<span>The word is "pitch", which is exactly that: How "high" or "low" a sound is.</span>
As stated in the statement, we will apply energy conservation to solve this problem.
From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as
![\Delta KE = \Delta PE](https://tex.z-dn.net/?f=%5CDelta%20KE%20%3D%20%5CDelta%20PE)
![\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv_f%5E2-%5Cfrac%7B1%7D%7B2%7D%20mv_0%5E2%20%3D%20mgh_2-mgh_1)
Where,
m = mass
= initial and final velocity
g = Gravity
h = height
As the mass is tHe same and the final height is zero we have that the expression is now:
![\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dv_f%5E2-%5Cfrac%7B1%7D%7B2%7D%20v_0%5E2%20%3D%20gh_2)
![\frac{1}{2} (v_f^2-v_0^2) = gh_2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%28v_f%5E2-v_0%5E2%29%20%3D%20gh_2)
![(v_f^2-v_0^2) = 2gh_2](https://tex.z-dn.net/?f=%28v_f%5E2-v_0%5E2%29%20%3D%202gh_2)
![v_f = \sqrt{2gh_2+v_0^2}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Csqrt%7B2gh_2%2Bv_0%5E2%7D)
![v_f = \sqrt{2(9.8)(23.5)+13.6^2}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Csqrt%7B2%289.8%29%2823.5%29%2B13.6%5E2%7D)
![v_f = 25.4m/s](https://tex.z-dn.net/?f=v_f%20%3D%2025.4m%2Fs)
Answer:
Air resistance
Answer B is correct
Explanation:
The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.
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Answer:
B = 9.16 10⁻² T
Explanation:
The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel
q v B = q E
v = E / B
B = E / v
Let's calculate
B = 4.4 10⁵ / 4.8 10⁶
B = 9.16 10⁻² T
Answer:
The correct answer is Option A.