The slope of an object is given relative to an origin.
The condition of earths atmosphere at a given time and place is the weather.
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 velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.
Explanation:
Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:
Dividing the second equation by the first one, we obtain:
And, since 
, then:
It means that the velocity at the bottom of the ramp is 1.81m/s.
We could use this data, plus any of the two initial equations, to determine the acceleration:
So the acceleration is 3.30m/s^2.
Answer:Negatively charged particle called Free Electrons
Explanation:
Current is the flow of charged particles called Free electrons. Electrons are free to move from one atom to another and we call them a sea of de-localized electrons. In absence of any externally applied emf, these electrons are randomly moving but with the onset of emf, these electrons flow in a particular direction.
