Physics Formulas (Situational)

0.0(0)
Studied by 0 people
call kaiCall Kai
Locked
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/96

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 2:19 AM on 7/19/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

97 Terms

1
New cards

Given: Initial velocity v0v_0 , final velocity vfv_f , and time tt . Find: Kinematic average displacement Δx\Delta x .

Δx=v0+vf2t\Delta x = \frac{v_0 + v_f}{2}t

2
New cards

Given: Change in velocity Δv\Delta v and time tt . Find: Acceleration aa .

a=Δvta = \frac{\Delta v}{t}

3
New cards

Given: Masses of multiple objects m1,m2,m_1, m_2, \dots and their positions x1,x2,x_1, x_2, \dots . Find: Center of mass xcmx_{cm} .

xcm=m1x1+m2x2+m1+m2+x_{cm} = \frac{m_1x_1 + m_2x_2 + \dots}{m_1 + m_2 + \dots}

4
New cards

Given: Force FF , lever arm distance rr , and angle θ\theta . Find: Torque τ\tau .

τ=rFsin(θ)\tau = rF\sin(\theta)

5
New cards

Given: Output force FoutF_{out} and input force FinF_{in} . Find: Mechanical advantage MAMA .

MA=FoutFinMA = \frac{F_{out}}{F_{in}}

6
New cards

Given: Initial kinetic and potential energy KEi,PEiKE_i, PE_i . Find: Final kinetic and potential energy KEf,PEfKE_f, PE_f in a closed system.

KEi+PEi=KEf+PEfKE_i + PE_i = KE_f + PE_f

7
New cards

Given: Kinetic energy KEKE and potential energy PEPE . Find: Total mechanical energy EE .

E=KE+PEE = KE + PE

8
New cards

Given: Change in total mechanical energy ΔE\Delta E . Find: Work done by non-conservative forces WncW_{nc} .

Wnc=ΔEW_{nc} = \Delta E

9
New cards

Given: Force F1F_1 applied to an area A1A_1 in a confined fluid. Find: Resulting force F2F_2 on a second area A2A_2 .

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

10
New cards

Given: Density of fluid ρ\rho , volume VV , and gravity gg . Find: Weight of the fluid WfluidW_{fluid} .

Wfluid=ρVgW_{fluid} = \rho V g

11
New cards

Given: Force FF applied perpendicular to a surface area AA . Find: Pressure PP .

P=FAP = \frac{F}{A}

12
New cards

Given: Original length LL , coefficient of linear expansion α\alpha , and change in temperature ΔT\Delta T . Find: Change in length ΔL\Delta L .

ΔL=αLΔT\Delta L = \alpha L \Delta T

13
New cards

Given: Original volume VV , coefficient of volumetric expansion β\beta , and change in temperature ΔT\Delta T . Find: Change in volume ΔV\Delta V .

ΔV=βVΔT\Delta V = \beta V \Delta T

14
New cards

Given: Constant pressure PP and change in volume ΔV\Delta V . Find: Work done by the gas WW .

W=PΔVW = P \Delta V

15
New cards

Given: Reversible heat transfer QQ at a constant temperature TT . Find: Change in entropy ΔS\Delta S .

ΔS=QT\Delta S = \frac{Q}{T}

16
New cards

Given: Amount of charge Δq\Delta q passing through a point over time Δt\Delta t . Find: Current II .

I=ΔqΔtI = \frac{\Delta q}{\Delta t}

17
New cards

Given: Voltage drops across individual components V1,V2,V_1, V_2, \dots in a series circuit. Find: Total voltage VsV_s .

Vs=V1+V2+V_s = V_1 + V_2 + \dots

18
New cards

Given: Voltage across one branch V1V_1 in a parallel circuit. Find: Total voltage VpV_p .

Vp=V1=V2=V_p = V_1 = V_2 = \dots

19
New cards

Given: Capacitances C1,C2,C_1, C_2, \dots connected in series. Find: Equivalent capacitance CeqC_{eq} .

1Ceq=1C1+1C2+\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots

20
New cards

Given: Capacitances C1,C2,C_1, C_2, \dots connected in parallel. Find: Equivalent capacitance CeqC_{eq} .

Ceq=C1+C2+C_{eq} = C_1 + C_2 + \dots

21
New cards

Given: Voltage across plates VV and distance between them dd . Find: Electric field strength in a capacitor EE .

