AAMC Physics QPack

Question 1

Units for electric field

Answer 1

V/m which is the same as N/C

1V=1J/C

Question 2

Frequency formula

Answer 2

f = c/λ

Question 3

Period formula

Answer 3

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.

Question 4

Fraction submerged formula

Answer 4

Wice = mg = pVg

Question 5

doppler effect formulas

Answer 5

f’ (perceived frequency) = f (emitted frequency) ((v+/-vd)/(v+/-vS))

Δf/f = -v/c

Δλ/λ = v/c

Question 6

doppler effect formulas

Answer 6

f’ (perceived frequency) = f (emitted frequency) ((v+/-vd)/(v+/-vS))

Δf/f = -v/c

Δλ/λ = v/c

Question 7

electrical field between two electrodes

Answer 7

E= (V-IR)/ L

Question 8

A charged particle ______ in an electric field

Answer 8

accelerates

Question 9

Electrical force depends on the particle’s charge and the strength of the electric field experienced by the particle, not on the particle’s ____.

Answer 9

speed

Question 10

Capacitance C is ______ to the plate area and ______ to the separation distance of the plates.

Answer 10

proportional….inversely proportional

Question 11

The force on the charge is ___ and force is also ___.

Answer 11

qE and ma

Question 12

Because there is a constant force qE on the negative particle, it will ______ toward the positive plate

Answer 12

accelerate

Question 13

organ pipe closed at one end

Answer 13

L = nλ/ 4

Question 14

human threshold of hearing

Answer 14

10–12 W/m2

Question 15

absolute value of the work done (pully)

Answer 15

mgΔh

Question 16

kinetic energy formula

Answer 16

1/2 mv2

Question 17

potential energy formula

Answer 17

mgh

Question 18

Snell’s law of refraction

Answer 18

nsin0(theta) = nsin0(theta)

Question 19

energy stored in a capacitor

Answer 19

½CV2

Question 20

pressure formula

Answer 20

density x g x h

Question 21

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?

Answer 21

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.

Question 22

work done by the force (pully)

Answer 22

mg x distance raised

Question 23

see-saw fulcrum

Answer 23

Fleft x Lleft = Fright x Lright

Question 24

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?

Answer 24

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.

Question 25

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?

Answer 25

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.

Question 26

refraction index

Answer 26

n=c/v

Question 27

volume blood flow formula

Answer 27

area x velocity

Question 28

A magnetic force acts on a moving charge in a direction that is _______ to both the velocity of the charge and the direction of the magnetic field.

Answer 28

perpendicular

Question 29

Which of the following statements best explains why the intensity of sound heard is less when a wall is placed between a source of sound and the listener?

Answer 29

part of the sound energy is reflected by the solid

Question 30

fusion means

Answer 30

melting