Is there a certain context to this question so that I can answer it better?
Since
<span>Vf^2 = 2*a*S </span>
<span>Given S=3.6m, thus </span>
<span>a = Vf^2/(2*3.6) </span>
<span>a = Vf^2/7.2 </span>
<span>Let d be the distance along the slope at which the velocity is 0.5Vf, then </span>
<span>(0.5Vf)^2 = 2*a*d </span>
<span>or </span>
<span>d = (0.5*Vf)^2/(2*a) </span>
<span>with a = Vf^2/7.2, we have </span>
<span>d = 0.9 m</span>
It shows that acceleration of particle M is directly proportional to its displacement and its direction is opposite to that of displament. Thus particle M performs simple harmonic motion but M is projection of particle performing U.C.M. hence S.H.M. is projection of U.C.M. along a diameter, of circle.
<em>Time = (the set distance) / (the object's traveling speed)</em>
Answer: F = 2.1 x 10^-4N
Explanation: Question is incomplete.
The complete question is; A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth’s magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east.
Given parameters; l = 2.5m, I = 1.5A, B = 0.55 guass = 0.55 x 10^-4 Tesla , theta = 90 (from West to East), F = ?
F = BILsin(theta)
F = 0.55 x 10^-4 x 1.5 x 2.5 x sin 90
F = 2.1 x 10^-4 N.
According to right hand rule, it's direction is upward.