E=VdE = \frac{V}{d}

22
New cards

Given: Current II in a long straight wire, permeability of free space μ0\mu_0 , and distance rr from the wire. Find: Magnetic field BB .

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

23
New cards

Given: Current II in a circular loop of radius rr , and the permeability of free space μ0\mu_0 . Find: Magnetic field at the center of the loop BB .

B=μ0I2rB = \frac{\mu_0 I}{2r}

24
New cards

Given: Charge qq moving with velocity vv at an angle θ\theta in a magnetic field BB . Find: Magnetic force FBF_B .

FB=qvBsin(θ)F_B = qvB\sin(\theta)

25
New cards

Given: Current II , length of wire LL , angle θ\theta , and magnetic field BB . Find: Magnetic force on the wire FBF_B .

FB=ILBsin(θ)F_B = ILB\sin(\theta)

26
New cards

Given: Frequency ff or period TT . Find: Angular frequency ω\omega .

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}

27
New cards

Given: Bulk modulus of the medium BB and its density ρ\rho . Find: Speed of sound vv .

v=Bρv = \sqrt{\frac{B}{\rho}}

28
New cards

Given: Power of the source PP and the area it is spread over AA . Find: Wave intensity II .

I=PAI = \frac{P}{A}

29
New cards

Given: Final intensity IfI_f and initial intensity IiI_i . Find: Change in sound level/decibels Δβ\Delta \beta .

Δβ=10log(IfIi)\Delta \beta = 10 \log\left(\frac{I_f}{I_i}\right)

30
New cards

Given: Length of the string/pipe LL and harmonic number n=1,2,3,n = 1, 2, 3, \dots . Find: Wavelength for string and open pipes λ\lambda .

λ=2Ln\lambda = \frac{2L}{n}

31
New cards

Given: Length of the closed pipe LL and odd harmonic number n=1,3,5,n = 1, 3, 5, \dots . Find: Wavelength for closed pipes λ\lambda .

λ=4Ln\lambda = \frac{4L}{n}

32
New cards

Given: Wave speed vv , length LL , and harmonic number n=1,2,3,n = 1, 2, 3, \dots . Find: Frequency for string and open pipes ff .

f=nv2Lf = \frac{nv}{2L}

33
New cards

Given: Wave speed vv , length LL , and odd harmonic number n=1,3,5,n = 1, 3, 5, \dots . Find: Frequency for closed pipes ff .

f=nv4Lf = \frac{nv}{4L}

34
New cards

Given: Slit width aa , order of minimum nn , and wavelength λ\lambda . Find: Angle to the dark fringe in single-slit diffraction θ\theta .

asin(θ)=nλa \sin(\theta) = n\lambda

35
New cards

Given: Distance between slits dd , order nn , and wavelength λ\lambda . Find: Angle to the dark fringe in double-slit interference θ\theta .

dsin(θ)=(n+12)λd \sin(\theta) = \left(n + \frac{1}{2}\right)\lambda

36
New cards

Given: Index of refraction of the first medium n1n_1 and second medium n2n_2 . Find: Critical angle for total internal reflection θc\theta_c .

θc=sin1(n2n1)\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)

37
New cards

Given: Index of refraction nn and radii of curvature R1,R2R_1, R_2 . Find: Focal length using the Lensmaker's equation ff .

1f=(n1)(1R11R2)\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

38
New cards

Given: Focal lengths of individual lenses in contact f1,f2,f_1, f_2, \dots . Find: Equivalent focal length feqf_{eq} .

1feq=1f1+1f2+\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \dots

39
New cards

Given: Powers of individual lenses in contact P1,P2,P_1, P_2, \dots . Find: Equivalent power PeqP_{eq} .

Peq=P1+P2+P_{eq} = P_1 + P_2 + \dots

40
New cards

Given: Magnifications of individual lenses m1,m2,m_1, m_2, \dots . Find: Total magnification MtotalM_{total} .

Mtotal=m1×m2×M_{total} = m_1 \times m_2 \times \dots

41
New cards

Given: Mass mm and the speed of light cc . Find: Equivalent rest energy EE .

E=mc2E = mc^2

42
New cards

Given: A parent nucleus ZAX^{A}_{Z}X . Find: The products of beta-positive (positron) decay.

ZAXZ1AY+e++νe^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y + e^+ + \nu_e

43
New cards

Given: An excited parent nucleus ZAX^{A}_{Z}X^* . Find: The products of gamma decay.

