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July 4, 2015

Posts by Elena


Total # Posts: 4,369

physics
Let a material point of mass m is at a distance r from the center of the Earth. If r >R (the Earth's radius), then F = GmM / r ² where G - gravitational constant , and M - the mass of the Earth. If r <R, then the Earth’s action exerted on the point can be ...
October 30, 2013

Physics
F=ma
October 30, 2013

Physics
b) Ff c) m
October 30, 2013

Physics
F=ma=mv²/R
October 29, 2013

PHYSICS
KE=mv²/2 => v=sqrt(2KE/m) = =sqrt(2•11•10⁶•1.6•10⁻¹⁹/1.67•10⁻²⁷)=4.6•10⁷ m/s F=qvBsinα sinα=1 q=e F=ma=mv²/R =evB B=mv/eR = =1.67•10⁻²⁷•4.6•10&#...
October 29, 2013

physics
U =U₀exp(-t/RC) U/U₀ =exp(-t/RC) ln(U/U₀) = -t/RC t= - RC ln(U/U₀) = = - 1•10⁶•2•10⁻⁶•ln(0.75•6/6)=0.58 s
October 29, 2013

physics
Time constant T=RC U =U₀exp(-t/RC) U/U₀ =exp(-t/RC) U₀/2U₀ =1/2 = exp(-t/RC) ln(1/2) = -t/RC) ln(1/2) =- ln2 ln2 = t/RC t=RC•ln2=T•ln2
October 29, 2013

science
a=v²/2s
October 29, 2013

physics
U =U₀exp(-t/T) U/U₀ =exp(-t/T) ln(U/U₀) = -t/T ln(0.01U₀/U₀) =ln0.01 = -t/T t=-Tln0.01=4.61 T
October 29, 2013

physics
U₀ is the max charge of capacitor. Discharging U=U₀exp(-t/T)= =U₀exp(-2T/T)= =U₀exp(-2)=0.135U₀ Charging U=U₀{1-exp(-t/T)}= = U₀{1-exp(-2)}= U₀(1-0.135)=0.865U₀
October 29, 2013

physics
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/DC-Current/RCSeries.html
October 29, 2013

physics
σ =Eε σ=Weight/A σ₁ =mg/πr²= E₁ε σ₂=Mg/πR² =E₂ε (mg/πr²):(Mg/πR²) = (E₁ε):( E₂ε) R =sqrt[ r²•M•E₁/m•E₂]=….
October 29, 2013

physics
KE(total)= KE(transl) +KE(rot) = =mv²/2 +Iω²/2= =mv²/2 +(mR²)v²/2R²=mv² v= sqrt{KE/m)=sqrt(0.2/0.04) =0.5 m/s
October 29, 2013

ET3034TUx Solar Energy
a) The majority of silicon crystals grown for device production are produced by the Czochralski process, (CZ-Si) since it is the cheapest method available and it is capable of producing large size crystals http://en.wikipedia.org/wiki/Silicon
October 29, 2013

PHYSICS
F=qvBsinα sinα=1 q=e=1.6•10⁻¹⁹ C F=evB=..... B↓v←F⊙ mv²/R=evB R=mv/eB=.....
October 29, 2013

PHYSICS
ma=qvBsinα sinα=1 q=e ma=evB B=ma/ev = =1.67•10⁻²⁷•2 •10¹³/1.6•10⁻¹⁹•1.2•10⁷= =0.0173 T F=q[v,B] => B is in –y direction
October 29, 2013

physics
B=0.014 T KE=670 eV =1.6•10⁻¹⁹•670 =1.072•10⁻¹⁶ J KE=mv²/2 v=sqrt{2•KE/m} = 2•1.072•10⁻¹⁶/9.1•10⁻³¹=1.5•10⁻⁷ m/s F(el) = F(mag) eE =evB E=vB =1.5•10...
October 29, 2013

Physics 101
L=I ω= (mr²/2) •2πf= mr²πf =59•0.15²• π•2.9=12.1 kg•m²/s
October 28, 2013

physics
m=6173 kg h=4.71•10⁵ m R=4.31•10⁶m T=1.6 h=1.6•3600=5760 s mg=? F=GmM/(R+h)² F=mv²/(R+H) GmM/(R+h)² =mv²/(R+H) G M/(R+h)² = v²/(R+H) G M/(R+h)= v² T=2π(R+h)/v => v=2π(R+h)/T => v²=4 π²...
October 28, 2013

physics
F=ma=mv²/R m=FR/ v²=…..
October 28, 2013

physics
a₁= ω²R₁ a₂=ω²R₂ a₁/a₂ = R₁/R₂ a₂=a₁R₂/R₁= =282•0.0636/0.0316 =567.6 m/s²
October 28, 2013

physics
a=v²/R
October 28, 2013

Physics
v=v₀ +at a=(v-v₀)/t=(32.8-13.5)/11 =1.75 m/s² F=ma=1532.6•1.75 =2682.05 N
October 28, 2013

