HomeWork

**Friday, November 13 Homework**
//Chapter 6 - Pg 111 "Multiple Choice" 1-10// //Chapter 10 - Pg 192 "Multiple Choice" 1-10//
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**Wednesday, November 11 Homework**
//Chapter 6 - Pg 108 "Review Questions"// 12-15 Possible Controlled Experiment Ideas
 * **Question 12:** **Can a machine multiply input force? Input distance? Input energy? (If your three answers are the same, seek help, because the last question is especially important.)** a) Yes, a good example would be the lever. b) Yes, if the machine is designed so that more force is required but the distance multiplies. An example would be moving the fulcrum of a lever closer to the end being pushed down (the empty end). c) No, the energy put into the machine will come out the same amount, but probably in a converted state; energy cannot be created nor destroyed.
 * **Question 13:** **If a machine multiplies force by a factor of four, what other quantity is diminished, and how much?** The distance will decreased, it'll be one fourth of it's original quantity before the force gets multiplied. The input and output energy has to be balanced as energy cannot be created or destroyed. In order for it to be balanced, the distance quantity has to be divided.
 * **Question 14:** **What is the efficiency of a machine that miraculously converts all input energy to useful output energy?** 100%, nothing goes to waste, all the energy that comes out is converted and used.
 * **Question 15: Is a machine physically possible that has an efficiency greater than 100%?** Energy cannot be created nor destroyed, no. A efficiency higher than 100% means that more output energy comes out than the input energy that comes in. This goes against the law stated above, energy cannot be created nor destroyed, but it can be converted. It's like putting in 100 N of force but the work achieved is equal to 1000 N, the energy can't come from mid air (except for gravity).
 * 1) Measure loudness of speaker when pitch of music is altered
 * 2) Measure loudness of speaker when speed of music is altered
 * 3) Generator Vs Motor?

**Monday, November 9 Homework**
//Chapter 6 - Pg 108 "Review Questions"// 6-10
 * **Question 6:** **A car is lifted a certain distance in a service station and therefore has potential energy relative to the floor. If it were lifted twice as high, how much potential energy would it have?** Twice as much potential energy, if the wire snaps, and the car drops, twice as much potential energy will impact the collision force.
 * **Question 7:** **Two cars are lifted to the same elevation in a service station. If one car is twice as massive as the other, how do their potential energies compare?** The car with twice as much mass will have bigger potential energy because weight is involved in this case,
 * **Question 8:** **How many joules of potential energy does a 1N book gain when it is elevated 4m? When it is elevated 8m?** a) 4 joules because 4 multiplied by 1 is 4 (force multiplied by distance. b) 8 joules because 8 multiplied by 1 is 8 (force multiplied by distance).
 * **Question 9: A moving car has kinetic energy. If it speeds up until it is going four times as fast, how much kinetic energy does it have in comparison?** The kinetic energy is 16 times as much, because in the equation, speed is squared. If the unit for the car's starting rate is one, and four times is four, one squared is one while four squared is 16. 16 is 16 times larger than 1; the kinetic energy is about 16 times larger.
 * **Question 10: Compared to some original speed, how much work must the brakes of a car supply to stop a car moving three times as fast? How will the stopping distance compare?** The same force is required to stop a moving car moving at three times regular speed; force provided by the brakes, it just requires more work. The car moves three times as fast, to fine kinetic energy, the speed has to be squared, 3 squared is 9; the car's stopping distance will be 9 times further as there is more kinetic energy in the car moving 3 times normal speed. It takes more work to stop this car using the same energy (friction) provided by the brakes.

**Wednesday, November 4 Homework**
//Chapter 6 - Pg 108 "Review Questions"// 1-5
 * **Question 1:** **Cite an example in which force is exerted on an object without doing work on the object.** Pushing a box against a wall.
 * **Question 2:** **Which requires more work - lifting a 50-kg sack a vertical distance of 2m or lifting a 25-kg sack a vertical distance of 4m?** The same
 * **Question 3:** **If both sacks in question two are lifted at the same time, how does the power required for each compare? How about for the case in which the lighter sack is move its distance in half the time?** a) Power stays the same because same amount of work is done, and at the same time. b) Power for lighter sack is two times more because same amount of work done in half the time.
 * **Question 4: Exactly what is that a body having energy is capable of doing?** A body that has energy is capable of doing work.
 * **Question 5: What are two main forms of mechanical energy?** Potential energy and kinetic energy.

