y component
y^2 = v^2 - x^2
y^2 = 49^2 - 28^2
y = 40 lb
=====
angle
x = 180 + arctan ( 40 / 28)
x = 235 counterclockwise from +x axis
=====
Comment before rate in case any doubt, will reply for sure.. goodluck
I need help with 4-7
thank you
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Three charges (q1 =
7 μC, q2 = -5.6 μC, and q3 = 2.9μC) are
located at the vertices of an equilateral triangle with side d =
8.5 cm as shown
A charge q4 = 2.9 μC is now added as shown.
3) 43 91 A charge q4-2.9 HC is now added as shown What is F2,x, the x-component of the new net force on q2? N Submit You currently have 7 submissions for this question. Only 10 submission are...
please answer all portions of the question, thank you(:
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Three charges (q1 = 7 μC, q2 = -5.6 μC,
and q3 = 2.9μC) are located at the vertices of an
equilateral triangle with side d = 8.5 cm as shown
A charge q4 = 2.9 μC is now added as shown.
What is F2,x, the x-component of the new net force on
q2?
3) 43 91 A charge q4-2.9 HC is now added as shown What is F2,x, the x-component of the new net force on q2? N Submit...
What is the answer to this question and how did you do
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