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Honors Physics problem that is confusing me! Please help me before midnight?

Honors Physics problem that is confusing me! Please help me before midnight? Topic: The problem solving process steps
June 26, 2019 / By Savanna
Question: A computer is reading data from a rotating CD-ROM. At a point that is 0.0245 m from the center of the disk, the centripetal acceleration is 106 m/s2. What is the centripetal acceleration at a point that is 0.0723 m from the center of the disc? The number of significant digits is set to 3; the tolerance is +/-1 in the 3rd significant digit Number:___________________ Units:_________ PLEASE SHOW ME THE ANSWER AND PLEASE EXPLAIN STEP BY STEP THE FORMULAS AND STEPS YOU USED TO SOLVE THIS. I HAVE A HONORS PHYSICS TEST IN A WEEK AND CIRCULAR MOTION CONFUSES ME.
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Best Answers: Honors Physics problem that is confusing me! Please help me before midnight?

Noreen Noreen | 3 days ago
Well, centripetal acceleration of a point in uniform circular motion is: ac = v^2 / R where v is the tangential velocity at that point. Note that the tangential velocity of a point in uniform circular motion is equal to the rotational velocity, w, of the object times the distance, r, from the center: v = w * r where w will be constant for all points on the disk. The problem gives you a point on the disk and asks you to find the properties of another point. To solve this problem, you need to use equation #1 to get the tangential velocity of the given point, then use equation #2 to find the CDs rotational velocity. Since the rotational velocity is constant for all points, you can use equation #2 to find the tangential velocity of the second point, then equation #1 to find the centripetal acceleration on the second point: POINT 1: 106 m/s^2 = v^2 / (0.0245 m) v = (106 * 0.0245)^(1/2) = 1.611 m/s 1.611 = w * 0.0245 w = 1.611 / 0.0245 = 65.776 rad/s POINT 2: v = 65.776 * 0.0723 = 4.755 m/s ac = (4.755)^2 / 0.0723 = 312.808 m/s^2 Thus, the centripetal acceleration at the second point is 312 m/s^2 (in significant figures) (HINT: I did the steps and calculations individually, but if you perform the same process using abstract algebra instead, you can get the following equation: ac2 = ac1 * (r2 / r1) or ac2 = (v1^2 / r1) * (r2 / r1) You may want to remember this equation for your test, as it will save you time)
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We found more questions related to the topic: The problem solving process steps


Noreen Originally Answered: Physics problem.?
Let B & G are the two points from where the boy and girl starts .As they are approaching each other their relative velocity will be the sum of their individual velocities. The total distance to be covered is the distance between them. For their rendezvous the time taken will be=total dist./rel.velocity t = 60/ (2+3)= 12 sec In this duration the dist. covered by boy= 12 x 2 = 24m ................. '........... " . . . . " . . . . . . . girl= 12 x 3 = 36m Hence, the meeting point will be 24m from the boy & . . . . . . . . . . . . . . . . . . . . . . . . . . . 36m from the girl. ======================================...
Noreen Originally Answered: Physics problem.?
the distance between them is closing at 5 m/s. (assuming they are traveling toward each other) at this rate it will take 12 seconds for them to meet. Now we know t=12 is 12 seconds, the girl travels 36m(that is 12 s*3m/s). The boy travels 24m (12m*2m/s).

Mable Mable
No, traveling down a slope doesn't necessarily mean you're accelerating. If the up-slope forces (due to friction) match the down-slope forces (due to gravity), then your net force will be zero, so your acceleration will be zero. At some point, she must have started moving, and must have accelerated from zero to her current speed. But the problem isn't concerned with that phase of her trip. It CAN happen in real life, under the right circumstances. It would depend on the slope angle, the coefficient of kinetic friction, and the force due to air resistance all having just the right values.
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Kelley Kelley
a = ω²R Since the angular speed is the same at all point of the rotating disk, a is proportional to R: a2/a1 = R2/R1 a2 = a1(R2/R1) a2 = 106×(0.0723/0.0245) = 312.8 m/s²
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Kelley Originally Answered: Is it possible to solve this physics problem? If so, can you show me how?
I did b) first and then a) b) vf^2 - vi^2 = 2ax At t1, the maximum height is .225 m At t2, the maximum height is .9 To find out its increasing rate, in other words, how to find the constant of proportionality, multuply what you originally found with a proportionality x = .9/.225 = 4. Exactly a) t = sqrt(2h/x) t1 = .21428 t2 = .4287 x = 2

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