ZAXZAX+γ^{A}_{Z}X^* \rightarrow ^{A}_{Z}X + \gamma

44
New cards

Given: A parent nucleus ZAX^{A}_{Z}X absorbing an inner-shell electron ee^- . Find: The products of electron capture.

ZAX+eZ1AY+νe^{A}_{Z}X + e^- \rightarrow ^{A}_{Z-1}Y + \nu_e

45
New cards

Given: Decay constant λ\lambda and number of undecayed nuclei NN . Find: The rate of nuclear decay ΔNΔt\frac{\Delta N}{\Delta t} .

ΔNΔt=λN\frac{\Delta N}{\Delta t} = -\lambda N

46
New cards

Given: Initial velocity ( v0v_0 ), acceleration ( aa ), and time ( tt ). Find: Final velocity ( vfv_f ) without knowing displacement.

vf=v0+atv_f = v_0 + at

47
New cards

Given: Initial velocity ( v0v_0 ), acceleration ( aa ), and displacement ( Δx\Delta x ). Find: Final velocity ( vfv_f ) without knowing time.

vf=v02+2aΔxv_{f}^{}=\sqrt{v_0^2+2a\Delta x}

48
New cards

Given: Initial velocity ( v0v_0 ), acceleration ( aa ), and time ( tt ). Find: Displacement ( Δx\Delta x ).

Δx=v0t+12at2\Delta x = v_0t + \frac{1}{2}at^2

49
New cards

Given: An object is launched at an angle θ\theta with initial velocity vv . Find: The vertical and horizontal components of velocity.

Vertical: vy=vsinθv_y = v\sin\theta ; Horizontal: vx=vcosθv_x = v\cos\theta

50
New cards

Given: The mass of an object and its acceleration. Find: The net force acting on the object.

Newton's Second Law: Fnet=maF_{net} = ma

51
New cards

Given: The coefficient of static friction ( μs\mu_s ) and the normal force ( FNF_N ). Find: The maximum force that can be applied before the object begins to move.

Maximum static friction: fs,max=μsFNf_{s, max} = \mu_s F_N

52
New cards

Given: Mass of an object ( m1m_1 ), mass of a planet ( m2m_2 ), and the distance between their centers ( rr ). Find: The gravitational force of attraction.

Newton's Law of Universal Gravitation: Fg=Gm1m2r2F_g = \frac{Gm_1m_2}{r^2}

53
New cards

Given: Mass ( mm ), velocity ( vv ), and radius of a circular path ( rr ). Find: The force keeping the object in uniform circular motion.

Centripetal Force: Fc=mv2rF_c = \frac{mv^2}{r}

54
New cards

Given: An object of mass mm on an inclined plane with angle θ\theta. Find: The components of gravity parallel and perpendicular to the plane.

Parallel to plane: Fg=mgsinθF_{g\parallel} = mg\sin\theta ; Perpendicular to plane (Normal Force if flat): Fg=mgcosθF_{g\perp} = mg\cos\theta

55
New cards

Given: Force applied ( FF ), distance traveled ( dd ), and the angle between the force and displacement vectors ( θ\theta ). Find: Work done.

W=FdcosθW = Fd\cos\theta

56
New cards

Given: The mass of an object (m) and its velocity (v). Find: The kinetic energy associated with its motion (KE).

Kinetic Energy: KE=12mv2KE = \frac{1}{2}mv^2

57
New cards

Given: Mass ( mm ), height relative to a datum ( hh ), and gravity ( gg ). Find: Energy associated with a objects position.

Gravitational Potential Energy: U=mghU = mgh

58
New cards

Given: The spring constant ( kk ) and the displacement from equilibrium ( xx ). Find: The energy stored in the spring.

Elastic Potential Energy: U=12kx2U = \frac{1}{2}kx^2

59
New cards

Given: The net work done on an object. Find: The change in the object's speed/kinetic energy.

Work-Energy Theorem: Wnet=ΔKE=KEfKEiW_{net} = \Delta KE = KE_f - KE_i

60
New cards

Given: Work done ( WW ) and the time it took to do it ( tt ), OR Force ( FF ) and velocity ( vv ). Find: Power output.

P=Wt=FvcosθP = \frac{W}{t} = Fv\cos\theta

61
New cards

Given: Heat added to a system ( QQ ) and work done by the system ( WW ). Find: The change in internal energy ( ΔU\Delta U ).

