Physics Flashcards
Kinematics
Pulley Mechanical Advantage
T=Mg/(every strand running between two pulleys)
A bobsled team of four men, each of mass m, pushes the sled of mass M over a horizontal displacement d before leaping into the sled and beginning to slide down the hill. The total length of the winding track from the top of the hill to the bottom is L, and the height difference between the top of the hill and the bottom is Δh. If the sled starts at rest and we assume the whole track is frictionless, how does the final kinetic energy of the sled at the bottom of the hill (prior to the brakes being applied) compare to the initial potential energy of the sled at the top of the hill?
KEf > PEi
With no friction, this is a case of conservation of mechanical energy after work is done on the combined mass by the running men. The conservation relation can be written KEfinal = PEinitial + Wexternal or, if you take the speed achieved by the bobsled team just as they leap into the sled, KEfinal = PEinitial + KEinitial. Either way it is clear that the final kinetic energy exceeds the initial potential energy of the sled.
An ideal projectile launched from the ground at an angle of 45° up from the horizontal has a total flight time of T. Which of the following correctly expresses a relationship between the displacement d, velocity v, and acceleration a at T / 2 compared to time T (the instant before it hits the ground)?
2 × |dT/2| > |dT|
|vT/2| = |vT|
|aT/2| > |aT|
vT/2 = 0
At half the total flight time, a projectile launched upward from even ground upon which it will eventually land will be at the apex of its flight. At this time, all of its velocity is in the horizontal direction (meaning it has slowed down, eliminating choice B, but is still moving, eliminating choice D). Its acceleration throughout the flight is a constant g downward, eliminating choice C. That only leaves answer choice A, which is correct: The magnitude of the displacement from the launch point to the apex is a diagonal line that is the hypotenuse of a right triangle with the max height of the projectile as one leg and half the range (R = |dT|) as the other. Twice this hypotenuse must therefore be greater than the total range dT.
Why is this possible: An object in motion for some period of time has zero average velocity but positive average speed, An object maintains constant speed but has nonzero acceleration, An object maintains constant speed but has nonzero acceleration.
an object traveling in a closed curved path (like a circle) will have zero displacement and therefore zero average velocity but not zero distance or average speed
uniform circular motion.
Why is this impossible?
An object in motion for some period of time has an average velocity with a greater magnitude than its average speed.
Average velocity has magnitude of displacement / time, whereas average speed equals total distance / time. The time is the same for both, and displacement ≤ total distance, so average velocity ≤ average speed. Put another way, there’s no shorter path than a straight line, for which the magnitude of displacement equals the total distance. For any other type of path, the displacement is shorter than the distance traveled.
Impulse
∆p = pf – pi = mvf – mvi
don’t forget to change sign of momenta if things bouce off. positive/negative directions of motion matter
What is conserved in a collision? Elastic Collision?
Mass, Momentum
Elastic: also, Kinetic Energy
Normal Force
It is attributable to the binding forces between atoms and molecules in solid matter.
it can contribute to the change in direction of a moving object i.e., banked circular track normal provides centripetal acceleration
Not always equal and opposite to weight
What phase is capable of permanently withstanding a force perpendicular to its surface? A force not perpendicular?
Solid, Liquid, and Gas can witstand perpendicular force. For example, a gas can withstand the force of a stretched balloon.
Only solids can withstand nonperpendicular force because bonds are not continually breaking and reforming.
Fluids; if high cross sectional area,
lower velocity (Q=vA)
Q=Constant Fluid Rate
Pressure increase
Where is pressure lowest in circulatory system (ignore gravity)
Vena Cava
Change in Pressure=QR (similar to V=IR)
Horizontal, viscous flow, pressure is analogous to voltage, both decrease in the direction of flow or current
(note, cross sectional area must be constant to apply the equation /\P=QR, but pressure drops steadily as we move away from the heart)
Flow Rate for a pipe that splits into two
Flow rate is constant if it continues to be one long tube
But if it splits, you have to add the flow rates
The continuity equation says Q = A*v This says that assuming the flow rate (Q) is the same for both parts you’re comparing, then as the area increases, the velocity decreases. You can basically write that as: Q1 = A1*v1 Q2 = A2*v2 And so if the Qs are the same for both points, we can set them equal to each other as: A1*v1 = A2*v2 But when we’ve done this, we have assumed the volume flow rate is the same at both places. If you’re talking about a single tube, then the volume flow rate is the same at all places…since if a certain amount of fluid enters the tube, that exact amount must flow through every place in between, and then exit the tube. But if you have a tube or pipe system that has branches…then you cannot assume that the flow rate is the same at each of the branches. If a certain amount of fluid flows in the opening, and then the tube branches into two sections…you can say that the fluid flowing through both branches combined must equal the fluid that came in the opening…but that does not mean that an equal amount of fluid went down each branch. What we can say for sure about this system is that if both are at the same height, and both are at the same pressure (since each spigot is opened to the atmosphere), then they must also have the same velocity (according to Bernoulli equation). Then…we can use the Q=A*v equation to say that if they both have the same velocity, then the wider spigot must have a larger flow rate.
Young’s Modulus
Stress/Strain
Greater Modulus, more it can stretch and be flexible
What changes the modulus of elasticity
Changing the temperature, because that changest how the object responds to stress