**Monday, November 2 Homework**
//Chapter 10 - Pg 191 "Think and Explain" 3, 6, and 13//
 * **Question 3:** In an electrical force, electrons move.
 * **Question 6:** 25 N
 * **Question 13:** A motor uses electrical energy (converting it into usable energy, a generator converts energy into electrical energy, or something else.

Wednesday, October 25 Homework
//Chapter 10 - Pg 175 - Pg 189// ~ 5 Essential Questions
 * 1) How did magnetic fields form?
 * 2) How are atoms made?
 * 3) Are all conversion of energy based on this concept?
 * 4) How does magnetic fields work in relation to other laws of physics?
 * 5) How severe can a magnet damage electronics?

Friday, October 23 Homework
//Chapter 9 - Pg 173 "Think and Compare" 1-4, "Think and Solve" 2-5//
 * Think and Compare**
 * **Question 1: Compare the circuits shown in the following figure. Rank them according to the brightness of the bulbs, from brightest to dimmest.** A, B, C (they all look the same?)
 * **Question 2: The bulbs shown in the following figure are identical. An ammeter is placed in different branches as shown. Rank the current readings in the ammeter from greatest to least.** A, B, C (they all look the same?)
 * **Question 3: All bulbs are identical in the circuits shown in the following figure. An ammeter is connected next to the battery as shown. Rank the current readings in the ammeter, from greatest to least.** A, B, C
 * **Question 4: All bulbs shown in the following figure are identical. (a) Rank the current through each, from greatest to least. (b) Rank the voltages across A, B, and C, from greatest to least.** a) The middle branch with one bulb, the top branch with two bulbs, the battery. b) The middle branch with one bulb tied with the top branch with two bulbs, along with the battery.
 * Think and Solve**
 * **Question 2: A radio speaker has a resistance of 8 ohms. When 12 V is applied across the speaker, show that the current flowing through the speaker is 1.5 A.** Current = Voltage/resistance. This means that Current = 12/8, 12/8 can be simplified to 3/2, which is equivalent to 1.5 A.
 * **Question 3: If the circuit shown in "think and Explain" Question 19 is a 6-V battery, and the voltage across lamp A is 2 V, show that the voltage across lamp B is 4 V.** The amount of electricity that goes out of the battery has to come back in. Voltage is like the pressure, lamp B is between Lamb A and the battery. The battery is 6-V while Lamp A is 2-V, because lamp B is like the intersection between these two, it's voltage is the difference of the two. 2 subtracted from 6 is 4, so the voltage across lamp B is 4-V.
 * **Question 4: Rearrange the equation current = voltage/resistance to express resistance in terms of current and voltage. Then solve the following: A certain device in a 120-V circuit has a current rating of 20 A. Show that the resistance of the device is 6 ohms.** Resistance = Voltage/Current. Resistance = 120/20, Resistance = 6. To verify, 20 = 120/6, 20 = 20.
 * **Question 5: Using the definition of power (current x voltage), show that the current that flows in a 6-W clock radio operating in a common household of 0.05 A.** 6 = 0.05 x Voltage. 6/0.05 = Voltage.Voltage = 120-V. If the voltage of this clock radio is 1120-V, then the current is 0.05 A.

