Adhesion is when forces of the molecules are more strongly drawn to each other than the molecules in other substances true or fa
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Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)
Answer:
368224.29906 m/s
Explanation:
M = Mass of Polonium nucleus = 214 u
V = Velocity of nucleus
m = Mass of Helium nucleus = 4 u
v = Velocity of alpha particle =
In this system the momentum is conserved
The recoil speed of the nucleus that remains after the decay is 368224.29906 m/s