<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s</span><span>
i hope I'm right</span>
Answer:
Explanation:
Check attachment for solution
Answer:
as the period decreases, the frequency and energy of the wave increase
Explanation:
Electromagnetic waves are oscillations of the electric and magnetic fields, described by maxwell's equations, the speed of the wave is called the speed of light
c = λ f
E = E cos (kx - wt)
Angular velocity is related to frequency and period.
w = 2π f = 2π / T
Let's analyze what happens when the wave period decreases, angular velocity and frequency increase.
This increase in frequency is reflected with the Planck equation in wave energy
E = h f
Therefore the wave carries more energy and can lead to stronger interactions with matter.
In summary, as the period decreases, the frequency and energy of the wave increase
Answer:
A = [kg]
B = [m/s²]
Explanation:
E = ½ Av² + Bmx
Substitute the units:
[J] = ½ A [m/s]² + B [kg] [m]
A Joule written in base units is:
1 J = 1 Nm = 1 kg m²/s²
Each term must have the same units.
[kg m²/s²] = A [m/s]²
[kg m²/s²] = A [m²/s²]
A = [kg]
[kg m²/s²] = B [kg] [m]
B = [m/s²]
