Answer:
Explanation:
We first identify the elements of this simple harmonic motion:
The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.
The angular frequency ω can be calculated with the formula:
Where k is the spring constant and m is the mass of the particle.
Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:
Finally, the equation of the motion of the system is:
or
Answer:
Im doing edge too rn-
Explanation:
Can you tell me the drop down options
Answer:
- 273.77 rad/s^2
Explanation:
fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps
f = 0
ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s
ω = 2 π f = 0
θ = 46 revolutions = 46 x 2π radian = 288.88 radian
Let α be the angular acceleration of the centrifuge
Use third equation of motion for rotational motion
α = - 273.77 rad/s^2
Answer:
q = 2,95 10-6 C
Explanation:
The magnetic force on a particle is described by the equation
F = q v x B
Where bold indicate vectors
Let's make the vector product
vxB =![\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C-3%264%2612%5C%5C0%260%26-0.13%5Cend%7Barray%7D%5Cright%5D)
v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]
vx B = 1.20 106 [0.52 i ^ - 0.39j ^]
As they give us the force module, let's use Pythagoras' theorem,
|v xB | =1.20 10⁶ √( 0.52² + 0.39²)
|v x B| = 1.20 10⁶ 0.65
v xB = 0.78 10⁶
Let's replace and calculate
2.30 = q 0.78 10⁶
q = 2.3 / 0.78 106
q = 2,95 10-6 C