Physics
V= 6.3m³= A•h =1m•1m• 6.3 m t=h/c=6.3/3•10⁸=2.1•10⁻⁸s. Energy hitting the base of 1m² is E₀=I/A=1.2•10³(W/m²)/1(m²)= = 1.2•10³ W = 1.2•10³ W J/s. E=E₀•t=1.2•10&#...
October 28, 2013

physics
P=F/A =mg/4A₀ A₀=mg/4F = 1660•9.8/4•2.1•10⁵= =1.94•10⁻² m²
October 28, 2013

PHYSICS (HELP)
http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W05D1-2.pdf
October 28, 2013

Physics
W(alc) =W(air)-F(buoyant) = W(air)-ρVg V={W(air)-W(alc)}/ρg= ={275-160}/0.7•9.8 =16.8 m³ W(air) =mg =ρ₀Vg ρ₀=W(air)/ Vg = 275/16.8•9.8 =1.67 kg/m³
October 28, 2013

physics(elena)
Problem 3 http://mikebloxham.com/H7A/HW7soln.pdf
October 28, 2013

physics
h=gt²/2 =>t=sqrt(2h/g) L=v(x)•t
October 28, 2013

electricity magnetism physics
ΔKE=W(electric field) =e•Δφ=e•E•x= =1.6•10⁻¹⁹•850•2.5 =3.4•10⁻¹⁶ J.
October 28, 2013

Physics Classical Mechanics
www.google.com.ua/url?sa=t&rct=j&q=&esrc=s&source=web&cd=8&cad=rja&ved=0CHAQFjAH&url=http%3A%2F%2Fweb.mit.edu%2F8.01t%2Fwww%2Fmaterials%2FProblemSolving%2FRaw%2Fsolution06.doc&ei=bphuUsyCEsejtAa31YHIDg&usg=AFQjCNF2N0eF8Wa-JIZTeUxi8Etaz_Z5dQ&sig2=ZWISf9N_URhkDrvW1b4cqw&bvm=bv....
October 28, 2013

Physics Classical Mechanics
http://www.learningace.com/doc/2339465/7e7ffe3cc6dd88b45b69cdef6aa9ec29/ic_sol_w07d3-2
October 28, 2013

Physics Classical Mechanics
mv=(2m)u u=v/2 KE(of two carts) =(2m)u²/2=mv²/4 =0.5(mv²/2) =0.5KE(cart A) Ans 1)
October 28, 2013

Physics Classical Mechanics
Pr 4 http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/momentum/momentum-and-impulse/MIT8_01SC_problems16_soln.pdf
October 28, 2013

Physics Classical Mechanics
http://www.masteringphysicssolutions.net/chapter-6-6-the-center-of-mass-of-the-earth-moon-sun-system/
October 28, 2013

Physics Classical Mechanics
http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W05D1-2.pdf
October 28, 2013

Physics Classical Mechanics
Pr/3 cw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/momentum/momentum-and-impulse/MIT8_01SC_problems16_soln.pdf
October 28, 2013

Physics Classical Mechanics
http://web.mit.edu/8.01t/www/materials/ExamPrep/practice_exam02_sol_f12.pdf
October 28, 2013

PHYSICS
v=(6i-3j+k) m/s B=(i+2j-3k) T F=e[v,B] i j k x₁ y₁ z₁ = x₂ y₂ z₂ i j k 6 -3 1 = 1 2 -3 =(y₁z₂-z₁y₂)i +(z₁x₂-x₁z₂)j +(x₁y₂-y₁x₂)k= =(7i+19j+15k). |[v,B] |=sqrt{7²+19...
October 28, 2013

PHYSICS
B=0.014 T KE=670 eV =1.6•10⁻¹⁹•670 =1.072•10⁻¹⁶ J KE=mv²/2 v=sqrt{2•KE/m} = 2•1.072•10⁻¹⁶/9.1•10⁻³¹=1.5•10⁻⁷ m/s F(el) = F(mag) eE =evB E=vB =1.5•10...
October 28, 2013

physics
f=N/t=10/30=0.33 s⁻¹ ω=2πf=2π•0.33=2.1 rad/s
October 27, 2013

physics
http://www.learningace.com/doc/2339465/7e7ffe3cc6dd88b45b69cdef6aa9ec29/ic_sol_w07d3-2
October 27, 2013

