Thursday
July 24, 2014

Posts by Elena


Total # Posts: 4,343

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

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

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...

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...

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

physics
http://www.learningace.com/doc/2339465/7e7ffe3cc6dd88b45b69cdef6aa9ec29/ic_sol_w07d3-2

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 pa...

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 pa...

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

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

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

PHYSICS
d=sqrt(s₁²+s₂²) =sqrt(900+1600)=50 km

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...

physics
2Tsin16°

Physics
L=W/cos21 R=Wsin21

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

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 we...

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

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

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) =A...

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&...

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

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) =A...

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=...

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) =A...

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= = ...

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

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

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

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= = ...

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...

physics
http://answers.yahoo.com/question/index?qid=20130318231622AAX71vd

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= = ...

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= = ...

Physics
http://www.algarcia.org/AnimationPhysics/BalanceTutorial.pdf

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

physics
mgh=mv²/2 v=sqrt{2gh) p=mv =m• sqrt{2gh)

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

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

Physics
L=Iω=(mR²/2)•2πf=…

physics
http://www.jiskha.com/display.cgi?id=1381249499

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

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-d...

Physical science
v=gt=9.8 x 3.5 =34.3 m/s

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

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-d...

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

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

physics
mv²/2= mgh h= v²/2g = 16²/2•1.62 =79 m

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

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

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

physics
âsin53/sin 90 = n/n(gl) n=n(gl) •sin53=1.47•0.8=1.17

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

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

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

Physics
q =E/U =0.1/19•10⁶ =5.26•10⁻⁹ C

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

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

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 =...

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 in...

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

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

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 in...

physics(help)
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 in...

PHYSICS
E=σ/2ε₀=2.05•10⁻⁶/2•8.85•10⁻¹²=1.16•10⁵

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 in...

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 in...

physics
θ=θ(max) •sin(ωt+α) dθ/dt =θ(max) •ω•sin(ωt+α) (dθ/dt)(max) = θ(max)•ω= θ(max)√(g/L) v(max) =(dθ/dt)(max) •L = L•θ(max)√(g/L)

Physics
R=0.5 m W= mg mg =F(buoyant) W = (1/2) •(4πR³/3) •ρ•g = =1000•4π•0.5³•10/2•3= =2618 N

physics
ma=T-mg a=(T/m)-g= 19.5 – 9.8 = 9.7 m/s² h=at²/2=> t=2h/a = 2•10/9.7 = 2.06 s.

physics
F=ma=mv/t =12.3•10⁻⁶•4/1•10⁻³ =49.2•10⁻³ N F/mg = 49.2•10⁻³/12.3•10⁻⁶•9.8 =408.2

physics
r =(0.020m/s³)t ³ i ^ +(2.2m/s)tj ^ −(0.060m/s² )t²k ^. v= dr/dt =(3•0.020m/s³ ) •t²• i ^ +(2.2m/s) • j ^ −2• (0.060m/s² )t•k ^. a=d²r/dt²=(6•0.020m/s ³) •t i ^ -2• (0.06...

physics
The ball moves in projectile motion. When it is moving horizontally, v(y) =0 +x is directed to the right, +y is directed upward, a(x)= 0, a(y) =-g v₀(y)=sqrt(2gh) = sqrt(2•9.8•4) = 8.85 m/s v= v₀(y)-gt At the top point v=0 => t= v₀(y)/g =8.85/9.8...

Physics
v(x) =v₀(x) = 50 m/s v₀=84 m/s cosα =v₀(x)/ v₀=50/84 =0.6 α=53.5⁰

Physics
F(fr) +mg= F(y) F(y) = Fsinα F(fr) =Fsinα-mg

Physics
W=PE+KE+W(fr) =mgh+mv²/2 +W(fr) = =m(gh+ v²/2)+W(fr) =1260(0.6•9.8 +3²/2)+2870 = =13078.8+2870 =15948.8 J

Physics
mv²/2 =W(fr) = μNs= μmgs μ=v²/2gs =5.8²/2•9.8•20 =0.086

Electricity and Electronics
15. A 19. D

physics
d(x)=dcos46⁰=110•cos46⁰=76.4 m. d(y) =- dsin46⁰=- 110•sin46⁰= - 79.1 m. v(x) = vcos28⁰=350•cos28⁰= 309 m/s v(y) = vsin28⁰=350•sin28⁰=164.3 m/s a(x) = 0 a(y) = - 9 m/s²

Physics
Magnetic field at the center of a circular current-carrying loop of N turns and radius r is B=Nμ₀I/2r….. (1) N=L/2πr ….(2) I=U/R=U/R₀L ….(3) Substitution (2) and (3) in (1) gives B=Nμ₀I/2r = (L/2πr) •(μ̀...

