°) Ty=Tcos(5°) X-Direction ∑Fx=0 Fe-Tx=0 kq^2/r^2=Tsin(5°) Y-Direction ∑Fy=0 mg-Ty=0 mg=Tcos(5°) Solve Equations Solve for T T=mg/cos(5°) T=0.005*9.8/cos(5°) Solve for q k*q^2/r^2=T*sin(5°) q^2=T*r^2*sin(5°)/k q=√T*r^2*sin(5°)/k Enter all variables into equation q=√(((0.005*9.8/cos(5°))*sin(5°)^2*sin(5°))/9E9) q ≅ 6.015180789E-8

First Problem Seems to have an error. E2 =500,000,000; not 55,555,555.56 E1=9E9*.04/0.8^2=562,500,000 E2=9E9*.02/0.6^2=500,000,000 Add Vectors Rx=562,500,000*cos(180°) +500,000,000*cos(35°)=-152,923,977.9 Ry=562,500,000*sin(180°) +500,000,000*sin(35°)=286,788,218.2 Answers Angle = 180-arctan(Ry/Rx)=180+arctan(286,788,218.2/-152,923,977.9)=118.0678612° Rnet= √(Rx^2+Ry^2)= √(286,788,218.2^2-152,923,977.9^2)= 325,012,653.8

I don't know how many people read these comments but, On the last problem sin(Ry) = -19890000 Q2=2C∠180° Q3=3C∠(180+arctan(3/4))° (I use exact numbers to avoid rounding error) Q4=4C∠270° F12=11,250,000 =(9*10^9 (1)(2))/40^2 F13=10,800,000 =(9*10^9 (1)(3))/(40^2+30^2) F14= 40,000,000 =(9*10^9 (1)(4))/30^2 √(FX^2+FY^2)=FNET FX=-46,480,000 = F12X+F13X+F14X = 11,250,000cos(180°)+10,800,000cos((180+arctan(30/40))°)+ 40,000,000cos(270°) FY=-19,890,000 = F12Y+F13Y+F14Y = 11,250,000sin(180°)+10,800,000sin((180+arctan(30/40))°)+ 40,000,000sin(270°) Fnet=50,556,923=√46,480,000^2+19,890,000^2 Angle =203.167° = arctan(FY/FX)+180° = arctan(-19,890,000/-46,480,000)+180°

Thanks.