Answer:
8 s
Explanation:
Damping factor, F is given by
where Q is quality factor
Also, we know that the settling time, T is given by
where W is the resonant frequency.and substituting the first equation here we get
Since it takes 2% of final value to settle and considering 100 rad/s for resonant frequency and taking quality factor of 100 then
Three units deformations observed in these bands.
<h3>What forces do a rubber band encounter?</h3>
Elastic force is the force that permits some materials to regain their former shape after being stretched or crushed. Thus, a stretched rubber band is subject to an elastic force.
The rubber band experiment uses a straightforward rubber band to show entropic force and a refrigeration cycle. The rubber band experiment involves stretching and then releasing a rubber band while measuring its temperature.
Always acting in the opposite direction of motion is friction. This indicates that if friction is there, it cancels out some of the force driving the motion (if the object is being accelerated). This results in a decreased acceleration and a smaller net force.
learn more about Elastic force refer
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How many meters per second was it traveling
Answer:
r₂ = 4 r
Explanation:
For this exercise let's use Newton's second law with the magnetic force
F = q v x B
bold letters indicate vectors, the magnitude of this expression is
F = q v B sin θ
in this case we assume that the angle is 90º between the speed and the magnetic field.
If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal
a = v² / r
let's use Newton's second law
F = ma
q v B = m v² / r
r =
Let's apply this expression to our case.
Proton 1
r = \frac{qB_1}{mv_1}
Proton 2
r₂ = 
in the exercise indicate some relationships between the two protons
* v₁ = 2 v₂
v₂ = v₁ / 2
* B₂ = 2B₁
we substitute
r₂ =
r₂ = 4
r₂ = 4 r
Answer:
500 Pa
Explanation:
Convert given units to SI:
100 dyne = 0.001 N
0.02 cm² = 2×10⁻⁶ m²
Pressure = force / area
P = 0.001 N / (2×10⁻⁶ m²)
P = 500 Pa