The answer is A.
Sy = 1650 x sin30.5 = 837.4 m toward south
Sx = 1650 x cos30.5 = 1421.7 m toward east
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]
Answer:0.253Joules
Explanation:
First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.
F = ke where;
F is the force
k is spring constant = 34N/m
e is the extension = 0.12m
F = 34× 0.12 = 4.08N
To get work done,
Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.
Work done = Force × Distance
Since F = 4.08m, distance = 0.062m
Work done = 4.08 × 0.062
Work done = 0.253Joules
Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules
Answer:
r = 9.92 mm
Explanation:
Given that,
Mass of oil drop, 
It acquires 2 surplus electrons, q = +2e 
Potential difference, V = 620 V
Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.
We need to find the distance between the plates.
At equilibrium,
mg = qE
Since, E = V/r (r is distance between plates)

So, the distance between the plates is 9.92 mm.