AAMC Physics QPack Flashcards
Units for electric field
V/m which is the same as N/C
1V=1J/C
Frequency formula
f = c/λ
Period formula
T = 1/f
If the first harmonic has a frequency of 100 Hz, then the second harmonic has a frequency of 200 Hz. The period corresponding to 200 Hz is 1/200 s-1 = 0.005 s.
Fraction submerged formula
Wice = mg = pVg
doppler effect formulas
f’ (perceived frequency) = f (emitted frequency) ((v+/-vd)/(v+/-vS))
Δf/f = -v/c
Δλ/λ = v/c
doppler effect formulas
f’ (perceived frequency) = f (emitted frequency) ((v+/-vd)/(v+/-vS))
Δf/f = -v/c
Δλ/λ = v/c
electrical field between two electrodes
E= (V-IR)/ L
A charged particle ______ in an electric field
accelerates
Electrical force depends on the particle’s charge and the strength of the electric field experienced by the particle, not on the particle’s ____.
speed
Capacitance C is ______ to the plate area and ______ to the separation distance of the plates.
proportional….inversely proportional
The force on the charge is ___ and force is also ___.
qE and ma
Because there is a constant force qE on the negative particle, it will ______ toward the positive plate
accelerate
organ pipe closed at one end
L = nλ/ 4
human threshold of hearing
10–12 W/m2
absolute value of the work done (pully)
mgΔh
kinetic energy formula
1/2 mv2
potential energy formula
mgh
Snell’s law of refraction
nsin0(theta) = nsin0(theta)
energy stored in a capacitor
½CV2
pressure formula
density x g x h
The glass that is used as a beam splitter is replaced with glass that is identical except that it has a 10% higher index of refraction. Which of the following changes will occur to the pinhole image?
it will move
The examinee is asked what an increase in the index of refraction of the beam splitter will do to the image of a pinhole on the viewing screen. The pinhole image is formed by light that: (1) passes through the glass of the beam splitter, (2) reflects off the retroflector, (3) returns to the surface of beam splitter nearest the retroflector, (4) reflects off that beam splitter surface, and (5) falls on the viewing screen, forming an image. The only refraction occurs in step (1), when the entering light is refracted. This refraction shifts the rays of light over a bit. If the index of refraction is increased, this increases the shift. This will cause a shift in the position of the light striking the retroflector which will persist through the steps (3), (4), and (5) above, causing the image on the screen to shift in position.
work done by the force (pully)
mg x distance raised
see-saw fulcrum
Fleft x Lleft = Fright x Lright
When a downward force is applied at a point 0.60 m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10–7 kg at a point 0.40 m to the right of the fulcrum. What is the magnitude of the downward force?
6.5 x 10^-7 N
The system will be in equilibrium only if the right and left forces exert torques that cancel each other. One force’s torque will attempt to twist the “see-saw” supported by the fulcrum clockwise, while the other force provides a counterclockwise torque. The torque of a force F applied at a distance L from the fulcrum is given by FLsinθ, with θ is the angle between the beam and the force. At equilibrium the left and right forces point vertically downward with the beam horizontal, hence θ = 90° and sin90° = 1. At equilibrium the torque magnitudes are equal so: Fleft Lleft = Fright Lright. The force at Lright = 0.40 m is the weight of the given mass, mg = (10–7 kg)(9.8 m/s2) = 9.8 × 10–7 N. The unknown force is at Lleft = 0.60 m, so Fleft (0.60 m) = (9.8 × 10–7 N)(0.40 m) which may be solved to yield Fleft = 6.5 × 10–7 m.
