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156 Cards in this Set
- Front
- Back
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Average Force=
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ΔP/ΔT
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J=
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FΔt
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J=
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ΔP
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ΔP=
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FΔt
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law of conservation of linear momentum
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the total linear momentum will remain constant
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total P before collisions=
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total P after collisions
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elastic collision
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KE conserved
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inelastic collision
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KE not conserved
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if objects are perpendicular, what must you do to get Ptotal?
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resolve triangle to get Ptotal
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x axis-time
y axis- force area=? |
impulse
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x axis-distance/displacement
y axis- force area=? |
work
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Fc=
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mv^2/r
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Fg=
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GMm/r^2
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Ac=
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v^2/r
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V=
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2pir/t
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gtan(theta)=
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v^2/r
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KE=
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.5mv^2
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PE=
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mgh
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PE1 + KE1 + Fnc=
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PE2 + KE2
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Fs=
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-kx
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PEs=
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.5kx^2
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average speed=
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distance/time
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average velocity=
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displacement/time
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x=
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v(avg)*t
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Δx=
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Vot + .5at^2
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Vf=
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Vo + at
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average acceleration=
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Δv/Δt
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Vf^2=
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Vo^2 + 2aΔx
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First Newton Law
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Inertia
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Second Newton Law
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f=ma
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Third Newton Law
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action-reaction
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pressure=
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force/area
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Ffk=
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Mk*Fn
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Ffs=
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Ms*Fn (max)
Fpull (less than max) |
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First Kepler Law
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path is elliptical
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Second Kepler Law
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Equal area in equal time
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Third Kepler Law
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T^2 is proportional to r^3
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t^2 is proportional to
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r^3
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G=
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6.67*10^-11
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w=
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fxcos(theta)
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w=
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ΔKE
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w=
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ΔPE
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x axis-distance
y axis- force concerns a spring slope= |
spring constant k
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conservation of energy
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total energy is conserved
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FnetΔt=
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mΔv=impulse
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torque=
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rFsintheta
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Pavg=
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W/Δt
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F1/A1=
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F2/A2=constant pressure
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p(liquid)=
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ρgh
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Fg
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ρVg
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p(total)=
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patm + ρgh
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a submerged object displaces a volume of water
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equal to its own volume
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flow rate=
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vA
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v1A1=
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v2A2
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P1 + ρgh1 + .5ρ(v1)^2
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P2 + ρgh2 + .5ρ(v2)^2
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v=
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root(2gh)
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ρ=
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m/v
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pressure=
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force/area
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specific gravity=
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ρsubstance/ρwater
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Fb=
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ρfluid * Vsub * g
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Vsub/V
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ρobject/ρfluid
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the pressure is lower where
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the flow speed is greater
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isochoric
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no change in volume
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isobaric
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no change in pressure
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isothermal
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no change in temperature (not as steep as adiabatic)
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adiabatic
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no change in Q (steeper than isothermal)
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W/Qh=
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1-Qc/Qh or Qh-Qc/Qh
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.0821
8.314 |
meters and atm
meters^3 and pascals |
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Fe=
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kQq/r^2
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E=
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Fe/q=KQ/r^2
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ΔV=
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-EΔd
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V=
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kQ/r
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Ue=
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kqq/r
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qΔV=
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ΔUe=W
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C=
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Q/V
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EoA/d=
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Q/V=C
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Cp=
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C1 + C2 + C3 + C4
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1/Cs=
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1/C1 + 1/C2 + 1/C3 + 1/C4
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positive particles go toward areas of
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lower potential
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negative particles go toward areas of
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higher potential
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electric field lines point toward areas of
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lower electric potential
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Ue=
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-qEΔr=-W
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Ucapacitor=
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.5QV=.5CV^2=Q^2/2C
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V=
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EΔr
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I (avg)=
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ΔQ/Δt
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V (potential difference)=
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I (current) * R (resistance)
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R (resistance)=
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PL/A (special constant)*(length)/(area)
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P (power- energy used per second)=
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IV=I^2R=V^2/R
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resistors in series share the same
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current
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resistors in parallel share the same
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voltage drop
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Rp=
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1/R1 + 1/R2 + 1/R3
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Rs=
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R1 + R2 + R3
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internal resistance
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resistance right next to battery (technically a part of it)
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terminal voltage
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lower than ideal voltage (which would have no internal resitance)... subtracts internal resistance
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the loop rule
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the sum of V's in a closed loop is 0
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the junction rule
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the total current that enters a junction is equal to the total current that leaves a junction
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when going across a resistor in the same direction as the current
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V drops by IR
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when going across a resistor in the opposite direction from the current
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V increases by IR
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when going from the negative to positive side of a battery
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add V
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when going from the positive to negative side of a battery
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subtract V
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capacitors in resistance circuits
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-when just closed have no effect
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-when closed for a long time
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they block current
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energy used by a resistor
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power*time
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magnetic field lines point away from
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north
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magnetic field lines point toward
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south
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X X X X X X X X X
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magnetic field going into the page
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. . . . . . . . .
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magnetic field going out of the page
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Fb=
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!q!vBsin(theta)
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if q is positive
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use right hand
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if q is negative
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use left hand
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right/left hand rule
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-thumb points in the direction of v
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Fb=0 when
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v=0
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magnetic forces can do what to speed and direction
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cant change speed
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Fc=
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mv^2/r=Fb=qvB
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qE=
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Fb=qvb
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Fb (wire)=
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BILsin(theta)
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right hand rule (wire)
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thumb points with the current
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B (wire)=
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ui/2pi(r)
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right hand rule (single straight wire)
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thumb points toward current
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potential difference of wire within magnetic field=
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EL=vBL
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motional emf
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E=
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P (that external agent supplies)=
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IV(ba)=IE(squiggly)=ILBv
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P (electrical power delivered to the circuit)=
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IV(ba)=IE(squiggly)=IvBL
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ΦB (magnetic flux)=
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BAcostheta
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Eavg=
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-ΔΦB/Δt
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RHR for induction
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thumb=velocity
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T=
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1/f
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f=
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1/T
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what is a period?
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the amount of time it takes to complete a cycle
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what is frequency?
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the amount of cycles completed in one unit of time
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f=
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-kx
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Us=
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.5kx^2
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f (of a spring)=
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(1/2pi)(k/m)^1/2
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T (of a spring)=
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(2pi)(m/k)^1/2
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period and frequency of simple harmonic motion are independent of
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amplitude
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equilibrium in a vertical spring is kx=
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mg
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f (of a pendulum)=
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(1/2pi)(g/L)^.5
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T (of a pendulum)=
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(2pi)(L/g)^1/2
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harmonic motion objects have maximum acceleration when
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potential energy is at a maximum
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simple harmonic motion occurs when
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the acceleration/restoring force is directly proportional to the displacement
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the velocity of motion is greatest at the
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middle part of motion
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F restoring (pendulum)=
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mgsin(theta)
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a (pendulum)=
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gsin(theta)=gs/l
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E(nuclear)=
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hf=pc
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Kmax (nuclear)=
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hf-(work function)
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wavelength (nuclear)=
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h/p
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ΔE (nuclear)=
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(Δm)c^2
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v(wave)=
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fd
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n=
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c/v
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n1sintheta1=
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n2sintheta2
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sintheta(c)=
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n2/n1
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1/si + 1/s0=
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1/f
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M=
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hi/h0=-si/s0
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f=
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R/2
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dsintheta=
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m*wavelength
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xm=
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m*L*wavelength/d
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