Physics Questions-Michigan State University .

You have been studying ideas of potential energy and kinetic energy (KE). Potential energy due to position in the earth’s gravitational field or the potential energy of a stretched or compressed spring are examples of mechanical potential energy (PE). Energy stored in a lump of coal that is released when it is burned is chemical potential energy and not mechanical potential energy. This is the subject of chemistry.

The sum of kinetic and mechanical potential energy is called total mechanical energy. If there is no friction. The total mechanical energy of a system is conserved or constant if forces such as gravity or spring forces are the only forces acting on the object. These are called conservative forces. When a rock is thrown up, down, horizontally, or at a weird angle and air resistance is ignored, the total mechanical energy of the rock stays constant, or KE + PE = constant. This also means KE can become PE and vice versa. If there is friction, the work done by friction “gobbles up” total mechanical energy. For example, if you jump off of a high diving board into a swimming pool, as you fall toward the water, your initial potential energy becomes kinetic energy, but your total mechanical energy is conserved until you hit the water.

Discussion Question: What happens to your total mechanical energy when you hit the water? Where does the lost mechanical energy go? Why do you fill a swimming pool with water before diving in? Can you think of another important example of non-conservation of total mechanical energy?

Now I want to make a very important point. Total mechanical energy is conserved if there is no friction or other non-conservative forces doing work. If non-conservative forces are doing work, then total mechanical energy is not conserved. Total energy is all energy, not just mechanical energy. Now here is the big lesson, total energy is always conserved!!! This is a fundamental law of nature.

Circuit Theory Problems-Abcott Institute .

Vc V1 R, RA Rs Rc RE R2 Vs RA VD 4 Ve Assumptions: 1. R = R = 900 2. 2. R2 = R3 = 1.100 2. 3. RA= R=2.000 12. 4. Re = 1.000 2. 5. Vs = 9 V. Tasks: (a) (5 points) Determine Ve. (b) (5 points) Determine Vp. (C) (10 points) Determine V. and V.2. (d) (5 points) Determine Vol – V.2.

Radioactive Decay Physics Questions-Boston College .

Radioactive decay


Gamma radiation can be significantly reduced by…





What form of radiation is emitted by nuclei with too many neutrons?





Gamma radiation is…

An electromagnetic wave

A helium nucleus

An electron


Which form of radiation has the highest ionising power?





What causes an atom’s nucleus to become unstable?

Too many or too few neutrons in the nucleus

More electrons than protons

Fewer electrons than protons


What is the activity of a source that emits 1,200 particles per minute?

1,200 Bq

20 Bq

72,000 Bq


Uranium 238 emits an alpha particle to become what nucleus?





An alpha particle is…

An electron

A helium nucleus

An electromagnetic wave


Which type of radiation can reach the furthest?





What fraction of a radioisotope has decayed after two half-lives? – Higher

One quarter

All of it

Three quarters

Electricity Batteries and Current Physics Lab Report-UArizona Global Campus .

Electricity, Batteries, and Current You can access the simulations via the title of each section. Simply hoover your mouse over the titles and press CTRL + Click to follow the links. Battery Voltage: • Do the small blue spheres represent positive or negative charges? • Which side of the battery is labeled positive, and which side is negative? • How can you determine which side of the battery is positive and negative just by the location of the blue charges? Resistance in a Wire: • What characteristics of a resistor are variable in this model? • How does each affect the resistance? (Will increasing or decreasing each make the resistance correspondingly increase or decrease?) • Explain your ideas about why they change the resistance. Ohm’s Law: • As you change the value of the battery voltage, how does this change the current through the circuit and the resistance of the resistor? If the current or resistance remains constant, why do you think? • As you change the value of the resistance of the resistor, how does this change the current through the circuit and the battery voltage? If the current or voltage remains constant, why do you think?

Friction Forces Questions-American Commercial College of Texas .

