Monday
September 22, 2014

Posts by Elena


Total # Posts: 4,331

physics
mv²/R =GmM/R² v² =GM/R M= v²R/G (the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²)
July 29, 2013

physics
F=ma=mω²R ω↑=> F↑
July 29, 2013

physics
mv₀²/2= mgh h= v₀²/2g mg(H+h) = mv²/2 mgH +mv₀²/2 = mv²/2 v=sqrt{ v₀² +gH} = = sqrt{1.7²+9.8•3.2} = 5.85 m/s
July 29, 2013

physics
The 1st ball h=v₀t₁+gt₁²/2 The 2nd ball x=v₀t₂ h = gt₂²/2 t₁<t₂
July 29, 2013

physics
s=vt=90000•0.5/3600 = 12.5 m
July 29, 2013

Physics
ℰ = -N•dΦ/dt = -N•d(B₀Scosωt)/dt = = NB₀Sωsinωt, ℰ max= NB₀Sω, Without the figure, I believe that S=πR² => ℰ max= NB₀πR²ω = =100•10⁻⁶•3.14•0.1&#...
July 29, 2013

physics
ma=mω²R, ω=sqrt(a/R) = =sqrt(g/R) = =sqrt(9.8/100)=0.313 rad/s
July 29, 2013

physics
T=2πsqrt(L/g), T₁=2πsqrt(L/g₁), (T/T₁)²=g₁/g, g₁ = g(T/T₁)² =…
July 28, 2013

physics
f=1/T =1/2πsqrt(L/g)=...
July 28, 2013

physics
m₁=10 kg, m₂=m₁+62=72 kg, f₁= ω₁/2π = sqrt(k/m₁)/2 π , f₂=ω₂/2π = sqrt(m₂/k)/2 π, f₂ =f₁•sqrt(m₁/ m₂)=3sqrt(10/72)=1.12 Hz.
July 28, 2013

physics
R=0.04 m t= 0.012 m The solution is correct.May be there is mistake in calculations (or in your answer)...
July 28, 2013

physics
Newton’s 2nd law vector(ma) = vector(mg ) +vector( N) +vector(Ffr) Projections: ma=F(fr) 0=N-mg mv²/r =μN N=mg mv²/r = kN = μmg μ=v²/gr. tanα =F(fr)/N = μmg/mg = μ = v²/gr. v=sqrt{gr•tanα} T=2πR/v= 2πR...
July 28, 2013

physics
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html η=1-Tc/Th =1- 253.15/323.15 =0.2166, η = W/Qh => Qh= W/ η =3.6•10⁷/0.2166 = 1.66•10⁸ J= =39648.4 kCal 10 kW•h•10 ¢/kW•h= 100 ¢ (1.66&#...
July 28, 2013

physics
T=2π•sqrt(m/k) T₁²=4π²m₁/k T₂²=4π²m₂/k T₁²/T₂²=m₁/m₂ m₂=m₁T₂²/T₁² Δm= m₂-m₁ = =m₁(T₂²/T₁²-1)= =0.46...
July 28, 2013

physics
T₀=2πsqrt(L/g) = =2πsqrt(0.81/9.8)= =1.8064 s. ΔL=α•L•ΔT=19•10⁻⁶•0.81•18 =2.77•10⁻⁴m L₁=L+ ΔL =0.81+2.77•10⁻⁴= =0.810277 T₁=2πsqrt(L₁/g)= =2π...
July 28, 2013

physic
PE=KE=KE(transl) +KE (rot), mgh=mv²/2 + Iω²/2 = =mv²/2 + mR²v²/4R² =3mv²/4, mgh=3mv²/4, gLsinθ=3v²/4, v=sqrt(4gLsinθ/3).
July 28, 2013

