Answer: How to solve for FX and FY?
to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.
Explanation:
Answer:
A. when the mass has a displacement of zero
Explanation:
The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

where
k is the spring constant
x is the displacement of the mass with respect to the equilibrium position of the spring
m is the mass
v is the velocity of the mass
Since the total energy E must remain constant, we can notice the following:
- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum
- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0
Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.
Answer:
the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.
and/or
Kinematics is the study of motion of a system of bodies without directly considering the forces or potential fields affecting the motion. In other words, kinematics examines how the momentum and energy are shared among interacting bodies.
Answer:
im pretty sure about C
Explanation:
a switch acts a resistor that can be turned on and off
Also it's the same for Automotive purposes
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA =
/ ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