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Cheat Sheet / Updated 03-28-2022
Solving physics problems correctly is a lot easier when you have a couple tricks under your belt. In fact, you can greatly improve your odds of getting the right answer if you make sure that what you calculated is plausible in the real world. Another trick is to draw your own visual when one isn’t provided for you — no artistic ability required. It also helps to have this handy reference for some of the most common unit prefixes and unit conversions you’re bound to encounter in your physics homework.
View Cheat SheetCheat Sheet / Updated 03-10-2022
Avoid difficulties when working on physics by knowing the common issues that can cause trouble in physics problems, understanding physical constants, and grasping principal physics equations.
View Cheat SheetCheat Sheet / Updated 02-18-2022
Optics covers the study of light. Three phenomena — reflection, refraction, and diffraction — help you predict where a ray or rays of light will go. Study up on other important optics topics, too, including interference, polarization, and fiber optics.
View Cheat SheetCheat Sheet / Updated 02-15-2022
Thermodynamics sounds intimidating, and it can be. However, if you focus on the most important thermodynamic formulas and equations, get comfortable converting from one unit of physical measurement to another, and become familiar with the physical constants related to thermodynamics, you’ll be at the head of the class.
View Cheat SheetCheat Sheet / Updated 02-11-2022
Here’s a list of some of the most important equations in Physics II courses. You can use these physics formulas as a quick reference for when you’re solving problems in electricity and magnetism, light waves and optics, special relativity, and modern physics.
View Cheat SheetCheat Sheet / Updated 01-24-2022
Physics involves a lot of calculations and problem solving. Having on hand the most frequently used physics equations and formulas helps you perform these tasks more efficiently and accurately. This Cheat Sheet also includes a list physics constants that you’ll find useful in a broad range of physics problems.
View Cheat SheetCheat Sheet / Updated 01-19-2022
String theory, often called the “theory of everything,” is a relatively young science that includes such unusual concepts as superstrings, branes, and extra dimensions. Scientists are hopeful that string theory will unlock one of the biggest mysteries of the universe, namely how gravity and quantum physics fit together.
View Cheat SheetArticle / Updated 10-29-2021
Any physicist knows that if an object applies a force to a spring, then the spring applies an equal and opposite force to the object. Hooke’s law gives the force a spring exerts on an object attached to it with the following equation: F = –kx The minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. The variables of the equation are F, which represents force, k, which is called the spring constant and measures how stiff and strong the spring is, and x, the distance the spring is stretched or compressed away from its equilibrium or rest position. The force exerted by a spring is called a restoring force; it always acts to restore the spring toward equilibrium. In Hooke’s law, the negative sign on the spring’s force means that the force exerted by the spring opposes the spring’s displacement. Understanding springs and their direction of force The direction of force exerted by a spring The preceding figure shows a ball attached to a spring. You can see that if the spring isn’t stretched or compressed, it exerts no force on the ball. If you push the spring, however, it pushes back, and if you pull the spring, it pulls back. Hooke’s law is valid as long as the elastic material you’re dealing with stays elastic — that is, it stays within its elastic limit. If you pull a spring too far, it loses its stretchy ability. As long as a spring stays within its elastic limit, you can say that F = –kx. When a spring stays within its elastic limit and obeys Hooke’s law, the spring is called an ideal spring. How to find the spring constant (example problem) Suppose that a group of car designers knocks on your door and asks whether you can help design a suspension system. “Sure,” you say. They inform you that the car will have a mass of 1,000 kilograms, and you have four shock absorbers, each 0.5 meters long, to work with. How strong do the springs have to be? Assuming these shock absorbers use springs, each one has to support a mass of at least 250 kilograms, which weighs the following: F = mg = (250 kg)(9.8 m/s2) = 2,450 N where F equals force, m equals the mass of the object, and g equals the acceleration due to gravity, 9.8 meters per second2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke’s law says: F = –kx Now, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k. Looking only at the magnitudes and therefore omitting the negative sign, you get Time to plug in the numbers: The springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter. The car designers rush out, ecstatic, but you call after them, “Don’t forget, you need to at least double that if you actually want your car to be able to handle potholes.”