Science. Check answers
1. True or False: If the net force of an object is 0 N, the forces are considered unbalanced. (True) 2. Which of the following forces causes a feather to fall slower than a blowing ball? gravity ← rolling friction sliding friction air resistance 3. A small fish in the ...
October 21, 2013

Science help! URGENT. Check answers
1. True or False: If the net force of an object is 0 N, the forces are considered unbalanced. (True) 2. Which of the following forces causes a feather to fall slower than a blowing ball? gravity ← rolling friction sliding friction air resistance 3. A small fish in the ...
October 21, 2013

Physics pls anyone can help me? its urgent
http://www.youtube.com/watch?v=N-WR1A7k6lo
October 20, 2013

physics
v=const => Net force =0 => Fcosα=F(fr) =μN=μmg μ= Fcosα/mg=89•cos33/71•9.8=0.1
October 18, 2013

PHYSICS(HELP!!)
http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/conservation-of-energy/mechanical-energy-and-the-simple-harmonic-oscillator/MIT8_01SC_problems15_soln.pdf
October 18, 2013

PHYSICS
d=sqrt(s₁²+s₂²) =sqrt(900+1600)=50 km
October 16, 2013

Science help! URGENT. Check answers
1. True or False? A model train traveling at a constant speed around a circular track has a constant velocity. (False) 2.Which of the following would be affected if a soaring eagle changed direction but remained at a constant speed? acceleration only velocity only acceleration...
October 15, 2013

physics
2Tsin16°
October 11, 2013

Physics
L=W/cos21 R=Wsin21
October 11, 2013

Physics
mg=F(buo) ρ(ice)V₁g = ρ(water)V₂g ρ(ice)•50•50•h•g = ρ(water) •50•50•(h-20) •g ρ(ice)• h = ρ(water) • (h-20) 917•h=1000(h-20) h=241 m Above the water surface 241-20 =221 m
October 11, 2013

Physics
V=0.3/4 m³ Helium ρ₁=0.2 kg/m³ Air ρ₂=1.33 kg/m³ the buoyant force F = ρ₂Vg mg+T =F W=mg = F-T = ρ₂Vg-T= =9.8•1.33•0.3/4 – 0.8=0.978-0.8=0.178 N If W₁ is the weight of helium and W₀ is the ...
October 11, 2013

Physics
If W₁ is the weight of helium and W₀ is the weight of balloon, W=W₁+W₀ W₁ =m₁g= ρ₁Vg=9.8•0.2• 0.3/4 =0.147 N Then W₀=W-W₁ = 0.178-0.147=0.031 N
October 11, 2013

Physics
V=0.3/4 m³ Helium ρ₁=0.2 kg/m³ Air ρ₂=1.33 kg/m³ the buoyant force F = ρ₂Vg mg+T =F W=mg = F-T = ρ₂Vg-T= =9.8•1.33•0.3/4 – 0.8=0.978-0.8=0.178 N
October 11, 2013

physics
ω₀=sqrt(k/m) = sqrt(20/6) =1.83 rad/s T=2π/ ω =2π/1.83 =3.43 s. (1) x=Acos (ω₀t +α) At t=0 -x₀=x v₀=v(x) v(x)= dx/dt = - ω₀Asin(ω₀t +α) … (1) If t=0 -x₀=Acos (ω₀t +α) =...
October 11, 2013

Physics
Coefficient of the kinetic friction μ₁ =0.3 Coefficient of the static friction μ ₂ =0.5 m₁= 2.5 kg, m₂=2.5 kg α=30° (A) m₁a= T- m₁gsinα- F(fr)… (1) 0=N-m₁gcosα………….(2) m₂a= m...
October 11, 2013

Physics
800•10⁶ N/m² ….1 m² mg……………πD²/4= π(6•10⁻³)²/4 mg=800•10⁶•π(6•10⁻³)²/4=22619 N m=22619/9.8=2308 kg
October 11, 2013

physics
ω₀=sqrt(k/m) = sqrt(20/6) =1.83 rad/s T=2π/ ω =2π/1.83 =3.43 s. (1) x=Acos (ω₀t +α) At t=0 -x₀=x v₀=v(x) v(x)= dx/dt = - ω₀Asin(ω₀t +α) … (1) If t=0 -x₀=Acos (ω₀t +α) =...
October 11, 2013

