Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
Answer:
this pdf should help you out
Explanation:
The answer is emagination emagination
Answer:
Explanation:
Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.
As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.