Friday, October 23 Homework
//Chapter 9 - Pg 172 "Think and Explain" 7-12, and 19//
 * **Question 7:** **Why is a good conductor of electricity also a good conductor of heat?** When electrons pass through a conductor, they will collide with each other, this is a cause of head. If a conductor is good with conducting electricity, then it means electrons can flow through it easily, when electrons flow through things, it converts energy into heat, making the conductor also good for heat.
 * **Question 8: What happens to the brightness of light emitted by a lamp when the current that flows in it increases?** The brightness of the light will also increase because electrons that pass through these lamps are responsible for making it light, the bigger the current, the more electrons will flow through the lamp and resulting in an increase in brightness. Basically, when electrons flow throw the circuit with a lamp, it will light the lamp, and if the flow increases, then there will be more energy to power up the lamp.
 * **Question 9: Your tutor tells you that an ampere and a volt really measure the same thing, and the different terms only make a simple concept seem confusing. Why should you consider getting a different tutor?** Ampere is the rate in which electrons flow per second (measured in coulombs). Voltage is the electrical pressure of an object. An unbalance in voltage and a circuit allows a current. Ampere is what measures this current, if the tutor does not know the difference between this terms, then it's safe to say that the tutor does not know much about the concept of electricity. Getting a new tutor would be highly recommended to prevent myself from learning the wrong idea of electricity.
 * **Question 10: In which of the circuits shown in the following figure does a current exist to light the light bulbs?** Figure 5, there's a full circuit where the wire touches the metal part of the light bulb and each end of the battery is in the circuit. This allows a current of electrons to go through from the + of the battery, to the light bulb, light it up as it passes, then out the other end, then back to the - part of the battery.
 * **Question 11: Does more current flow out of a battery than into it? Does more current flow into a light bulb than out of it?** No, the current flow out of the batter and into it stays the same unless there is a short circuit. The electrons go from parts of high pressure (high voltage), to low pressure (low voltage, - part of battery), a circuit is where the flow goes from one end to another. Same goes with the light bulb, in a circuit, with the absence of a short circuit, the electrons will go from one point to another. The electrons goes through the light bulb and while it's passing through the filament, it charges up the particles around it along it's way. But the flow does not change, if not there won't be a closed circuit.
 * **Question 12: Only a small percentage of the electric energy going into a common light bulb is transformed into light. What happens to the rest?** The rest stays as it is and goes back to the original, they stay as electric energy rather than transforming into light energy (used to light the light bulb). These electrons continue the circuit all the way to the other end without losing much of it's charge.
 * **Question 19: In the circuit shown, how does the brightness of each identical light bulb compare? Which light bulb draws the most current? What will happen with bulb A is unscrewed? If bulb C is unscrewed?** Bulb C should be brighter than A and B because it's on it's own circuit and it's the only resistance in the circuit. Bulb C draws the most current because it's circuit has the least amount of resistance (only one light bulb in the current's path). If Bulb A is unscrewed, then Bulb B will not light because the circuit breaks at Bulb A, creating an incomplete circuit. If Bulb C is unscrewed, it won't impact the other bulbs because it's on it's own path. Bulb A and B will continue to light.

Wednesday, October 17 Homework
//Chapter 9 - Pg 154 - Pg 169// ~ 5 Essential Questions This isn't really a scientific question but doesn't the whole concept of science feel like Magic? The basis of it, like gravity, what makes gravity? How does it exist? How was it created? I'm not saying it's magic, I'm just wondering how it all started... Even with the big bang theory, having all these concepts working together seems magnificent. Cells, particles, atoms, molecules, physics, motion, biology...etc... All working alongside each other.
 * 1) Would one be better off not knowing the dangers of electrical shocks? (paranoia vs safety)
 * 2) How did scientists first discover the concept of particles and all the units they've already discovered?
 * 3) What causes an electron to be an electron? (gives it it's negative charge)
 * 4) Why do opposites attract? What causes them to do so?
 * 5) If a charge we produce by rubbing objects together is much smaller compared to a coulomb, is there still a way to produce electrical current without harming the environment?

