Enable contrast version

Tutor profile: Ryan C.

Inactive
Ryan C.
Certified Math and Science Teacher
Tutor Satisfaction Guarantee

Questions

Subject: Physics (Waves and Optics)

TutorMe
Question:

A 2.5 cm tall object is placed 6.0 cm in front of a thin convex lens with a focal length of 8.0 cm. What will be the characteristics of the resulting image?

Inactive
Ryan C.
Answer:

In order to answer this question, we first need to figure out where the image will appear. To do this, we'll use the thin lens equation: 1/do + 1/di = 1/f f is the focal length and do and di are the object distance and image distance respectively (sometimes p and q are used instead, but I prefer do and di because they are clearer.) Our focal length is 8.0 cm and since it is a convex lens, f will be positive (if it were concave, it would be negative). The object distance is how far away our object is from the lens, which is 6.0 cm. The object distance is always positive. The image distance will be how far the image is from the lens, which is what we want to know right now. So we have: do = 6.0 cm di = we want to know f = 8.0 cm Plugging these into our equation, we get: 1/6.0 + 1/di = 1/8.0 Subtract 1/6.0 from both sides to get: 1/di = -0.042 I rounded here to write the number down, but you're better off leaving the whole thing in your calculator before doing the next part. We currently have 1/di, but we want di. To get that, we need to raise both sides to the negative first power, which essentially flips the numerator and denominator. Most calculators have a x^-1 button you can use to do this. Doing so should result in: di = -24 The fact that our image distance is negative reveals the first characteristic. Real images have positive image distances and virtual images have negative ones. So our image is going to be virtual. For a lens, this means the image will be on the same side of the lens as the object and since it's not a real image, we would not be able to project it on a screen, since light is not actually going there; it only appears to be. For the last two characteristics, we're going to need a second equation: the magnification equation. This one looks like: M = hi/ho = -di/do M is the magnification, which is defined as the ratio of the image height to the object height (hi and ho). This is conveniently equal to the ratio of the image distance to the object distance (with a negative sign thrown in). We could use our object height of 2.5 cm here to find out how tall our image is going to be, but we can answer the question about the characteristics with just the magnification. Plugging into the magnification equation gives us: M = -(-24)/6.0 The two negatives cancel and make a positive and leave us with: M = 4.0 This magnification tells us a few things. The fact that it is positive means the image is upright. If it were negative, the image would be flipped upside down, or "inverted." Additionally, the magnification has an absolute value greater than 1. This means the image is bigger than the object (4 times bigger, in fact). We can say our image is "enlarged" rather than "diminished." We can now answer the question by saying the image will be VIRTUAL, UPRIGHT, and ENLARGED. Before we go, let's do a quick sanity check on our answer. A magnifying glass is an optical instrument made from a convex lens. When we put something close to a magnifying glass (within its focal length) we obviously would in fact expect an enlarged image (that's kind of the point of them). We would also expect an upright image (it would be pretty annoying if magnifying glasses flipped things). Finally, magnifying glasses produce virtual images, because we could not put a sheet of paper where the image appears to be and see it projected on the paper as a real image would be. So we can be confident that our answer of virtual, upright, enlarged matches what we would expect from reality.

Subject: Physics

TutorMe
Question:

If a ball is dropped from a height of 22 meters, how long will it take to reach the ground? Assume you are on Earth and can ignore air resistance.

Inactive
Ryan C.
Answer:

A good first step when solving any problem is identifying your knowns and unknowns. This particular problem is a free fall problem, where an object is falling under nothing but the influence of the force of gravity. Such problems only have 5 factors for us to consider. Those are vertical displacement, initial velocity, final velocity, acceleration, and time. So which of these factors were given to us in the problem and which are we looking for? Let's start with our unknown. To find out which factor you want to be looking for, just look at what kind of question was asked. In this case, we are being asked "how long will it take?". That sort of question is asking for an amount of time. So we now know that time is our unknown factor. At first glance, you might be worried that there is only one number given to us in the problem. How can we possibly have enough information to solve this? Don't worry though. If we look closely at the problem, we can get everything we need out of it. Let's start with that one number they do give us. The ball is dropped from a height of 22 meters, but which factor does that match up with? Well, vertical displacement means a change in vertical position (in other words, how far something fell). Therefore, 22 meters is related to our vertical displacement. However, there is still one change we have to make. Displacement is a vector quantity, which means it has a direction associated with it. In a one-dimensional problem like this, we only have two options for direction: up or down. We can represent this by assigning positive values to vectors pointing up and negative values to vectors pointing down. Therefore, since our ball falls down 22 m, our vertical displacement is going to be -22 m. Next let's try to see if we are given any information about the ball's initial velocity. Initial velocity means how fast the ball is moving at the start. If you need help remembering this, think about how your INITIALS are what your name STARTS with. It might seem like we don't know anything about the ball's initial velocity, but if we look closely, we see the detail that it was dropped. It wasn't thrown up or down, it was simply dropped. When you drop something, at the moment you let go, it isn't moving at all yet. Therefore our ball's initial velocity is zero. What about the final velocity? That is how fast the ball is moving at the end (you take FINALS at the END of a class). You might be thinking this is also zero, since the ball will land on the ground and stop moving. However, once the ball touches the ground, this is no longer a free fall problem and our equations won't work. The final velocity is how fast the ball is moving at the end of free fall, just before it touches the ground. We don't know how fast this is (that's a different problem we could solve), but it isn't zero. It's okay that we don't know the final velocity, because the kinematic equations are set up so that you only need to know 3 of the 5 factors in order to get a solution. Our last factor to consider is the ball's acceleration. Acceleration is the rate at which an object's velocity increases or decreases. That might seem hard to determine, especially from the information given, but fortunately this is the easiest one as long as you know one thing: All objects falling near the surface of the Earth accelerate at the same rate (as long as you ignore air resistance). That rate is the acceleration due to gravity, which is 9.8 meters per second per second (m/s/s) downward. Since our problem told us it was okay to make those assumptions, we know our acceleration is -9.8 m/s/s. Let's summarize what we've figured out so far and create variables for our factors so we don't have to keep writing out their full names. Vertical displacement will be y (we could just use a d for displacement, but once we get into problems with 2 dimensions of motion, it will be nice to use x for horizontal things and y for vertical things, so let's just get in that habit now). Initial velocity and final velocity will be vi and vf respectively, written with the i and f as subscripts. Acceleration will be a (although you will often see g used for acceleration due to gravity) and time will be t. This gives us: y = -22 m vi = 0 m/s vf = we don't know and don't care a = -9.8 m/s/s t = we want to know Each kinematic equation has 4 of the 5 factors in it. We want to use the one that has the 3 factors we know as well as the one factor we are looking for. That would be: y = vi*t + 1/2*a*t^2 Plugging in our values from above gives us: -22 = 0*t + 1/2*-9.8*t^2 We can simplify this to: -22 = -4.9*t^2 Next we can divide both sides by -4.9 to get t squared by itself: 4.5 = t^2 Finally, we can take the square root of both sides to solve for t, rounding our answer to two significant figures: t = 2.1 We can now answer the question by saying it will take the ball about 2.1 seconds to reach the ground.

Subject: Algebra

TutorMe
Question:

42 people went to the movies. The movie theater sold $300 worth of tickets. An adult ticket costs $10 and a children's ticket costs $6. How many children's tickets were sold? How many adult tickets were sold?

Inactive
Ryan C.
Answer:

Given a word problem like this, where we have two things we don't know (a.k.a. variables), what we want to do is set up a system of equations. As long as we have as many unique equations as we have variables, we can solve for each of those variables. One thing we know is that 42 people went to the movies. Assuming all of those people qualify as adults or children (at least when it comes to buying tickets) this gives us the equation: A + C = 42 where A represents the number of adult tickets sold and C represents the number of children's tickets sold. For our second equation, we can look at the money. The total amount of money made is going to be the adult price times the number of adult tickets plus the children's price times the number of children's tickets. In other words: 10A + 6C = 300 Now that we have our system of two equations, we have a couple ways we can solve them. These are known as the substitution and elimination methods. I'm going to use the substitution method here, since it's slightly more convenient for this problem, but either would work. First we'll take our first equation and subtract C from both sides. This is because when solving an algebra problem, you want to do the opposite operation from what was being done in order to isolate your variable (C was being added to A, so we're subtracting it) and whatever we do, we have to do it from both sides - otherwise the equation is no longer balanced and therefore untrue. This gives us: A = 42 - C Now we have an expression for A that we can substitute into the second equation any time we see A there. The benefit of this is that we now have an equation with just one variable and those are much easier to solve. Our new equation is: 10(42-C) + 6C = 300 Remembering to use the distributive property, we can simplify this to: 420 -10C + 6C = 300 Then we can combine like terms to get: 420 - 4C = 300 Then subtract 420 from both sides: -4C = -120 And finally divide both sides by negative 4 to solve for C: C = 30 This means 30 children's tickets were sold and we've answered the first half of the question. Fortunately, the second half will be much faster now that we have that information. We can simply use the fact that C = 30 and plug that into our rearranged first equation from earlier, giving us: A = 42 -30 = 12 This means 12 adult tickets were sold and we've completely solved the problem.

Contact tutor

Send a message explaining your
needs and Ryan will reply soon.
Contact Ryan

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.