Geschreven door studenten die geslaagd zijn Direct beschikbaar na je betaling Online lezen of als PDF Verkeerd document? Gratis ruilen 4,6 TrustPilot
logo-home
Interview

Formula physics

Beoordeling
-
Verkocht
-
Pagina's
11
Geüpload op
04-05-2023
Geschreven in
2022/2023

Interview study book NCERT Solutions Physics Class 11th of Nipendra Bhatnagar - ISBN: 9789351416340 (Physics formula)

Instelling
Vak

Voorbeeld van de inhoud

Formulae S he e t f o r P h ys i c s w w w .c o n c e p t s - o f -p h y s i c s . c om | pg. 1

Physics formulas from Mechanics, Waves, Optics, Heat and Motion in a straight line with constant a:
Thermodynamics, Electricity and Magnetism and Modern
Physics. Also includes the value of Physical Constants. Helps v = u + at, s = ut + 21 at2 , v 2 − u2 = 2as
in quick revision for CBSE, NEET, JEE Mains, and Advanced.
Relative Velocity: ~vA/B = ~vA − ~vB
0.1: Physical Constants

Speed of light c 3 × 108 m/s
u
Planck constant h 6.63 × 10−34 J s y




u sin θ
hc 1242 eV-nm H
x
Projectile Motion:
Gravitation constant G 6.67×10−11 m3 kg−1 s−2 θ
Boltzmann constant k 1.38 × 10−23 J/K O u cos θ
Molar gas constant R 8.314 J/(mol K) R

Avogadro’s number NA 6.023 × 1023 mol−1
Charge of electron e 1.602 × 10−19 C x = ut cos θ, y = ut sin θ − 21 gt2
Permeability of vac- µ0 4π × 10−7 N/A2 g
y = x tan θ − 2 x2
uum 2u cos2 θ
Permitivity of vacuum 0 8.85 × 10−12 F/m 2u sin θ u2 sin 2θ u2 sin2 θ
T = , R= , H=
Coulomb constant 1
4π0 9 × 109 N m2 /C2 g g 2g
Faraday constant F 96485 C/mol
Mass of electron me 9.1 × 10−31 kg 1.3: Newton’s Laws and Friction
Mass of proton mp 1.6726 × 10−27 kg
Mass of neutron mn 1.6749 × 10−27 kg Linear momentum: p~ = m~v
Atomic mass unit u 1.66 × 10−27 kg
Atomic mass unit u 931.49 MeV/c2 Newton’s first law: inertial frame.
Stefan-Boltzmann σ 5.67×10−8 W/(m2 K4 )
Newton’s second law: F~ = d~
p
dt , F~ = m~a
constant
Rydberg constant R∞ 1.097 × 107 m−1 Newton’s third law: F~AB = −F~BA
Bohr magneton µB 9.27 × 10−24 J/T
Bohr radius a0 0.529 × 10−10 m Frictional force: fstatic, max = µs N, fkinetic = µk N
Standard atmosphere atm 1.01325 × 105 Pa
v2 v2
Wien displacement b 2.9 × 10−3 m K Banking angle: rg = tan θ, rg = µ+tan θ
1−µ tan θ
constant
mv 2 v2
Centripetal force: Fc = r , ac = r
2

1 MECHANICS Pseudo force: F~pseudo = −m~a0 , Fcentrifugal = − mv
r

Minimum speed to complete vertical circle:
1.1: Vectors
p p
vmin, bottom = 5gl, vmin, top = gl
Notation: ~a = ax ı̂ + ay ̂ + az k̂
q
Magnitude: a = |~a| = a2x + a2y + a2z θ
l
q
l cos θ
Conical pendulum: T = 2π θ T
Dot product: ~a · ~b = ax bx + ay by + az bz = ab cos θ g


ı̂ mg
a × ~b
~ ~b
Cross product:
θ k̂ ̂
~
a
1.4: Work, Power and Energy
~a ×~b = (ay bz − az by )ı̂ + (az bx − ax bz )̂ + (ax by − ay bx )k̂ Work: W = F~ · S
~ = F S cos θ, W =
R
F~ · dS
~

|~a × ~b| = ab sin θ Kinetic energy: K = 12 mv 2 = p2
2m

Potential energy: F = −∂U/∂x for conservative forces.
1.2: Kinematics
Ugravitational = mgh, Uspring = 21 kx2
Average and Instantaneous Vel. and Accel.:

~vav = ∆~r/∆t, ~vinst = d~r/dt Work done by conservative forces is path indepen-
dent and depends only on initial and final points:
~aav = ∆~v /∆t ~ainst = d~v /dt H
F~conservative · d~r = 0.

