Answer:
The distance by which the spring stretches is 1.48 cm.
Explanation:
Given that,
An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s, ![\omega=25.7\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D25.7%5C%20rad%2Fs)
We know that angular frequency in SHM is given by :
![\omega=\sqrt{\dfrac{k}{m}} \\\\\omega^2=\dfrac{k}{m}\\\\\dfrac{k}{m}=(25.7)^2.............(1)](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D%20%5C%5C%5C%5C%5Comega%5E2%3D%5Cdfrac%7Bk%7D%7Bm%7D%5C%5C%5C%5C%5Cdfrac%7Bk%7D%7Bm%7D%3D%2825.7%29%5E2.............%281%29)
When the object is allowed to hang stationary from it, the force due to spring is balanced by its weight. Such that :
![kd=mg\\\\d=\dfrac{g}{(k/m)}](https://tex.z-dn.net/?f=kd%3Dmg%5C%5C%5C%5Cd%3D%5Cdfrac%7Bg%7D%7B%28k%2Fm%29%7D)
From equation (1) :
![d=\dfrac{9.8}{(25.7)^2}\\\\d=0.0148\ m\\\\d=1.48\ cm](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7B9.8%7D%7B%2825.7%29%5E2%7D%5C%5C%5C%5Cd%3D0.0148%5C%20m%5C%5C%5C%5Cd%3D1.48%5C%20cm)
So, the distance by which the spring stretches is 1.48 cm.
Convection Zone: Just beneath the photosphere, and extending inward to about 0.7 Rsun, is the convection zone. Energy generated in the core of the Sun moves outward through this layer by a boiling motion in which hot plasma rises, releases some of its energy, cools, and then sinks again.
Answer:
A) 2.5 * 10^22 molecules
B) 3.33 * 10^-9 m
C) 1134 miles/hour
D) average spacing = 2.1 m , New mass = 17 kg
Explanation:
Given data :
diameter of atom = 10^-10 m
volume of air = 1 liter
molar masses : 28 g/mol of N2, 32 g/mol of O2
atmospheric pressure = 10^5 Pa
attached below is the detailed solution
Velocity is your answer (: