Velocity = distance / time = ( 2 * pi * r ) / t = 20.583 m/s
<span>x component = sine ( 32 ° ) * 20.583 = 10.91 m/s
hope this helps :)
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Integrating the velocity equation, we will see that the position equation is:

<h3>How to get the position equation of the particle?</h3>
Let the velocity of the particle is:

To get the position equation we just need to integrate the above equation:


Then:


Replacing that in our integral we get:


Where C is a constant of integration.
Now we remember that 
Then we have:

To find the value of C, we use the fact that f(0) = 0.

C = -1 / 3
Then the position function is:

Integrating the velocity equation, we will see that the position equation is:

To learn more about motion equations, refer to:
brainly.com/question/19365526
#SPJ4
Answer:
Level 4 to level 2
Explanation:
Electrons in an atom are contained in specific energy levels (1, 2, 3, and so on) having different distances from the nucleus. When light is emitted by electrons from one energy level to a lower level, level 4 to level 2 has the greatest energy.
Hence, the correct option is "Level 4 to level 2".
Answer:
the moment of inertia with the arms extended is Io and when the arms are lowered the moment
I₀/I > 1 ⇒ w > w₀
Explanation:
The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,
L₀ = L_f
I₀ w₀ = I w
w =
w₀
where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms
I₀ = I_cylinder + 2 m r²
where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.
If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be
I <I₀
I₀/I > 1 ⇒ w > w₀
therefore the angular velocity (rotations) must increase
in this way the skater can adjust his spin speed to the musician.
Answer:
q₃ = -4.81 nC
Explanation:
We can use the Gauss Law here:
∅ = q/∈₀
where,
∅ = Net Flux = - 216 N.m²/C
q = total charge enclosed inside sphere = ?
∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²
Therefore,
- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²
q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)
q = - 1.91 nC
So, the total charge will be sum of all three charges:
q = q₁ + q₂ + q₃
- 1.91 nC = 1.74 nC + 1.16 nC + q₃
q₃ = - 1.91 nC - 1.74 nC - 1.16 nC
<u>q₃ = -4.81 nC</u>