Physics
ma=F mv²/R =qvBsinα. v⊥B => sinα = 1. mv/R =qB, R = mv/ qB, T=2πR/v=2π m /qB= = 2π/B(q/m) =2π/0.69•5.7•10⁸0.69= =1.6•10⁻⁸ s

Physics
http://einstein.drexel.edu/~wking/courses/phys101_w07/hwk/hwk4.pdf

physics
(a) the work done by the applied force W=F(x) •s=F•cos α•s= 18.6•cos23.7•2.09 = …. (b) the magnitude of the normal force exerted by the table N =mg+F(y) =mg+Fsin α = 2.4•9.8 +18.6•sin23.7 = … (c) the magnitude of the force...

physics
d=17.7 m vₒ =15.4 m/s α =38.7⁰ =>2 α =77.4⁰ L=vₒ²•sin2α/g=15.4²•sin77.4/9.8 =23.6 m t= 2vₒ•sinα/g=2•15.4•sin38.7/9.8 = 1.97 s. s=(L-d)/t =(23.6-17.7)/1.97 =3 m/s

Physicis
Make the route plan s₁= 300 =x₁+x₂ s₂ =395, α =26.5⁰ s₃ = 150 x₃ connects the point of s₂ and s₃ intersection and the point on s₁ which divides s₁ by x₁ and x₂ (x₃ ⊥ s₁) s̀...

Physics
(A) L=18.3 m, h=0.809 m h=gt²/2 t=sqrt(2h/g) = .... L=vt => v=L/t=.... (B) h=gt²/2 t=sqrt(2h/g) = … L₁=v₁t =... L₂=v₂t =....

physics
d(x)=dcos46⁰=110•cos46⁰=76.4 m. d(y) =- dsin46⁰=- 110•sin46⁰= - 79.1 m. v(x) = vcos28⁰=350•cos28⁰= 309 m/s v(y) = vsin28⁰=350•sin28⁰=164.3 m/s a(x) = 0 a(y) = - 9 m/s²

Physics
(A) v=gt t=v/g h=gt²/2 =gv²/2g²=v²/2g (B) mv²/2 = mgh h= v²/2g

physics
g=1.6 m/s² Upward motion v=v₀-gt v=0 t= v₀/g=5/1.6 =3.125 s Downward motion t₁=15-3.125 =11.875 m/s v=gt₁=1.6•11.875=19 m/s

Physics
F₁ =qvBsinα₁ F₂ =qvBsinα₂ F₂=2.4F₁ F₁/F₂=1/2.4 =qvBsinα₁/qvBsinα₂ sin α₂=2.4•sin α₁ =2.4•sin19⁰=0.78 α₂=51.4⁰

physics urgent please help!
v=81.7 km/h =81700/3600=22.7 m/s d= v²/2a=22.7²/300=1.72 m

PHYSICS HELP PLEASE!!
v=81.7 km/h =81700/3600=22.7 m/s d= v²/2a=22.7²/300=1.72 m

Physics
6.7 lb = 3.04 kg mg₁=3.04•9.8 =29.8 N mg₂=mg₁/6 =3.04•9.8/6=4.97 N

phusics
a=v²/2s F=ma=m v²/2s =2.1•10⁵•30²/2•120 = 787500 N

phys223
http://answers.yahoo.com/question/index?qid=20130422100626AAOAlzD

physics
v =at => a=v/t P=W/t = Fs/t=Fv = ma•v= mv²/t

physics
KE=PE mv₀²/2=mgh=mgxsinα x= v₀²/2gsinα=3²/2•9.8•sin22=1.23 m. a₁=v₀²/2x =9/2•1.23=3.66 m/s² Upward v= v₀-a₁t₁ v=0 t₁=v₀/a₁=3/3.66=0.82 s. Downward PE=KE mgxsinα= ...

physics
http://answers.yahoo.com/question/index?qid=20110214141546AAp0rSP

Physics
An initial velocity of the projectile v₀ = the velocity of the plane. v₀ (x) = v₀cosα, v₀ (y) = v₀sinα, h= v₀ (y) •t+ gt²/2= =v₀sinα•t + gt²/2. v₀={h- gt²/2}/ sinα•t= ={790 –...

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