First Law of Thermodynamics: ΔU=QW\Delta U = Q - W

62
New cards

Given: Mass of a substance ( mm ), its specific heat capacity ( cc ), and a change in temperature ( ΔT\Delta T ). Find: Heat gained or lost ( qq ) without a phase change.

q=mcΔTq = mc\Delta T

63
New cards

Given: Mass of a substance ( mm ) and its latent heat of fusion or vaporization ( LL ). Find: Heat required to change its phase.

Phase change heat: q=mLq = mL (Remember: Temperature does not change during a phase change).

64
New cards

Relationship: If a gas expands and does work on its surroundings, what happens to the sign of Work ( WW ) in the First Law equation?

Work done by the system is positive ( +W+W ). Expansion = positive work. Compression = negative work.

65
New cards

Given: The density of a fluid ( ρ\rho ), gravity ( gg ), and depth below the surface ( hh ). Find: The absolute pressure at that depth. (Hint: You need to know atmospheric pressure)

Hydrostatic Pressure: P=P0+ρghP = P_0 + \rho gh (where P0P_0 is surface/atmospheric pressure).
Atmospheric pressure = 1 atm = 760 mmHg = 760 torr = 1 × 10^5 Pa

66
New cards

Given: The density of a fluid ( ρfluid\rho_{fluid} ), and the volume of the object submerged ( VsubV_{sub} ). Find: The upward force on the object.

Buoyant Force (Archimedes' Principle): FB=ρfluidVsubgF_B = \rho_{fluid} V_{sub} g

67
New cards

Given: Cross-sectional area of a pipe ( A1A_1 ) and fluid velocity ( v1v_1 ) at point 1, and the area ( A2A_2 ) at point 2. Find: Velocity at point 2.

Continuity Equation: A1v1=A2v2A_1v_1 = A_2v_2

68
New cards

Given: Pressures, heights, and fluid velocities at two different points in a closed pipe system. Find: A missing pressure or velocity.
(Hint: You must know the density of the liquid)

Bernoulli's Equation: P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2

Density (ρ\rho) = ρwSG\rho_{w}\cdot SG (SG: Specific Gravity Ratio)

Density of Water (ρw\rho_{w} ): 1000 kg/m³ or 1 kg/L

69
New cards

Relationship: Fluid flows through a rigid pipe. If the radius of the pipe is reduced by half, what happens to the flow rate ( QQ ) assuming constant pressure difference?

Flow rate decreases by a factor of 16. (Poiseuille's Law: Qr4Q \propto r^4 ).

70
New cards

Given: The magnitude of two point charges ( q1,q2q_1, q_2 ) and the distance between them ( rr ). Find: The electrostatic force between them.
(Hint: You need to know Coulomb’s Constant (k) )

Coulomb's Law: Fe=kq1q2r2F_e = \frac{k\vert{}q_1q_2\vert{}}{r^2}

Coulomb’s Constant (k) = 9 × 109 N */ C²

71
New cards

Given: A source charge ( QQ ) and the distance from it ( rr ). Find: The magnitude of the electric field created by the charge. (Hint: You need Coulomb’s Constant)

Electric Field: E=kQr2E = \frac{k\vert{}Q\vert{}}{r^2}

Coulomb’s Constant (k) = 9 × 109 N * m2 / C2

72
New cards

Given: The electric field strength ( EE ) and a test charge ( qq ) placed within it. Find: The force acting on the test charge.

Fe=qEF_e = qE

73
New cards

Given: A source charge ( QQ ) and distance ( rr ). Find: The electrical potential (Voltage) at that point in space. (Hint: you need coulomb’s constant)

Electrical Potential: V=kQrV = \frac{kQ}{r}

Coulomb’s constant (k) = 9 × 109 N * m² / C²

74
New cards

Given: A charge ( qq ) and the electrical potential/voltage ( VV ) it experiences. Find: The electrical potential energy.

U=qVU = qV

75
New cards

Given: The current ( II ) flowing through a resistor and its resistance ( RR ). Find: The voltage drop across it.

Ohm's Law: V=IRV = IR

76
New cards

Given: The resistivity of a material ( ρ\rho ), its length ( LL ), and cross-sectional area ( AA ). Find: Its resistance.

R=ρLAR = \frac{\rho L}{A}

77
New cards

Given: Voltage ( VV ) and Current ( II ), OR Current and Resistance ( RR ). Find: The power dissipated by a resistor.

