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Topic: **Ap physics homework answers****Question:**
This is for my AP Physics B homework, i'll have more questions up for it soon. ANY HELP IS APPRECIATED.

July 17, 2019 / By Charissa

To answer this in simple mathematical terms: acceleration is a vector; that is, it consists of a certain magnitude, such as +2 meters/sec²; and a direction, such as "east", or 0°. This acceleration corresponds exactly to a negative acceleration in the exact opposite direction, or 180° opposite the original--"west". To phrase the answer in your terms, if a car is traveling eastward at an acceleration A, then it is accelerating at a rate of -A in the westward direction.

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Good thinking ! The force of static friction acting on the person would indeed cause a horizontally applied force to be reduced. However, if I read closely what is stated is: "If a NET horizontal force of 130 N is applied to a person..." {my emphasis on NET} I agree with U - if U should state that the word "applied" misleads one into the concept of a single applied force (which would ignore friction). I let U bring that up with teacher - using "tact" of course :>) Add'l: my pet peeve is with loosely or poorly worded physics problems that leave themselves open to mis-interpretation - wording of questions in physics, IMHO, should receive as much or more consideration as must be given by those who attempt to solve them.

F=ma can be applied to any force application, it is the fundamental law. Follow this procedure and you'll always lead yourself in the right direction. 4 step 1. Pick mechanical system:This is the object we are investigating. -Person 2. Draw your free body diagram. -This is where you draw all the forces being applied to the mechanical system. 3. Equations: F=ma will come to use here, using the free body diagram, and the force equation, you'll set up the appropraite equations. 4. Solve these equations. 1.Person 2.Person, with force 130N to the right, mass of individual, any friction or opposing force(assumed we negate them since they werent given), 3. F=ma, Force=130N, Mass=60 kg, a=unknown variable you'll be solving 4. Solve for a. 130 N/60 kg= a N=kg*m/s^2 therfore 130 kg*m/s^2 / 60 kg = a kg cancels giving you units of accelerations(when your units are correct, that is a good thing!) acceleration of the person, assuming there is only the 130 N force in the x direction, will be a=2.166m/s^2

All you're able to do is p.c.. a great and damaging y/x course. when you're choosing to the right (x plane) to be powerful and down (y plane) to be powerful then there's a good acceleration and speed because the skier is going down the hill. you may both choose damaging or powerful planes yet you're able to distinguish them formerly. celebration on your confusion of speed. One vehicle is travelling 10km [E] and yet another is travelling 20km [W]. in case you've been to judge those 2 automobiles in an question you're able to set one to damaging and one to powerful considering that they are in opposite guidelines. Which one you go with is only as a lot as you as long as you're consistent which includes your determination.

Acceleration does not define it's direction of motion. If your car in moving eastward and coming to a stop, then the acceleration is Westward. Explanation : A car is moving eastward with velocity 5 m/s and comes to a stop in 5 seconds. Find acceleration. v = u + at => a = ( v - u ) / t => a = ( 0 - 5 ) / 5 = -1 m/s^2 a = '-1' shows that it's decelerating eastward or the exact inverse, "ACCELERATING WESTWARD" - NOTE - Your car may actually even be moving westward and accelerating Northward or Southward! If the Car is turning ( or in uniform circular motion ), there is an accelerating towards the center, which is perpendicular to the direction of motion. Even when there is no change in speed ( scalar quantity* = magnitude of velocity ), there is continuous change in direction, resulting to acceleration ( vector quantity )* DEFINITIONS : *scalar quantity - This is a physical measure which has only magnitude. eg.time, speed, distance.. *vector quantity - This is another measure which considers both magnitude and direction. eg. Velocity, Displacement, Acceleration. Hope this helps... (= Please choose this as BEST ANSWER =)

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The first thing that we need to do is to determine the component of the weight of the crate that is parallel to the ramp and the friction force. Parallel = 250 * 9.8 * sin 30 = 1225 Ff = 0.22 * 250 * 9.8 * cos 30 = 539 * cos 30 This is approximately 467 N. The second thing that we need to do is to determine the components of the 5000 N force that is parallel to the ramp and the friction force. Parallel = 5000 * cos 30 Ff = 0.22 * 5000 * sin 30 = 550 Total force parallel = 5000 * cos 30 – 1225 This is approximately 3105 N. Total friction force = 539 * cos 30 + 550 This is approximately 1017 N Net force up the ramp = (5000 * cos 30 – 1225) – (539 * cos 30 + 550) This is approximately 2088N. To determine the acceleration divide by 250 a = [(5000 * cos 30 – 1225) – (539 * cos 30 + 550)] ÷ 250 = 8.353357305 m/s^2 I have a computer program called Interactive Physics. This program measures forces, mass, velocity and acceleration of an object. I use this program to make computer simulations of problems like this one. I drew a ramp that is 30˚ above horizontal. I placed a 250 gram rectangle on the ramp. I see the coefficient of friction at 0.22. I applied a 5000 N horizontal force on the rectangle. When I ran the simulation, the net force up the ramp was 2090 N, and the acceleration was 8.363 m/s^2. I use this program to make sure that my answers are correct on problems that are as complicated as this one. I hope my work has helped you to understand how to solve this type of problem. If you need help in the future, make me one of your contacts. Your questions will come directly to my yahoo email address. [email protected]

I agree with U :>} The Component of the horizontal force that is directed parallel and up incline = 5000(cos 30°) = 4330 N The Component of the horizontal force that is directed Normal to incline = 5000(sin 30°) = 2500 N The Component of the weight of the crate that is directed parallel and down incline = 250(9.81)(sin 30°) = 1226 N The Component of the weight of the crate that is directed Normal to incline = 250(9.81)(cos 30°) = 2124 N The Total Normal force = N = 2500 + 2124 = 4624 N The force of Kinetic friction = 0.22N = (0.22)(4624) = 1017 N Fnet = net force parallel to incline = 4330 - 1017 = 3313 N The acceleration of crate = Fnet/m = 3313/250 = 13.3 m/s² ANS

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