Answer:
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
The damping constant can be found by
Finally, the mass m can be found by
Where g is approximately 32 ft/s²
Therefore, the required differential equation is
The initial position is
The initial velocity is
I believe the answer is C, 3
Answer:
D
Explanation:
When the spring is wound, then it gathers potential energy in the form of tension energy. As it slowly unwinds, the potential energy is converted to kinetic energy of the hands' movements of the clock. This energy is channeled through the use of cogs/gears in the clock.
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
