Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
1 answer:
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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)
The concepts required to solve this problem are those related to the conservation of the angular momentum and the moment of inertia of the disk. We will begin by calculating the moment of inertia of the disc, then the moment of inertia of the disc after the two two blocks hits and sticks to the edges of the turn table. In the end we will apply the conservation theorem.
The radius is given as,
When a block falls from above and sticks to the turn table, the moment inertia of the turntable increases.
Since two blocks are stick to the turn table, the total final moment of inertia of the turntable is the sum moment of inertias of individual turntable, and two blocks.
The moment of inertia of each block is
Total moment of inertia of two block is
The final moment of inertia of the turn table is
From the conservation of the angular momentum, the initial angular momentum of the system is equal to final angular momentum of the system,
Rearrange the equation we have that
The magnitude of the turntable's angular velocity is 66.67rpm