Answer:
A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)
Explanation:
It's a Newton Meter Those two are multiplied to get a joule.
<span>Answer: Force = 81.6 N
Explanation:
According to Newton's Second law:
F = ma --- (1)
Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2
Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>
Potential and kinetic energy both decrease with the acorn's falling potential and kinetic energy.
The acorn's potential energy is at its peak when it reaches the top of the tree, yet its kinetic energy is zero (i.e., it is not accelerating).
The height of the ball reduces along with the potential energy as the acorn tumbles down the tree, but the kinetic energy rises (energy due to motion)
The height will be 0 and the kinetic and potential energy will be zero at the ground. This demonstrates that as an item falls, both potential and kinetic energy are lost.
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Answer: If x + y = a, xxy = b and x • a = 1 , then 2 (a~ - l)a- a x b (b2 ... xy-plane, then the vector in the same plane having projections