Friction forces Introduction The aim of this experiment is to determine the sliding friction and static friction denoted by Fk and Fs. The experiment clearly shows the comparison between the sliding friction force and rolling friction force denoted as Fk and Fr respectively, as a weight function. The experiment demonstrated that the friction force do not affect the coefficient of friction, µ and the friction force, F. Friction force is taken as the force that will always oppose the object motion. In regard to velocity, friction can either be classified as sliding force, where velocity is not equal to zero, or static force when the velocity is zero. The force of static friction is the force that acts between two objects that do not move in relative to one another. The object in conduct with one another and experiencing static friction force must be overcome to enable an object to move on the surface. Sliding friction force occurs when two bodies move relative to one another, when one object is sliding over another surface and there is opposition on the motion of the body. The two types of frictions are described based on two different coefficients of friction, µ, the coefficient of sliding friction and coefficient of static friction represents respectively the two different coefficient values that will be included in the calculation of both the force of static friction and the force of sliding friction. The coefficient of friction will describe how one object will move with an easy on a surface as compared to the other objects. Therefore, a high friction coefficient will mean that there is a lot of friction force acting between two bodies that are in contact. In a case where a force Fp is used in pulling a block in parallel direction to horizontal surface, friction force, Ff, will be acting in the opposite direction. The value of the friction is supposed to be less than the pulling force to enable the object to move or else the object will not move. Resultant force = pulling force – force of friction = Fp – Ff Coefficient of static and sliding friction 𝐹𝑓 𝜇 = 𝑁𝑓 = 𝐹𝑠 𝜇s = 𝑊 𝐹𝑠 𝜇k = 𝑊 𝐹𝑝 𝑊 Objectives 1. To apply the frictional force formula 𝑓 = 𝜇 ⋅ 𝑁 2. To understand the difference between static and kinetic friction 3. To understand the difference between static friction force and maximum static friction force 4. To understand the limit of the oversimplified friction model 𝑓 = 𝜇 ⋅ 𝑁 5. To cultivate the critical thinking habit of not blindly following formulas 6. To cultivate the habit of keeping all experimental data in a well-organized manner Methodology Procedure 1. The weight of the wooden block and metal block were recorded 2. The blocks were placed at a flat wooden plane and pulled using a pulley connected to a weighted hanger 3. The set up was used to calculate the pulling force and the coefficient of frictions 4. The procedure was repeated Data and calculations 0≤ 𝐹 fs ≤ 𝐹 𝑚𝑎𝑥 fs = 𝜇 sFN Ffk = 𝜇 kFN Trial mB mE MT 1 Surface Area type Wood Large 0.132 0 kg 2 Wood Large 0.132 0 kg 3 Wood Large 0.132 0.5 kg 4 Wood Large 0.132 0.5 kg 5 Wood Large 0.132 1 kg 6 Wood Large 0.132 1 kg 0.132 kg 0.132 kg 0.632 kg 0.632 kg 1.132 kg 1.132 F= M Tg 1.29 N 1.29 N 6.19 N 6.19 N 11.09 N 11.09 Fts Ftk µs µk 0.46 N 0.46 N 1.56 N 1.43 N 2.44 0.23 0.356 0.178 0.12 0.356 0.093 1.09 0.252 0.168 0.89 0.231 0.136 1.87 0.220 0.169 2.08 1.43 0.182 0.129 7 Felt Large 0.132 1 kg 8 Felt Large 0.132 1 kg 9 Wood Small 0.132 1 kg 10 Wood Small 0.132 1 kg kg 1.132 kg 1.132 kg 1.132 kg 1. 132 kg N 11.09 N 11.09 N 11.09 N 11.09 N 2.08 2.09 0.188 0.188 2.29 1.65 0.149 0.149 2.53 1.2 0.108 0.108 1.71 0.57 0.154 0.051 Table 1: Determining the values of static coefficient and kinetic coefficient Average 1 and 2 Average 3 and 4 Average 5 and 6 Average 7 and 8 Average 9 and 10 Surface type Wood Area MT (kg) 𝜇̅ s 𝜇̅ k 𝜇̅ s > 𝜇̅ k Large 0.132 0.356 0.136 Yes Wood Large 0.632 0.94 0.152 Yes Wood Large 1.132 0.204 0.149 Yes Felt Large 1.132 0.169 0.169 Same Wood Small 1.132 0.131 0.0795 Yes Table 2: Comparing static coefficient and kinetic coefficient of friction Exploration 1. Your friction curves may look different, but should share the same general trend of the curve shown in Figure 1. Do they? Yes they showed the same trend as in figure 1 2. For the same contact surface, will the values of the coefficient 𝜇s and 𝜇k change significantly when the normal force at the contact surface changes substantially? The coefficient themselves do not change, however when more weight is applied, the higher the friction force goes 3. If you repeat exactly the same experiment parameters (same contact materials, same normal force, same contact area, etc.) a few times, do the values of 𝜇s and 𝜇k remain consistent or vary significantly? µk + µs remained fairly consistent 4. For the same contact materials with the same normal force but different contact areas, as shown to the right, will the values of 𝜇s and 𝜇k obtained remain consistent or vary significantly? Remains consistent 5. Under exactly the same experiment conditions, is 𝜇s always greater than 𝜇k? Yes, though occasionally we measured them as identical 6. When you use the force sensor to pull the block in order to measure the frictional force, will it make a difference if the pulling force is not parallel with the surface as shown to the right? Why? Absolute yes, when the forces is not parallel it alters how much force is actually applied as the angle throws it off Application Surface type chosen: wooden Area size chosen: Large 𝜇 k (average) = 0.159 𝜇 s (Average) = 0.159 Trial mB mE MT 1 Surface Area type Wood Large 0.132 2 Wood Large 0.132 3 Wood Large 0.132 1.02 kg 1.19 kg 1.20 1.15 kg 1.32 kg 1.20 F= M Tg 11.3 N 13.0 N 11.8 Fts Ftk mfs mfk 2.11 1.91 1.15 1.23 2.44 1.49 0.813 1.956 2.20 1.24 0.677 0.796 Average Wood 0,132 kg 1.22 kg kg 1.22 kg N 12.0 N 2.14 1.55 0.88 0.94 Table 3: Determining the mass of kinetic friction and static friction mfs (average) = 0.88 mfs (average) = 0.94 Mmeasured = 0.955 % error = 4.9% We will use kinetic friction when there is sliding between interfaces and static friction when there is no sliding. The average coefficient of kinetic friction and static friction were used. Conclusion The coefficient of a sliding friction was determined to be mainly influenced by the angle of inclination but the surface area had no effect. An object could on move against a contact surface when the pulling or pushing force could exceed static friction force. Reference Riedo, E., Gnecco, E., Bennewitz, R., Meyer, E. and Brune, H., 2003. Interaction potential and hopping dynamics governing sliding friction. Physical review letters, 91(8), p.084502. Dai, Z., Gorb, S.N. and Schwarz, U., 2002. Roughness-dependent friction force of the tarsal claw system in the beetle Pachnoda marginata (Coleoptera, Scarabaeidae). Journal of Experimental Biology, 205(16), pp.2479-2488. Johannes, V.I., Green, M.A. and Brockley, C.A., 1973. The role of the rate of application of the tangential force in determining the static friction coefficient. Wear, 24(3), pp.381-385.