Wednesday, September 17 Homework
//Chapter 4 - Pg 72 "Think and Explain" 10,11,12,16-18, "Multiple Choice" 1-10//
 * **Question 10:** **Suppose two carts, one twice as massive as the other, fly apart when the compressed spring that joins them is released. How fast does the heavier cart roll compared to the lighter cart?** When rolling, the heavier car will accelerate twice as slow as the lighter cart, because acceleration = force/mass, so when the mass is 1, the acceleration will the same as the net force. When the mass is 2 (heavier cart), the acceleration is divided by 2 as the net force is divided by 2 and acceleration and force are directly proportional. When the heavier cart accelerates slower than the lighter cart, and both carts start at a still position, then the lighter cart will travel faster than the heavier cart.
 * **Question 11: If you exert a horizontal force of 200N to slide a crate across a factory at a constant velocity, how much friction does the floor exert on the crate? Is the force of friction equal and oppositely directed to your 200N push? If the force of friction isn't the reaction force to your push, what is?** If I need 200N to slide a crate across at a constant velocity, it means I've already skipped the acceleration part, therefore, the 200N I exert is to zero-out the force from friction. This means the floor exerts 200N friction on the crate, this force is equal and opposite to my 200N push as it's going the opposite direction in which I am exerting the 200N to. The friction isn't the reaction of my push, it was already there to start with, I needed to exert force to get the crate moving at a constant velocity. The reaction is the crate pushing against my hand. The crate will be exerting 200N to my hand but the reaction on me is not enough to push my backwards.
 * **Question 12: If a massive truck and a small sports car have a head-on collision, on which vehicle is the impact force greater?** **Which vehicle experiences the greater acceleration? Explain your answers.** The impact force on both vehicle will be the same, for ever action, there is an equal and opposite reaction.Even if the truck exerted a larger force, the car will react to it and the same amount of force will be directed back at the truck, only with different effects. The small sports car will experience a greater acceleration because acceleration = net force/mass. The net force of both cars will be the same, they will most likely have the same amount of friction and because acceleration is directly proportional to the net force, when the net force is divided by a smaller number, the acceleration will then be bigger than dividing by a bigger number. In this case, the small sports car is the smaller number, so it will experience greater acceleration.
 * **Question 16: A stone is shown at rest on the ground. a) The vector shows the weight of the stone. Complete the vector diagram by showing another vector that results in zero net force on the stone. b) What is the conventional name of the vector you have drawn?** a) See notebook b) The conventional name of the vector I have drawn would be the support force (of the table). If the weight and gravity pushing down is greater than the support force of the table, then the table would break or collapse.
 * **Question 17: Here a stone is suspended at rest by a string. a) Draw force vectors for all the forces that act on the stone. b) Should your vectors have a zero resultant? c) Why or why not?** a) See notebook b) Yes c) Mainly because the rock isn't accelerating, when an object isn't accelerating, the net force is 0. For the net force to be 0, the forces acting on the stone should cancel out each other, for example, if the downward arrow was 10N, then the upwards arrow should be 10N if those are the only two fores that impacted on the stone. The downward vector will be canceled out by the upward arrow and have a zero resultant; no acceleration or deceleration occurs with a zero resultant.
 * **Question 18: Here the same stone is being accelerated vertically upward. a) Draw force vectors to some suitable scale showing the relative forces acting on the stone. b) Which is the longer vector, and why?** a) See notebook b) The upwards pull vector should be the longer vector (the one pointing up), because the stone is accelerating upwards. In order for an object to accelerate, the net force has to be non-zero. In this case, the non-zero net force is going in the upwards direction because the stone is accelerating upwards. The force up has to be greater than the force down in order for it to dominate and for the stone to accelerate upwards.
 * +++ Multiple Choice (pg. 73) +++**
 * **1)** D
 * **2)** B
 * **3)** B
 * **4)** C
 * **5)** B/C
 * **6)** A
 * **7)** A
 * **8)** C
 * **9)** A
 * **10)** D

