Physics Formulas

Physics Formulas – List of all Physics Formulas

Physics is that branch of science that basically deals with the prime constituents of matter, its characteristics and behaviour and different related elements of force and energy. Physics formulas are not meant to be memorised but conveyed in real-time and space. The application of the formulas of physics includes the concepts of mathematics and its formulas.

It is also important to remember that if one is unable to crack the theoretical part of physics, that is, understanding the theories properly, one can never find any relation between the formulas. In order to make the understanding easier, it is also recommended to know the S.I. units of Physics well.

List of All Physics Formulas

Given below is the list of all Physics formulas:

Important Physics Formulas

Given below is the most important Physics formulae list:

• Planck constant h = 6.63 × 10−34 J.s = 4.136 × 10-15 eV.s
• Gravitation constant G = 6.67×10−11 m3 kg−1 s−2
• Boltzmann constant k = 1.38 × 10−23 J/K
• Molar gas constant R = 8.314 J/(mol K)
• Avogadro’s number NA = 6.023 × 1023 mol−1
• Charge of electron e = 1.602 × 10−19 C
• Permittivity of vacuum 0 = 8.85 × 10−12 F/m
• Coulomb constant 1/4πε0 = 8.9875517923(14) × 109 N m2/C2
• Faraday constant F = 96485 C/mol
• Mass of electron me = 9.1 × 10−31 kg
• Mass of proton mp = 1.6726 × 10−27 kg
• Mass of neutron mn = 1.6749 × 10−27 kg
• Stefan-Boltzmann constant σ = 5.67 × 10−8 W/(m2 K4)
• Rydberg constant R∞ = 1.097 × 107 m−1
• Bohr magneton µB = 9.27 × 10−24 J/T 
• Bohr radius a0 = 0.529 × 10−10 m 
• Standard atmosphere atm = 1.01325 × 105 Pa 
• Wien displacement constant b = 2.9 × 10−3 m K .
• Wave = ∆x ∆t wave = average velocity ∆x = displacement ∆t = elapsed time.
• Vavg = (vi + vf*)2

Vavg = The average velocity 

vi = initial velocity 

vf = final velocity

• a = ∆v ∆t,

a = acceleration 

∆v = change in velocity 

∆t = elapsed time.

• ∆x = vi∆t + 1/2 a(∆t)2

∆x = the displacement 

vi = the initial velocity 

∆t = the elapsed time 

a = the acceleration 

• ∆x = vf∆t − 1/2 a(∆t)2

∆x = displacement 

vf = is the final velocity 

∆t = elapsed time

a = acceleration 

• F = ma 

F = force 

m = mass 

a = acceleration

• W = mg 

W = weight 

m = mass 

g = acceleration which is due to gravity.

• f = µN 

f = friction force 

µ = coefficient of friction 

N = normal force 

• p = mv
• W = F d cos θ or W = F!d 

W = work t

F = force 

d = distance 

θ = angle between F and the direction of motion

• KE  = 1/2 mv2 K

KE = kinetic energy

m = mass

v = velocity

• PE = mgh 

PE = potential energy 

m = mass 

g = acceleration due to gravity 

h = height

• W = ∆(KE) 

W = work done 

KE = kinetic energy. 

• P = W ∆t 

P = power

W = work

∆t = elapsed time

Solved Examples

Q.1. Calculate the dc voltage drop if the circuit length is 500 cms and in it 10 A of current flows in 20 s ?

Solution: 

The DC voltage drop is given as : V=L×I/T

where, I = current through the circuit in Amperes.

L = Length of the circuit in metres

T = time for which the current has flowed through the circuit in seconds

V = Voltage in Volts.

So, 

V=500×10/20

V = 0.25 Volts.

Q.2. The spring constant of a stretched string is 50Nm−1 and displacement is 20 cm. Compute potential energy stored in the stretched string.

Solution: 

Given parameters are,

k=50Nm−1

x = 20 cm = 0.2 m

Potential energy will be:

P.E=1/2k×x2

=½ X 50×(0.2)2

= 1 J

Q.3. A body moves along the x- axis according to the relation x=1–2t+3t2x=1–2t+3t2x = 1 – 2 t + 3t^{2}, where x is in metres and t is in seconds. Find the Acceleration of the body when t = 3s

Solution: 

We have,

x=1–2t+3t2x=1–2t+3t2x = 1 – 2 t + 3t^{2}then;

Velocity v= dx/dt= −2+6t

v= dx/dt= −2+6t

Acceleration:  v=dv/dt =6(m/s2)

Q.4. Calculate the weight of an object on the moon which weighs 50kg on earth.

Solution: 

Here weight = mass x gravitational acceleration

= m x g

= 50 kg x 1.6 m/s2

= 80 Kg m/s2

Q.5. Find the displacement covered by an object which accelerates from rest to 60 m/s in 3s. 

Solution: 

Initial velocity = 0 and final velocity = 60 m/s

Time taken = 3s

Therefore, acceleration = 60/3 = 20m/s2.

Displacement (S) = ut + ½ at2

= 90 m

Q.6. A person goes from Point A to Point B in 10s and returns back in 8s. If the distance between A and B is 36m, find the average speed of the person.

Solution: 

Here total distance covered = 72m

Total time taken = 18s

Therefore, average speed  = Total distance covered/Total time taken

=72/18

Total distance covered total time taken=7218

= 4 m/s.

Q.7. If an object is moving with a velocity of 5 m/s and has a kinetic energy of 100J, Find its mass.

Solution: 

We know that KE = ½ mv2

100 = ½ x m x 5 x 5

100 = 25 m/2

m = 100×2/25 = 200/25

= 8 kg

Q.8. A long thin rod of length 50 cm has a total charge of 5 mC uniformly distributed over it. Find the linear charge density.

Solution:

q = 5 mC = 5 × 10–3 C, l= 50 cm = 0.5 m. λ=?

λ=ql

=(5×10^−3)/0.5

=(5×10^−3)/0.5

=10^-2Cm^−1

Q.9. A car is travelling with a velocity of 10 m/s and it has a mass of 250 Kg. Compute its Kinetic energy?

Solution: 

As given here,

Mass of the body, m = 250 Kg,

Velocity of it, v = 10 m/s,

So Kinetic energy will be as:

E = 1/2mv^2

E = 1/2×250×10^2

= 125 ×100

E = 12500 joules

Q.10. The heat needed for a phase transfer of a 2 kg substance is 400k.cal. Determine its latent heat.

Solution: 

Given parameters are,

Q = 400 k.cal

M = 2 kg

The formula for latent heat is given by,

L = Q / M

L = 400 / 2

L = 200 k.cal/kg