Physics
v=89 km/h=89000/3600= 24.7 m/s. The distance covered during the reaction time is s₁=v₀t₁=24.7•0.2=49.4 m The distance covered during braking s₂ s₂=v₀t₂-at₂²/2 v=v₀-at₂ v=0 => t₂=v₀/a s₂=...
October 11, 2013

Physics
ω₀=sqrt(k/m) = sqrt(20/6) =1.83 rad/s T=2π/ ω =2π/1.83 =3.43 s. (1) x=Acos (ω₀t +α) At t=0 -x₀=x v₀=v(x) v(x)= dx/dt = - ω₀Asin(ω₀t +α) … (1) If t=0 -x₀=Acos (ω₀t +α) =...
October 10, 2013

Physics
m₁a=m₁gsinα-T-F(fr)… (1) 0=N-m₁gcosα………….(2) m₂a=T-m₂g…………….(3) F(fr)=μN=μm₁gcosα sum of (1) and (3): a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g= = ...
October 10, 2013

Physics
T₁=T₂=F/2cos(α/2) =173/2•cos52=140.5 N
October 10, 2013

Physics
v(south) =12.9cos23 -6.6 =5.27 m/s v(west) =12.9sin23=5.04 m/s v=sqrt{v(west)²+v(south)²} = =sqrt{5.04²+5.27²} =7.3 m/s tanα =5.04/5.27 =0.95 α=43.7º
October 9, 2013

physics
F=mgsinα –F(fr)= = mgsinα - μmgcosα a=F/m =g(sinα – μcosα) =9.8(sin14-0.57•cos14)= -3.04 m/s² v²=v₀²+2ad v=sqrt{ v₀²+2ad} = =sqrt{18²+2• (-3.04) •25}=13.1 m/s
October 9, 2013

physics (elena)
m₁a=m₁gsinα-T-F(fr)… (1) 0=N-m₁gcosα………….(2) m₂a=T-m₂g…………….(3) F(fr)=μN=μm₁gcosα sum of (1) and (3): a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g= = ...
October 9, 2013

physics
Normal (or centripetal) acceleration is a(n) = v²/R = 31.7²/153=6.57 m/s² Tangential acceleration a(τ)= -5.35 m/s² Acceleration a=sqrt{ a(n)²+ (5.35)²} =8.47 m/s² tanφ=5.35/6.57=0.814 φ=31.2º α=90+φ=90 º +31...
October 9, 2013

physics
http://answers.yahoo.com/question/index?qid=20130318231622AAX71vd
October 9, 2013

physics
m₁a=m₁gsinα-T-F(fr)… (1) 0=N-m₁gcosα………….(2) m₂a=T-m₂g…………….(3) F(fr)=μN=μm₁gcosα sum of (1) and (3): a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g= = ...
October 9, 2013

physics
m₁a=m₁gsinα-T-F(fr)… (1) 0=N-m₁gcosα………….(2) m₂a=T-m₂g…………….(3) F(fr)=μN=μm₁gcosα sum of (1) and (3): a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g= = ...
October 9, 2013

Physics
http://www.algarcia.org/AnimationPhysics/BalanceTutorial.pdf
October 9, 2013

Physics
http://www.asu.edu/courses/kin335/documents/CM%20Lab.pdf
October 9, 2013

physics
mgh=mv²/2 v=sqrt{2gh) p=mv =m• sqrt{2gh)
October 9, 2013

physics
http://ca.answers.yahoo.com/question/index?qid=20110225140613AAXANMW
October 9, 2013

Physics
http://physics.ucsd.edu/students/courses/summer2009/session1/physics2b/CH29.pdf Problem 6
October 8, 2013

Physics
L=Iω=(mR²/2)•2πf=…
October 8, 2013

physics
http://www.jiskha.com/display.cgi?id=1381249499
October 8, 2013

Physics
ρ(g)=19•ρ(w). The buoyance force F =ρ(w) •V•g= =ρ(g)•V•g/19. The weight of the crown W = ρ(g)•V•g = 32 N => ρ(g)•V•g/19=32/19 =1.68 N. Upward force =W-F=32 – 1.68 =30.32 N
October 8, 2013

physics
Tension at the ends of the rope. We believe the rope as the point mass suspended by two massless strings that make the angles ϑ with respect the trees. => there are 3 forces: mg, T₁ and T₂ . T₁=T₂=T. The system is in equilibrium and mg is in y-...
October 8, 2013

