<p>A particle moves in a circle in such a way that the x ¬and y coordinates of its motion are given in meters as functions of time t in seconds by:
x = 5cos(3t) y = 5 sin (3t)</p>
<p>What is the period of revolution of the particle?
(A) 1/3 s (B) 3 s (C) 2pi/3 s (D) 3pi/2 s (E) 6pi s</p>
<p>Which of the following is true of the speed of the particle?
(A) It is always equal to 5 m/s. (B) It is always equal to 15 m/s.
(C) It oscillates between 0 and 5 m/s. (D) It oscillates between 0 and 15 m/s.
(E) It oscillates between 5 and 15 m/s.</p>
<p>Explain please!!
Answer: C,B</p>
<p>Also, since when did Physics C have parametrics???</p>
<ol>
<li><p>The period is always 2π/coefficient of the sinusoidal variable, so it is C</p></li>
<li><p>v = √((dy/dt)^2 + (dx/dt)^2)
are u sure that answer B isn’t 15√2?</p></li>
</ol>
<p>for the first one, since sin and cos have a period of 2pi, you want to make whatever is inside equal 2pi, therefore you get 3t=2pi and then divide both sides by 3 and get 2pi/3</p>
<p>for the second one, it is easiest just to take the derivative, then use the pythagorean theorem to get the total velocity and just plug in points</p>
<p>EDIT: i just got owned, see buddy’s post above :p</p>
<p>@buddy its definitely just 15, if you graph it, you get a straight line v(t)=15</p>
<p>also sqrt((-15sin(3t))^2+(15cos(3t))^2)) = sqrt(225((sin3t)^2+(cos3t)^2)) and since (sin3t)^2+(cos3t)^2=1 (trig identity) you get just sqrt(225) = 15</p>
<p>It goes in a circle, so if it starts at t=0 when it’s at the point (5, 0) it’s going to end up back at (5, 0) after a whole period. This is going to happen when 3t = 2pi (because cos 0 = cos 2pi and sin 0 = sin 2pi), so t=2pi/3 s.</p>
<h1>2</h1>
<p>First find v<em>x and v</em>y by differentiating x and y, so we get v<em>x = -15sin(3t) and v</em>y = 15cos(3t). They want speed. Speed = sqrt[ (v<em>x)^2 + (v</em>y)^2 ], you then factor out a 225, leaving you with sqrt[ 225(sin^2(3t) + cos^2(3t))], and after using some trigonometry magic we’re left with speed = sqrt(225) = 15.</p>
<p>EDIT: I am too slow.</p>
<p>This is the type of question where if I saw it on a calculus test, I’d get the answer immediately, but if I saw it on a physics test, I pause for about a half second then realize what I should be doing.</p>
<p>The basic idea is that the electric potential difference (delta V) on either side of the ammeter is zero, coming from both the left, and the right.</p>
<p>Suppose that a hole is drilled through the center of Earth to the other side along its axis. A small object of mass m is dropped from rest into the hole at the surface of Earth, as shown above. If Earth is assumed to be a solid sphere of mass M and radius R, and friction is assumed to be negligible, correct expressions for the kinetic energy of the mass as it passes the center include which of the following?</p>
<p>I. 1/2<em>MgR
II. 1/2</em>mgR
III. (GmM)/(2R)</p>
<p>(a) I only
(b) II only
(c) III only
(d) I and III only
(e) II and III only</p>
<p>I thought it would be GmM/R, which isn’t even there…</p>
<p>Two blocks are pushed along a horizontal frictionless surface by a force of 20 newtons to the right, as shown above. The force that the 2 kilogram block exerts on the 3 kilogram block is
(A) 8 newtons to the left (B) 8 newtons to the right (C) 10 newtons to the left
(D) 12 newtons to the right (E) 20 newtons to the left</p>
<p>Answer A
Someone please explain?</p>
<p>A rock is lifted for a certain time by a force F that is greater in magnitude than the rock’s weight W. The change in kinetic energy of the rock during this time is equal to the
(A) work done by the net force (F -W) (B) work done by F alone
(C) work done by W alone (D) difference in the momentum of the rock before and after this time (E) difference in the potential energy of the rock before and after this time</p>
<p>Thank you so much for the answer. I understand everything you said except for the relationship between the two currents. How did you know that I<em>1 = 2I</em>2?</p>
<p>@math Since the total force applied to the objects together (5kg) produces an acceleration of 4m/s^2, you then find that Fnet (for block 1) = 20 - Fblock2 = 3kg*4m/s^2 which means that Fblock = 8N</p>
<p>For the second question you know that change in kinetic energy is always equal to the work done, therefore, the change in kinetic energy is the work done by the net force, which is F-W</p>
<p>^That question requires that you know that for circular orbits mechanical energy = 1/2 potential = -K. Since U of small mass = mgR, ME =1/2mgR. Choice 2.</p>
<p>Also, K =1/2 mv^2. V= root Gm/R (another thing you should know). If you plug v in, K = 1/2mMG/R. Choice 3.</p>
<p>@ fluffy sorry if this is a stupid question but how is that circular orbit (i honestly dont know, im not doubting your answer (cause its correct lol :P))</p>
<p>and in the second one isnt the m in the v equation capitalized? like v=sqrt(GM/R)</p>
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