Answer:
233.3N
Explanation:
Given:
radius 'r'= 1.5m
mass 'm'= 165kg
time 't'=2 sec
angular speed 'ω'= 0.6 rev/s
the magnitude of a torque is given by:
τ = F . r = I . α
where, 'α' is the angular acceleration and 'I' is the rotational inertia
F= I . α/ r =>[ (
. m .
)(2π . ω/t) ]/r
F=[(
. 165 .
)(2π . 0.6/2)]/1.5
F= 233.3N
increase air resistance, which decreases gravitational acceleration
<span>So the question is what gases form the atmosphere of Uranus. So Uranus and Neptune are classified as "ice giants". They have a similar atmosphere to "gas giants" Saturn and jupiter and Neptunes atmosphere is primarily composed out of hidrogen and helyum. So the correct answer is a.</span>
You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final²=(V)initial²+2a(Δx)
Given:
a=0.500m/s²
Δx=4.75m
(V)intial=0m
(V)final= UNKNOWN
(V)final= 2.179m/s
Answer:
a)
, b)
,
, ![\phi = -0.473\pi](https://tex.z-dn.net/?f=%5Cphi%20%3D%20-0.473%5Cpi)
Explanation:
a) The system mass-spring is well described by the following equation of equilibrium:
![\Sigma F = k\cdot x - m\cdot g = m\cdot a](https://tex.z-dn.net/?f=%5CSigma%20F%20%3D%20k%5Ccdot%20x%20-%20m%5Ccdot%20g%20%3D%20m%5Ccdot%20a)
After some handling in physical and mathematical definition, the following non-homogeneous second-order linear differential equation:
The solution of this equation is:
![x (t) = A\cdot \cos \left(\sqrt{\frac{k}{m} } \cdot t + \phi\right)](https://tex.z-dn.net/?f=x%20%28t%29%20%3D%20A%5Ccdot%20%5Ccos%20%5Cleft%28%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%5Ccdot%20t%20%2B%20%5Cphi%5Cright%29)
The velocity function is:
![v(t) = \sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi \right)](https://tex.z-dn.net/?f=v%28t%29%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%5Ccdot%20A%5Ccdot%20%5Csin%20%5Cleft%28%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%5Ccdot%20t%20%2B%5Cphi%20%20%5Cright%29)
Initial conditions are:
![x(0\,s) = 0.25\,ft, v(0\,s) = -5\,\frac{ft}{s}](https://tex.z-dn.net/?f=x%280%5C%2Cs%29%20%3D%200.25%5C%2Cft%2C%20v%280%5C%2Cs%29%20%3D%20-5%5C%2C%5Cfrac%7Bft%7D%7Bs%7D)
Equations at
are:
![0.25\,ft = A\cdot \cos \phi\\-5\,\frac{ft}{s} =\sqrt{\frac{k}{m} }\cdot A\cdot \sin \phi](https://tex.z-dn.net/?f=0.25%5C%2Cft%20%3D%20%20A%5Ccdot%20%5Ccos%20%5Cphi%5C%5C-5%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%5Ccdot%20A%5Ccdot%20%5Csin%20%5Cphi)
The spring constant is:
![k = \frac{11\,lbf}{0.333\,ft}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B11%5C%2Clbf%7D%7B0.333%5C%2Cft%7D)
![k = 33\,\frac{lbf}{ft}](https://tex.z-dn.net/?f=k%20%3D%2033%5C%2C%5Cfrac%7Blbf%7D%7Bft%7D)
After some algebraic handling, amplitude and phase angle are found:
![\phi = -0.473\pi](https://tex.z-dn.net/?f=%5Cphi%20%3D%20-0.473%5Cpi)
![A = 2.845\,ft](https://tex.z-dn.net/?f=A%20%3D%202.845%5C%2Cft)
The position can be described by this function:
![x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)](https://tex.z-dn.net/?f=x%28t%29%20%3D%202.845%5Ccdot%20%5Ccos%20%5Cleft%281.732%5Ccdot%20t%20-0.473%5Cpi%20%5Cright%29)
b) The period of the motion is:
![T = \frac{2\pi}{\sqrt{\frac{k}{m} } }](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%7D)
![T = 3.628\,s](https://tex.z-dn.net/?f=T%20%3D%203.628%5C%2Cs)
The amplitude is:
![A = 2.845\,ft](https://tex.z-dn.net/?f=A%20%3D%202.845%5C%2Cft)
The phase of the motion is:
![\phi = -0.473\pi](https://tex.z-dn.net/?f=%5Cphi%20%3D%20-0.473%5Cpi)