View ArticleArticle / Updated 10-29-2021
In physics, when you’re given the vector components, such as (3, 4), you can easily convert to the magnitude/angle way of expressing vectors using trigonometry. For example, take a look at the vector in the image. Suppose that you’re given the coordinates of the end of the vector and want to find its magnitude, v, and angle, theta. Because of your knowledge of trigonometry, you know Where tan theta is the tangent of the angle. This means that theta = tan–1(y/x) Suppose that the coordinates of the vector are (3, 4). You can find the angle theta as the tan–1(4/3) = 53 degrees. You can use the Pythagorean theorem to find the hypotenuse — the magnitude, v — of the triangle formed by x, y, and v: Plug in the numbers for this example to get So if you have a vector given by the coordinates (3, 4), its magnitude is 5, and its angle is 53 degrees. Sample question Convert the vector given by the coordinates (1.0, 5.0) into magnitude/angle format. The correct answer is magnitude 5.1, angle 79 degrees. Apply the Pythagorean theorem to find the magnitude. Plug in the numbers to get 5.1. Apply the equation theta= tan–1(y/x) to find the angle. Plug in the numbers to get tan–1(5.0/1.0) = 79 degrees. Practice questions Convert the vector (5.0, 7.0) into magnitude/angle form. Convert the vector (13.0, 13.0) into magnitude/angle form. Convert the vector (–1.0, 1.0) into magnitude/angle form. Convert the vector (–5.0, –7.0) into magnitude/angle form. Following are answers to the practice questions: Magnitude 8.6, angle 54 degrees Apply the equation to find the magnitude, which is 8.6. Apply the equation theta = tan–1(y/x) to find the angle: tan–1(7.0/5.0) = 54 degrees. Magnitude 18.4, angle 45 degrees Apply the equation to find the magnitude, which is 18.4. Apply the equation theta = tan–1(y/x) to find the angle: tan–1(13.0/13.0) = 45 degrees. Magnitude 1.4, angle 135 degrees Apply the equation to find the magnitude, which is 1.4. Apply the equation theta = tan–1(y/x) to find the angle: tan–1(1.0/–1.0) = –45 degrees. However, note that the angle must really be between 90 degrees and 180 degrees because the first vector component is negative and the second is positive. That means you should add 180 degrees to –45 degrees, giving you 135 degrees (the tangent of 135 degrees is also 1.0/–1.0 = –1.0). Magnitude 8.6, angle 234 degrees Apply the equation to find the magnitude, which is 8.6. Apply the equation theta = tan–1(y/x) to find the angle: tan–1(–7.0/–5.0) = 54 degrees. However, note that the angle must really be between 180 degrees and 270 degrees because both vector components are negative. That means you should add 180 degrees to 54 degrees, giving you 234 degrees (the tangent of 234 degrees is also –7.0/–5.0 = 7.0/5.0).
View ArticleArticle / Updated 10-07-2021
Pressure and force are related, and so you can calculate one if you know the other by using the physics equation, P = F/A. Because pressure is force divided by area, its meter-kilogram-second (MKS) units are newtons per square meter, or N/m2. In the foot-pound-second (FPS) system, the units are pounds per square inch, or psi. The unit newtons per square meter is so common in physics that it has a special name: the pascal, which equals 1 newton per square meter. The pascal is abbreviated as Pa. You don’t have to be underwater to experience pressure from a fluid. Air exerts pressure, too, due to the weight of the air above you. Here’s how much pressure the air exerts on you at sea level: The air pressure at sea level is a standard pressure that people refer to as 1 atmosphere (abbreviated atm): If you convert an atmosphere to pounds per square inch, it’s about 14.7 psi. That means that 14.7 pounds of force are pressing in on every square inch of your body at sea level. Your body pushes back with 14.7 psi, so you don’t feel any pressure on you at all. But if you suddenly got transported to outer space, the inward pressure of the air pushing on you would be gone, and all that would remain would be the 14.7 pounds per square inch your body exerted outward. You wouldn’t explode, but your lungs could burst if you tried to hold your breath. The change in pressure could also cause the nitrogen in your blood to form bubbles and give you the bends! Here’s a pressure example problem using water pressure. Say you’re in your neighbor’s pool, waiting near the bottom until your neighbors give up trying to chase you off and go back into the house. You’re near the deep end of the pool, and using the handy pressure gauge you always carry, you measure the pressure on the back of your hand as What force does the water exert on the back of your hand? The back of your hand has an area of about You reason that if P = F/A, then the following is true: F = PA Plugging in the numbers and solving gives you the answer: Yikes. A thousand newtons! You whip out your underwater calculator to find that’s about 230 pounds. Forces add up quickly when you’re underwater because water is a heavy liquid. The force you feel is the weight of the water above you.
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