I think I understand. But if ##\frac {\partial x}{\partial u}=\frac{1}{\frac {\partial u}{\partial x}}##, then doesn't ##\frac {\partial x}{\partial u}\frac {\partial u}{\partial x}=1##? Then how do you resolve the contradiction that ##1 = \frac{\partial x}{\partial x} = \frac{\partial...
I got that ##{x_u}{y_v}-{x_y}{y_u}=####\frac{1}{\frac{1}{{x_u}{y_v}}-\frac{1}{{y_u}{x_v}}}##. But this implies that ##{x_u}{x_v}{y_u}{y_v}=-1## and I don't see how that is true?
Thank you so much for your answer! So I gather that my error was thinking of r as a scalar instead of a vector.
So does that mean that ##\frac{\iint r^2\,dr\,d\theta}{\iint_A dA}## is the average distance of a point in the sector from the origin, but the average displacement of a point in the...
If I use cartesian co-ordinates, I get:
##\bar{x}=\frac{1}{A}\iint x\, dA=\frac{1}{A} \iint r^2\cos\theta\, dr\, d\theta= \frac{2a\sin\theta}{3\theta}##
##\bar{y}=\frac{1}{A}\iint y\, dA=\frac{1}{A}\iint r^2\sin\theta\, dr\, d\theta= \frac{2a(1-\cos\theta)}{3\theta}##
But if I use polar...
I learned that there are two different definitions for the coefficient of restitution: e = final relative velocity / initial relative velocity and e = √(final KE/initial KE). However, I don't understand how these two definitions will always give the same value.
If one particle with mass m...
So the forces acting on the mast are the tensions of AC, AB and AO and the weight.
The equations are:
Pcos20=10cos70+Tcos40
Tsin40=20+10sin70+Psin20
P=72.3 N?
θ is the angle between P and the horizontal and φ is the angle between Q and the horizontal.
I got those equations by resolving the forces vertically and horziontally (as the system is in equilibrium so net force = 0)
Ok. I think that |P| = |tension in AC| and |Q| = |tension in BC| (because the length of the strings has to stay equal). Then I get the equations:
Psinθ-Psin20°+Tsin40°-20-10sin70°+10sinφ=0
Pcosθ+Pcos20°-Τcos40°-10cos70°-10cosφ=0.
I think maybe Tsin40°=20?
Homework Statement
On a model ship, the mast OC has length 50 cm and weight 20 Newtons. The mast is hinged to the deck at O, so that it can rotate in the vertical plane of the ship. Small smooth rings are fixed at points A and B on the deck in this plane such that AO=OB=50 cm. Threads from C...