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.
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<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:
The effect of this rotation on the person on the surface is the sky is moving, like an apparent diurnal motion.
Explanation:
<em>For an observer at a fixed position on Earth, the rotation of earth makes it appear as if the sky is revolving around the earth. In other words, if a person is standing for long enough in a field at night, it looks like the sky is moving, not the person. This motion is called "apparent diurnal motion." </em>
<em>Diurnal means having to do with a day, in a sense of a 24- hour period.</em>
Answer:
false
Explanation:
I had the same answer for this question on my schoolwork
:) HOPE THIS HELPS
We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is
Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)
Part A ) Replacing our values,
Part B ) To find the torque we apply the equation as follow,
