Explanation:
(D) i think there you go have a good day
Gravity increased the downward speed (or decreases the upward speed) by 9.8 m/s every second.
21.2/9.8 = 2.2 seconds
Answer:
Object should be placed at a distance, u = 7.8 cm
Given:
focal length of convex lens, F = 16.5 cm
magnification, m = 1.90
Solution:
Magnification of lens, m = -
where
u = object distance
v = image distance
Now,
1.90 = 
v = - 1.90u
To calculate the object distance, u by lens maker formula given by:
u = 7.8 cm
Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.
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]