The wind is blowing with a force of 48 N against a kite that Taylor is flying, so that
the tail of the kite makes an angle of 24° to the ground.
I am just really confused on this, I attacted the solution would anyone mind explaining it? I have no idea on how to get this
Homework Statement
Click to enlarge
Homework Equations
K= 1/2 mv12 + 1/2mv22
p= m1v1 +m2v2
The Attempt at a Solution
I have pages and pages of trying to simplify/work backwards and nothing is working!
Ok Thanks so Much :) . I think it asking for it A, or else It would be asking for too much.
I have to hand this question in. I already found a cubic equation and found the force to be 1.47N... but when I use a tangent line to find the force I get 1.66N. Which one do you think would be more...
The slope of the line at any given point, is the derivative of u(x) at that point. But seeing how the slope is constantly changing how would I estimate the Force.
And what I was saying is that it looks like the graph can be modeled by a cubic equation. If I take 4 points off the graph, I can...
Homework Statement
Click to make larger
Homework Equations
-du/dx = F (in x direction)
The Attempt at a Solution
Would I have to use system equations to estimate a cubic equation, and just take the negative derivative of it? Cause it turns out to be a mess and I end up getting a...
[X, Y]= YX- XY
Thus [Y, X]= XY- YX
then -[Y, X]= -(XY- YX)
then -[Y, X]= -XY +YX
Like that? So i don't have to use matricies with variables, just use X and Y to represent a matrix?
Homework Statement
Homework Equations
The Attempt at a Solution
-(y, x) = -(YX-XY)
= XY-YX
Can I do this or would I have to define a matrix X= ( a b c d ) Y= ( e f g h)...
Part A: A block of mass M resting on a 18.8° slope is shown. The block has coefficients of friction μs=0.620 and μk=0.314 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.97 kg. What is the minimum mass M1 that will stick...