Work-energy theorem: W = ∆K



Get Formulas www.concepts-of-physics.com Get Our Book
c 2020 by Jitender Singh Ver. 2020 1

, Formulae S he e t f o r P h ys i c s w w w .c o n c e p t s - o f -p h y s i c s . c om | pg. 2

Mechanical energy: E = U + K. Conserved if forces are Rotation about an axis with constant α:
conservative in nature.
ω = ω0 + αt, θ = ωt + 21 αt2 , ω 2 − ω0 2 = 2αθ
Power Pav = ∆W
∆t , Pinst = F~ · ~v

mi ri 2 , r2 dm
P R
Moment of Inertia: I = i I=
1.5: Centre of Mass and Collision
P R
R xdm
2
Pxi mi ,
1
2 mr m(a +b )
2 2
Centre of mass: xcm = xcm = mr 2 1 2 2 2 2 2 1 2 mr 2
mi dm 2 mr 3 mr 5 mr 12 ml 12


b
a
CM of few useful configurations: ring disk shell sphere rod hollow solid rectangle

m1 r m2
1. m1 , m2 separated by r: C
m2 r m1 r Ik Ic
m1 +m2 m1 +m2 2
Theorem of Parallel Axes: Ik = Icm + md d
cm
h
2. Triangle (CM ≡ Centroid) yc = 3 h
C
h
3
z y
Theorem of Perp. Axes: Iz = Ix + Iy
2r
x
3. Semicircular ring: yc = π
C
2r
r π
p
Radius of Gyration: k = I/m
4r
4. Semicircular disc: yc = 3π C 4r
r ~ = ~r × p~, ~ = I~
3π Angular Momentum: L L ω
r y
5. Hemispherical shell: yc = 2 C r ~ P θ ~
r 2
Torque: ~τ = ~r × F~ , ~τ = dL
dt , τ = Iα F
~
r x
O
3r
6. Solid Hemisphere: yc = 8 C 3r
r 8 ~ ~τext = 0 =⇒ L ~ = const.
Conservation of L:
7. Cone: the height of CM from the base is h/4 for Equilibrium condition:
P~
F = ~0,
P
~τ = ~0
the solid cone and h/3 for the hollow cone.
Kinetic Energy: Krot = 12 Iω 2

Motion of the CM: M =
P
mi Dynamics:
P
mi~vi F~ext ~τcm = Icm α
~, F~ext = m~acm , p~cm = m~vcm
~vcm = , p~cm = M~vcm , ~acm = 1 2 1 2 ~
M M ~ + ~rcm × m~vcm
K = 2 mvcm + 2 Icm ω , L = Icm ω

Impulse: J~ = F~ dt = ∆~
R
p
1.7: Gravitation
Before collision After collision
Collision: m1 m2 m1 m2 m1 F F m2
Gravitational force: F = G mr1 m
2
2

v1 v2 v10 v20 r
Momentum conservation: m1 v1 +m2 v2 = m1 v10 +m2 v20
2
Elastic Collision: 12 m1 v1 2+ 12 m2 v2 2 = 12 m1 v10 + 12 m2 v20
2 Potential energy: U = − GMr m
Coefficient of restitution: GM
Gravitational acceleration: g = R2
−(v10 − v20 )

1, completely elastic
e= = h

v1 − v2 0, completely in-elastic Variation of g with depth: ginside ≈ g 1 − R

2h

Variation of g with height: goutside ≈ g 1 −
If v2 = 0 and m1
m2 then v10 = −v1 . R

If v2 = 0 and m1
m2 then v20 = 2v1 . Effect of non-spherical earth shape on g:
Elastic collision with m1 = m2 : v10 = v2 and v20 = v1 . gat pole > gat equator (∵ Re − Rp ≈ 21 km)

Effect of earth rotation on apparent weight:
1.6: Rigid Body Dynamics
∆θ dθ
Angular velocity: ωav = ∆t , ω= dt , ~ × ~r
~v = ω
∆ω dω
Angular Accel.: αav = ∆t , α= dt , ~ × ~r
~a = α



Get Formulas www.concepts-of-physics.com Get Our Book
c 2020 by Jitender Singh Ver. 2020 1

Gekoppeld boek

Geschreven voor

Instelling
Middelbare school
School jaar
1

Documentinformatie

Geüpload op
4 mei 2023
Aantal pagina's
11
Geschreven in
2022/2023
Type
INTERVIEW
Bedrijf
Onbekend
Persoon
Onbekend

Onderwerpen

€7,10
Krijg toegang tot het volledige document:

Verkeerd document? Gratis ruilen Binnen 14 dagen na aankoop en voor het downloaden kun je een ander document kiezen. Je kunt het bedrag gewoon opnieuw besteden.
Geschreven door studenten die geslaagd zijn
Direct beschikbaar na je betaling
Online lezen of als PDF

Maak kennis met de verkoper
Seller avatar
opgoggle123

Maak kennis met de verkoper

Seller avatar
opgoggle123 Exam Questions
Volgen Je moet ingelogd zijn om studenten of vakken te kunnen volgen
Verkocht
-
Lid sinds
2 jaar
Aantal volgers
0
Documenten
1
Laatst verkocht
-

0,0

0 beoordelingen

5
0
4
0
3
0
2
0
1
0

Populaire documenten

Recent door jou bekeken

Waarom studenten kiezen voor Stuvia

Gemaakt door medestudenten, geverifieerd door reviews

Kwaliteit die je kunt vertrouwen: geschreven door studenten die slaagden en beoordeeld door anderen die dit document gebruikten.

Niet tevreden? Kies een ander document

Geen zorgen! Je kunt voor hetzelfde geld direct een ander document kiezen dat beter past bij wat je zoekt.

Betaal zoals je wilt, start meteen met leren

Geen abonnement, geen verplichtingen. Betaal zoals je gewend bent via iDeal of creditcard en download je PDF-document meteen.

Student with book image

“Gekocht, gedownload en geslaagd. Zo makkelijk kan het dus zijn.”

Alisha Student

Veelgestelde vragen