Tuesday, September 8 Homework
//Chapter 4 - Pg 57-6//4
 * **How do these concepts relate to the motion of your car?** The simplest answer to this question would be that my car moves with wheels and the wheels push backwards against the ground while the ground pushes the car forward as a reaction. But there's more to it than this, Newton's third law is that //for every action, there's an equal and opposite reaction//, The wheels of my car is powered by the motor which has has it's tip covered with tape and stuck into the hole area of the car's wheel. When the motor spins (action), the wheel will spin (reaction), they will both exert the same force, but the wheel is touching against the floor and it's attached to an object with a decent mass. The mass it's attached to will lower it's acceleration, so will the friction that goes the opposite direction that the car is moving. When the wheel spins as a reaction due to the force exerted by the motor, it pushes backwards against the floor and the floor pushes the car forwards. If I look closely at the bottom part of the wheel, the wheel is actually rotating the opposite direction compared to the car that's attached to it. If the wheel spun clockwise, the car will move forwards while the bottom part of the wheel (the part that is touching the ground), is moving backwards; pushing against the ground. Then friction has to be subtracted by the applied force and mass has to be put into the equation. This will get me the acceleration, and because my car is powered by a motor, it will keep accelerating until it reaches a point where the force exerted by the motor is balanced out and the reaction //affect// will be as fast as it can be. To summarize it all, the force exerted by the spinning motor caused a reaction from the wheel in which the motor is attached too; which is to spin. The motor spins the wheel, the wheel spins the motor but also pushes backwards on the ground simultaneously.

Friday, September 4 Homework
//Chapter 3 - Pg 44-52//
 * **Spend another 30-40 minutes studying the rest of chapter 3. While you study, think about how the concept here relate to the motion of your car. Write a few sentences with your thoughts on this.** Acceleration is talked about a lot in this section, it also included the concept of air resistance and friction. When my car is going down elevated slopes, it's experiencing less friction as less gravity pushes on it, but the difference here with the friction isn't very big because the car isn't very heavy. My car also made connections towards' Galileo's work with letting go of a ball and letting it drop from different inclined surfaces. His work shows that the more inclined a surface is, the faster an object rolling down it will accelerate, I'm glad to see this in my results, when the car rolled from the 30 degrees slope, it is accelerating much faster than the 10 degrees slope. The only part that doesn't apply here is the free fall, air drag, and terminal velocity, as the car wasn't drop from above. But there may have been little air drag as the car traveled down the incline surface through an area with oxygen. But even though not tested, it seems very likely that the car will continue to accelerate faster at higher inclined surfaces and may travel faster if the surface itself with change; resulting in less friction.

Wednesday, September 2 Homework
//Chapter 3 - Pg 38-43//
 * **Explain how to use your data to determine the acceleration of the cars you filmed; Write your best estimate for the car's maximum acceleration.** To determine the acceleration of the car I filmed, I would use the logger pro program to process some of my data and create a graph showing acceleration, with m/s squared on the y-axis while the time on the x axis, giving me the option to look at the car's acceleration at each time. The maximum acceleration of my car would be 2.8 meters per second(squared) which was achieved during the 30 degrees slope part of my lab. I found this by looking at the graph and searching for the highest point.

Monday, August 31 Homework
//Chapter 2 - Pg 27-33 (Questions on Page 36)//
 * **Question 14: A different scaffold that weighs 300 N supports two painters, one 250 N and the other 300 N. The reading on the left scale is 400 N. What is the reading on the right scale?** The downward forces would all have to calculated first in order to figure out the readings of upward forces. Part of the downward force includes the two painters and the scaffold, the two painters have a combined weight of 550 N. When the scaffold is added, there will be a total of 850 N pushing downwards, forming the downwards force. Because we are assuming that the scaffold is equilibrium, the upward force would have to cancel out or equal to the downwards force. Lets say the downwards force is -850 N, we know the left rope of the scaffold supports 400 N, to find the right part, I minus the downwards force by the left rope, getting 450 N. 400 N + 450 N equals 850 N, canceling out the downward force. Therefore, the reading on the right scaled would be 450 N.
 * **Question 15: Nellie Newton brings at rest from the ends of the rope as shown. How does the reading on the scale compare to her weight?** Nellie is holding onto two ropes, both of which supports her weight, and because she is in the center and not tilting towards her left or towards her right, the scale would read the same number as her weight because the upwards support force is divided between the two ropes. But the two ends of the rope actually connects, it's just one rope partially wrapped around a pulley that does not lower the amount of support/upward force required (both sides of the rope is connected to the scale). Nellie isn't trying to propel herself upwards, leading more the the possibility of getting a more accurate reading of her weight. The reading of the scale compared to her weight should be pretty much the same.
 * **Question 16: Harry the painter swings year after year from his bosun's chair. His weight is 500 N and the rope, unknown to him, has a breaking point of 300 N. Why doesn't the rope break when he is supported as shown at the left? One day Harry is painting near a flagpole, and, for a change, he ties the free end of the rope to the flagpole instead of to his chair as shown at the right. Why did Harry end up taking his vacation early?** A rope has two end points, when Harry used both end points to hook himself onto the rope, his weight is pretty evenly distributed between the two ends, one sustaining 250 N and the other also 250 N. 300 N being the breaking point, the rope wouldn't break in this formation. But when Harry decided to hook his entire weight to one side of the rope, and a seemingly light flag, 500 N is placed on one end of the rope, exceeding it's limit by 200 N, causing the rope to break and Harry to take an early vacation. Lets just assume the rope snapped and Harry had an unpleasant fall.

