Exam review

on the last couple of days we've been doing the questions from the exam review, we started on thursday, we did Waves and nature of science on thursday. On friday we continued working on it and we went over mechanics, mrs K had students come up on board and do the questions, Field is the only unit left and Mrs K said we will go over it on monday. Don't forget exam is on tusday from one to four oclock, study hard and good luck on your exams

Electric Fields

January 12, 2010

Today in class we read the Electric Fields handout, highlighted a few notations, and elaborated on new diagrams.

• In a diagram of a positive point charge, the arrows, or electric field line, they point outwards of the point charge. In a negative point charge, they point inwards towards the point charge.
• The density of the electric field line patterns designate the strength of the electric field.
• When the anode meets the cathode, or positive point charge meets the negative point charge, the electric field lines of the positive point charge are directed towards the negative point charge.
• In between the point charges, no field exists.
• We were assigned Transparency 20-1 and Transparency 21-1, as well as finishing up the previous days assignment: The study guide of chapter 21, fill in the blanks.

Accuracy and Precision

Accuracy and precision: In the fields of science, engineering, industry and statistics, the accuracy of a measurement system is the degree of closeness of measurements of a quantity to its actual value.

The precision of a measurement system, also called reproducibility or repeatability, is the degree to which repeated measurements under unchanged conditions show the same results. Although the two words can be synonymous in colloquial use, they are deliberately contrasted in the context of the scientific method.

A measurement system can be accurate but not precise, precise but not accurate, neither, or both. For example, if an experiment contains a systematic error, then increasing the sample size generally increases precision but does not improve accuracy. Eliminating the systematic error improves accuracy but does not change precision.

Accuracy versus precision; the target analogy

The analogy used here to explain the difference between accuracy and precision is the target comparison. In this analogy, repeated measurements are compared to arrows that are shot at a target. Accuracy describes the closeness of arrows to the bullseye at the target center. Arrows that strike closer to the bullseye are considered more accurate. The closer a system's measurements to the accepted value, the more accurate the system is considered to be.

High accuracy, but low precision

High precision, but low accuracy

To continue the analogy, if a large number of arrows are shot, precision would be the size of the arrow cluster. (When only one arrow is shot, precision is the size of the cluster one would expect if this were repeated many times under the same conditions.) When all arrows are grouped tightly together, the cluster is considered precise since they all struck close to the same spot, even if not necessarily near the bullseye. The measurements are precise, though not necessarily accurate.
However, it is not possible to reliably achieve accuracy in individual measurements without precision—if the arrows are not grouped close to one another, they cannot all be close to the bullseye. (Their average position might be an accurate estimation of the bullseye, but the individual arrows are inaccurate.) See also circular error probable for application of precision to the science of ballistics

yesterday we did a lab on "Experiment- petermination of the earth's gravitation field strength ."
so if you didn't do it, ask mr.vicent for help :)

Dynamics Review Answers

Here are the answers for the Dynamics Review:

Dynamics review answers

Have a good weekend.

Mr. V

Dynamics Test Review

We went over the Dynamics Test Review in class today and Mr. Vincent will be posting the answers to the review sometime on Sunday. For fun, I will answer some random review questions and show you how to get the answer.

Practice Problems

5. A grapefruit of mass 0.5 kg is accelerated at 10 m/s². How large is the force acting on it?

Given:

(m) Mass = 0.5 kg

(a) Acceleration = 10 m/s²

(F) =?

F = ma

F = (0.5 kg) (10 m/2²)

F = 5 kg•m/s²

F = +5 N

8. A box is pushed across the floor with a force of 120 N, but experiences a frictional force of 55 N. If the acceleration that results is 1.1 m/s², what is the mass of the box?

Since friction is a resistance to force, the 55 N is subtracted from the applied force to give us the net force.

F_net = F_Applied + F_Friction

F_net = 120N + (-55N)

F_net = 65 N

We now have force, are given acceleration (1.1 m /s²), and can solve the equation for mass.

F = ma

m = F / a

m = 65 N / 1.1 m/s²

m = 59 kg (2 significant digits)

Conceptual Review

7. A 600-kg car and an 1800-kg truck go from zero to 100 km/h in 10 seconds flat. Calculate the acceleration of each vehicle, including units with your answer. Do they have the same acceleration? Is the same force required to cause this amount of acceleration in each vehicle? Explain using Newton's laws of motion.

First thing to do is convert 100 km/h to meters per second. This is done by dividing by 3.6.

100 km/h

------------ = +27.8 m/s

3.6

Both vehicles reach this speed in 10 seconds, so their acceleration must be the same:

+27.8 m/s ÷ 10s = 2.78 m/s²

Now that we have both vehicles accelerations (2.78 m/s²) and masses, we can solve for the force both vehicles are exerting.

Do they have the same acceleration? Yes, 2.78 m/s

Is the same force required to cause this amount of acceleration in each vehicle? Explain using Newton's laws of motion.

The truck requires more force to accelerate at that rate because it has more mass.

The explanation would come from solving the force for each vehicle using Newton's 2nd law, Force = mass x acceleration:

Truck (1800-kg):

F = ma

F = (1800 kg) (2.78 m/s²)

F = 5004 N

Car (600-kg):

F = ma

F = (600 kg) (2.78 m/s²)

F = 1668 N

The truck needs almost 3x more force than the car to move at 2.78 m/s².

December 7th/2010

we have a hand out called "Free-body Diagrams," in this booklet we are learning about free body diagrams (duh!!) ;D, anyways we did the first 8 examples in class and the rest we are suppost to do on our own. it's only 4 questions and it's towards the end of the booklet. also i'm pretty sure the teacher will do a homework check tomarrow (i think).