Physical science
v=gt=9.8 x 3.5 =34.3 m/s
October 8, 2013

physics
T₁=T₂=F/2cos(α/2) =173/2•cos52=140.5 N
October 8, 2013

Physics
Tension at the ends of the rope. We believe the rope as the point mass suspended by two massless strings that make the angles ϑ with respect the trees. => there are 3 forces: mg, T₁ and T₂ . T₁=T₂=T. The system is in equilibrium and mg is in y-...
October 8, 2013

physics mechanical energy and power
P(i)=F(d) •v =300 •30=9000 W=12.07 hp P=[P(d)+mgsinα] •v= =(300+1850•9.8•sin12) •30 = =18231.3 W=24.45 hp
October 7, 2013

physics mechanical energy
W(fr)=PE-KE = mgh - mv²/2 W(fr) =F(fr) •s=F(fr)•h/sinα KE=W(fr)₁ mv²/2 =F(fr)₁ •s₁ F(fr)₁= mv²/2•s₁
October 7, 2013

physics
mv²/2= mgh h= v²/2g = 16²/2•1.62 =79 m
October 7, 2013

Physics
v=gt t=v/g =20/9.8 =2.04 s h=gt²/2 =9.8•2.04²/2 =20.4 m h=(v²-v₀²)/2g v=sqrt{2gh+ v₀²} =sqrt{2•9.8•20.4 +20²}= =28.3 m/s
October 7, 2013

Physics
(1) x₁=11.8-9.8 = 2 cm V=kx₁²/2 2.2V =kx₂²/2 2.2V/V = 2kx₁²/2 kx₂² 2.2= x₁²/ x₂² x=sqrt{2.2•2²} =2.97 cm L=9.8+2.97 =12.77 cm (2) x=sqrt{2.3•2²} =3.03 cm L=9.8-3.03=6.77 cm
October 7, 2013

Physics
x: mgsinα = F(fr) y: N=mgcosα + F mgsinα = F(fr) =μN =μ(mgcosα + F), F =mgsinα/μ - mgcosα = =mg(sinα/μ – cosα)
October 7, 2013

physics
âsin53/sin 90 = n/n(gl) n=n(gl) •sin53=1.47•0.8=1.17
October 7, 2013

Physics/Electricity
http://answers.yahoo.com/question/index?qid=20120314175225AAwIJ7M
October 6, 2013

Physics/electricity
T= 2πR/v, I=e/T=ev/2πR= =1.6•10⁻¹⁹•5.48•10⁵/2•π•2.12•10⁻¹º= =6.58•10⁻⁵A
October 6, 2013

Physics/electricity
resistivity at 20℃ is ρ=22•10⁻⁸ Ohm•m R= ρL/A= 4ρL/πD²= =2•22•10⁻⁸•19/ π(1.024•10⁻³)²= =5.076 Ohms
October 6, 2013

Physics
q =E/U =0.1/19•10⁶ =5.26•10⁻⁹ C
October 6, 2013

Physics
http://www.physics.brocku.ca/Courses/1P22_DAgostino/samples/Ch21P.pdf SP37 page 24
October 6, 2013

physical science
mgh/2=mcΔT For iron c=444 J/kg•K ΔT=gh/2c = 9.8•90/444•2=0.99℃
October 6, 2013

college physics
E=k2λ/r E=9•10⁹•2•87•10⁻⁶/2•0.1 =... E=9•10⁹•2•87•10⁻⁶/2•0.26 =... E=9•10⁹•2•87•10⁻⁶/2•1.4 =...
October 6, 2013

physics
The time to rich the highest point is tₒ= 2vₒ•sinα/2g =2•20•sin35/2•10 =1.15 s The time for covering the distance to the wall ‚d’ is t₁=d/vₒ•cosα =15/20•cos35= 0.92 s At horizontal distance d from the ...
October 6, 2013

physics
v⒳=v₀•cosα=10.02m/s v⒴=v₀•sinα=29.44 m/s a(x)=0 a(y)= -g
October 6, 2013

Physics
v=0.197m/s (?!?!) => t=s/v =5000/0.197 =25381 s=7.05 h (?!?!)
October 6, 2013

PHYSICS
The time to rich the highest point is tₒ= 2vₒ•sinα/2g =2•20•sin35/2•10 =1.15 s The time for covering the distance to the wall ‚d’ is t₁=d/vₒ•cosα =15/20•cos35= 0.92 s At horizontal distance d from the ...
October 6, 2013

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