Wave on A String Simulation Project-University of Central Florida .

Please download the Wave on a String handout. Use the simulation according to the instructions in the Wave on a String handout and answer the accompanying questions in the document. Upload your completed document here.

Physics Questionnaire.

| Constant Speed 1 Constant Speed Accelerating 60 Parachute opens 50 40 Deceleration velocity (m/s) 30 20 Freefali 10 0 0 10 20 30 40 50 60 time (s) 1) What is the skydiver’s terminal velocity while in feefall? m/s 2) What is the skydivers constant speed while under canopy (with parachute open)? m/s 3) How many seconds did it take for the parachute to open? This is a Parachute Harness. 4) If someone invents a “quick opening” parachute that opens in only 1 second, do you think that it would sell well? Yes or No.

Lens Equation Magnification Question-Boston College.

A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm. Determine the image distance and the image size

Physics Worksheet-ACC .

The parallel metal plates of large area spaced at a distance of 1cm from each other in air and a p.d of 5000V is maintained between them. If a sheet of glass 0.5cm thick and having a relative permittivity of 6 is introduced between the plates, what will be the maximum electric stress and where will it occur?


Electric Stress Physics Worksheet-ACC .

A capacitor, formed by two parallel plates of large area, spaced 2cm apart in air, is connected to a 10,000V d.c supply. Calculate the electric stress in the air when a flat sheet of glass of thickness 0.5cm and relative permittivity 5 is introduced between the plates.