Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0
Explanation:
False
electron and proton attract each other
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:

<em>Where</em>:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:

<em>Where:</em>
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
Now, we can find the work:
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
I think that the oceanic water particles mainly move in circles greater in the oceans surface because of how big the waves can be and how wind and air impact the motion. The water particles move more on the surface because of the other factors that impact it such as people, wind, air, etc...
Answer:
(a) A = 1 mm
(b) 
(c) ![a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D606.4%20m%2Fs%5E%7B2%7D%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EDistance%20moved%20back%20and%20forth%20%3D%202%20mm%20%3C%2Fp%3E%3Cp%3EFrequency%2C%20f%20%3D%20124%20Hz%3C%2Fp%3E%3Cp%3ESo%2C%20amplitude%20is%20the%20half%20of%20the%20distance%20traveled%20back%20and%20forth.%20%3C%2Fp%3E%3Cp%3E%28a%29%20So%2C%3Cstrong%3E%20amplitude%2C%20A%20%3D%201%20mm%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%28b%29%20Angular%20frequency%2C%20%CF%89%20%3D%202%20%CF%80%20f%20%3D%202%20x%203.14%20x%20124%20%3D%20778.72%20rad%2Fs%20%3C%2Fp%3E%3Cp%3EThe%20formula%20for%20the%20maximum%20speed%20is%20given%20by%20%3C%2Fp%3E%3Cp%3E%5Btex%5DV_%7Bmax%7D%3D%5Comega%20%5Ctimes%20A)


(c) The formula for the maximum acceleration is given by


[tex]a_{max}=606.4 m/s^{2}/tex]