Dr.Neutron

1st: Split the two angle from the ground to the spire atop into two right triangles 2nd: take tangent of the two angles and multiply them --> tan(17)time tan(20.9)=-6.71696 3rd: 9.28x10^2=928 4th: multiply 928time-6.71696= -6233.3 5th: htan (17)-htan(20.9...

1st: equation1: 5A+6B=60C equation2: 30C+D=A 2nd: divide equation1 by2-->((5A+6B=60C)/2)=5/2A+6/2B=30C now reduce itto 5/2A+3B=30C 3rd: subtract30C on both side of the equal sign on equation2 and get: D=A-30C 4th:subtitute for equation1 into the Nnew equa...

V1: v1x=18.69m/s;v1y=14.08m/s V2: v2x=0m/s; v2y=8.20m/s V3: v3x=-13.80m/s:v3y=0m/s V4: v4x= -5.08m/s;v4y=-20.38m/s

1st: To find R for part a), f1x,f1y,f2x, f2y, f3x, f3y. 2nd: f1x=0,f1y= -1.81, f2x=-.866,f2y=.500,f3x=1.00,f3y=1.73 3rd: To finf R, add all the fx together on one side and fy together on the other side.-->R=(0+-.866+1.00)i+(-1.81+.500+1.73)j 4th: answer o...

1st: Part a):To find R, first find f1x,f1y,f2x; f2y, f3x, f3y. 2nd: f1x is 0i because it is at the origin,f1y is-1.81j because that was giving as the arrow point toward the negative direction. For f2x, f was giving as 1f take it and multiply with cos(30)...

Rx=(sin(30)-.2*cos(30))=.32679F1

Ftotal/m-->(0i+226.75j)N/23.11kg=0i+9.81j m/s^2

a)=1.5 b)rP=1.22AU b) rA=1.70AU c)d=.308AU

a=1.5 b rP=1.22AU and rA=1.70AU c=0.308AU

First: m1=4kg, m2=2kg, T? Second: m2gsin(55)-T= m2a + m1gsin(35)+T=m1a ---------------- =(m2gsin(55)-m1gsin(35)/(m1+m2) Third: a=(2kg*9.81m/s^2sin(55)-4kg*9.81m/s^2sin(35))/(2kg+4kg a=1.109m/s^2 Fourth: plug into equation m1--> T=4kg*9.81m/s^2sin(35)+4kg*...

First: m1=4kg, m2=2kg, T? Second: m2gsin(55)-T= m2a + m1gsin(35)+T=m1a ---------------- =(m2gsin(55)-m1gsin(35)/(m1+m2) Third: a=(2kg*9.81m/s^2sin(55)-4kg*9.81m/s^2sin(35))/(2kg+4kg a=1.109m/s^2 Fourth: plug into equation m1--> T=4kg*9.81m/s^2sin(35)+4kg*...

1st: convert 30gto kg=.03kg 2nd: .03kg*9.81m/s^2=.294kg*m/s^2 3rd: answer .(94kg*m/s^2)÷(.5kg/s)=.589m/s