Wednesday, August 26 (Thurs. Aug 27) Homework
//Chapter 2 - Pg 19-26 (Questions on Page 35)//
 * **Question 4: If a huge bear were chasing you, its enormous mass would be very threatening. But if you ran in a zigzag pattern, the bear's mass would be your advantage. Why?** Changing direction would require some force, and since the bear has an enormous mass compared to me, it'd take the bear more time to turn around at each angle of the zigzag pattern I ran in. Or even if the bear runs at a straight line while I run in a zigzag pattern, it'd have to aim at my direction, whether it was left or right, he'd had to turn around, and because he has an enormous mass, it's more challenging for the bear to stop and face my direction (in that time, I'd have been gone far long), or to change direction while running, because of inertia, it's not easy for a bear of that mass to stop at 0 seconds, it'll take some force to do so. While for me, if I ran fast, it will take some time to stop the moving force, but not as much is required to turn the direction I'm running in, and since I'm the person being chased, then turning around is the only obstacle. Changing direction will cause the bear to slow down quite a bit while it won't impact me as much.
 * **Question 6: Consider a ball at rest in the middle of a toy wagon. When the wagon is pulled forward, the ball rolls against the back of the wagon. Interpret this observation in terms of Newton's first law.** According to Newton's first law, an object will //"continue in a state of rest or of uniform speed in a straight line unless acted on by a nonzero net force"//. The ball is in a state of rest before the toy wagon was pulled forward, and it stayed this way until the wagon was pulled. When a force pulled the wagon forward, the ball rolls against the back of the wagon because it isn't attached to the wagon, the back of the wagon might actually be pushing the ball forwards. Like the example about the plates, the ball isn't moving, it's staying at it's state of rest, but as the wagon moves, it looks like the ball moved because the wagon is like the table cloth in the analogy. The friction between the ball and the wagon caused the ball to move, but not as fast as the wagon itself, so it is caught at the back of the wagon, or maybe when the wagon is pulled forward, it had a slight tilt and since gravity pushes downwards, the ball i pushed to the lower part of the slope. But my main reasoning would be that the ball stays at rest because there is no force that keeps it moving, and when the back of the wagon touches the ball, it pushes the ball but no faster than the speed of the wagon.
 * **Question 7: Why do you lurch forward in a bus that suddenly slows? Why do you lurch backwards when it picks up speed? What laws applies here?** I lurch forward when the bus stops because there is no force to stop me except when my hands push against the chair in front of me or the seatbelt that my body clashes into. I lurch backwards when it picks up speed because the bus has a motor that helps propel it forward while the force pushing me forward would be the bus, and because I need a nonzero force to get me moving (the bus), I lurch backwards since the bus starts moving before I do. The law that applies here would be inertia, I keep moving unless a nonzero force inteferes with it, and I stay in a state of rest until a force pushes me forwards.