P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}

78
New cards

Relationship: You have three resistors ( R1,R2,R3R_1, R_2, R_3 ). How do you calculate total equivalent resistance if they are in Series vs. Parallel?

Series: Req=R1+R2+R3R_{eq} = R_1 + R_2 + R_3 ; Parallel: 1Req=1R1+1R2+1R3\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

79
New cards

Given: The charge stored on a capacitor ( QQ ) and the voltage across it ( VV ). Find: Capacitance.

C=QVC = \frac{Q}{V}

80
New cards

Given: The area of capacitor plates ( AA ), distance between them ( dd ), the permittivity of free space (ϵ0\epsilon_0), and dielectric constant ( κ\kappa ). Find: Capacitance (C).

C=κϵ0AdC = \frac{\kappa \epsilon_0 A}{d}

81
New cards

Given: Capacitance ( CC ) and Voltage ( VV ). Find: Energy stored in the capacitor.

U=12CV2U = \frac{1}{2}CV^2

82
New cards

Given: The frequency ( ff ) and wavelength ( λ\lambda ) of a wave. Find: Wave propagation speed.

v=fλv = f\lambda

83
New cards

Relationship: What is the relationship between the period of a wave ( TT ) and its frequency ( ff )?

Inverse: f=1Tf = \frac{1}{T} or T=1fT = \frac{1}{f}

84
New cards

Given: The velocity of a sound source ( vsv_s ), the velocity of the observer ( vov_o ), and the speed of sound ( vv ). Find: The perceived frequency ( ff' ).

Doppler Effect: f=f(v±vovvs)f' = f \left(\frac{v \pm v_o}{v \mp v_s}\right) (Top signs when moving toward, bottom signs when moving away).

85
New cards

Relationship: A sound source moves away from you. What happens to the perceived distance between wave peaks?

Wavelength increases (sound perceived drops to a lower frequency/pitch).

86
New cards

Given: The intensity of a sound wave ( II ). Find: The sound level ( β\beta ) in decibels (dB).

β=10log(II0)\beta = 10 \log\left(\frac{I}{I_0}\right) (where I0=1012W/m2I_0 = 10^{-12} W/m^2 ).

87
New cards

Given: The speed of light in a vacuum ( cc ) and the speed of light in a specific medium ( vv ). Find: The index of refraction ( nn ).

n=cvn = \frac{c}{v}

88
New cards

Given: The index of refraction of medium 1 ( n1n_1 ), medium 2 ( n2n_2 ), and the angle of incidence ( θ1\theta_1 ). Find: The angle of refraction ( θ2\theta_2 ).

Snell's Law: n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2

89
New cards

Given: The distance to the image ( did_i ) and the distance to the object ( dod_o ). Find: The focal length of a lens/mirror ( ff )

Thin Lens/Mirror Equation: 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

90
New cards

Given: Image distance ( did_i ) and object distance ( dod_o ). Find: Magnification ( mm ).

m=didom = -\frac{d_i}{d_o}

91
New cards

Relationship: If the magnification ( mm ) of a lens is a negative number, what does that tell you about the image?

The image is inverted and real. (Positive mm = upright and virtual).

92
New cards

Given: The focal length of a lens ( ff ) in meters. Find: The power of the lens ( PP ) in diopters.

P=1fP = \frac{1}{f}

93
New cards

Given: Planck's Constant (hh) and the frequency of a photon ( ff ) or its wavelength ( λ\lambda ). Find: The energy of the photon.
(Hint: You might need the speed of light)

E=hf=hcλE = hf = \frac{hc}{\lambda}
Speed of light (cc) = 3 × 108 m/s

94
New cards

Given: The energy of an incident photon ( E=hfE = hf ) and the work function of a metal ( WW ). Find: The maximum kinetic energy of the ejected electron.

Photoelectric Effect: Kmax=hfWK_{max} = hf - W

95
New cards

Relationship: An atom undergoes Alpha Decay. What happens to its mass number and atomic number?

Mass number decreases by 4. Atomic number decreases by 2. (Ejects a Helium nucleus).

96
New cards

Relationship: An atom undergoes Beta-Minus Decay. What happens to its mass number and atomic number?

Mass number remains unchanged. Atomic number increases by 1. (A neutron is converted into a proton)

97
New cards

Given: Displacement of spring (x) and spring constant (k) Find: Spring Force (Fs)

Hooke’s Law: Fs = kx-kx