physics
x is the length of unloaded spring F=k•Δx F=mg => m₁g=k(0.19-x) m₂g=k(0.75-x) m₁/m₂=(0.19-x)/(0.75-x) x=[0.19m₂ - 0.75m₁]/(m₂ -m₁) = =[0.19•2.8 – 0.75•0.3]/(2.8-0.3) = 0.1228 m k= m₁g/(0.19-x) = ...
July 28, 2013

physics
(a) F=k•Δx F=mg => mg=k•Δx k= mg/Δx =165/0.015 = 11000 N/m (b) m₁a=N-m₁g N=m₁(g+a) k•Δx₁= m₁(g+a) Δx₁=m₁(g+a)/k= =94(9.8+1.25)/11000=0.094 m
July 28, 2013

physics
PE=KE m₁gh =m₁v₁²/2 v₁ =sqrt(2gh) = sqrt(2•0.03•0.05) = =0.055 m/s. m₁v₁ = m₁u₁ + m₂u₂ u₂=2m₁v₁/(m₁+m₂)=2•0.03•0.055/(0.03+0.02) = 0.066 m/s. 0= u₂ -at a= u...
July 28, 2013

physics
Mg•L=2mg•x 70•0.65 = 2•30x x=70•0.65/60=0.75 m
July 28, 2013

physics
(a) mv₁²/2=0.22•N₁•Q N₁=mv₁²/2•0.22•Q = =2•10⁵•55²/2•0.22•1.3•10⁸ =10.6 gal (b) 0.22•N₂•Q= mv₂²/2 -mv₁²/2 +mgh, N₂ = m(v₂² -v...
July 28, 2013

physics
average SPEED = distance/time taken =(32.5+78)/30•60=0.062 m/s displacement = sqrt(32.5²+78²)=84.5 average VELOCITY = displacement/time taken =84.5/30•60=0.047 m/s
July 27, 2013

physics
88 km/h =88000/3600= 24.44 m/s ------ average SPEED = distance/time taken =(32.5+78)/30•60=0.062 m/s displacement = sqrt(32.5²+78²)=84.5 average VELOCITY = displacement/time taken =84.5/30•60=0.047 m/s
July 27, 2013

Physics
You can find the solution in http://www.cabrillo.edu/~cfigueroa/11/11prob_sets/Solutions/11prob_sol_5%20.pdf
July 27, 2013

MITx: 2.01x
The above answers are correct because I got a green chek and not a red X.
July 26, 2013

MITx: 2.01x
Q2_1_1 TXC=-t_0*L Q2=1_2 a) 0<=x<L ===>> (2*t_0*L)/(pi*G_0*R4) b)L<=x<=3L =>>(2*t_0*(2*L-))/pi*G_0*R^4) c) x=2*L
July 26, 2013

Physics Help ASAP
I can’t understand the direction of Observer motion (towards or from the car?). If the car and observer move towards each other, their relative speed is 25-2 =23 m/s, then f=f₀/[1-v/c)] = =520/[1-23/334] = 558.5 Hz If they move in the same direction v =25+2 = 27 m/s...
July 25, 2013

Physics Help ASAP
http://en.wikipedia.org/wiki/Doppler_effect http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html
July 25, 2013

physics
ΔL = αLΔT α =12 •10⁻⁶ °C⁻¹ ΔL = 12 •10⁻⁶•65•( -37-22) = - 0.046 m
July 25, 2013

physics
ΔL = αLΔT ΔL = 12•10⁻⁶•3•2.4•33 = 2.85•10⁻³ m ΔL/2 = 2.85•10⁻³ /2 = 1.425•10⁻³ m
July 25, 2013

physics
The volume of liquid that spills over the can is the difference between the increase in the volume of the liquid and that of the aluminum can ΔV=β₁VΔT - β ₂VΔT β₁=(ΔV/VΔT) + β ₂ = =(6.1•10⁻⁶/...
July 25, 2013

general physics
Q₁=m₁cΔT₁ Q₂=m₂cΔT₂ m₁cΔT₁=m₂cΔT₂ m₁ΔT₁=m₂ΔT₂ m₁ΔT₁=m₂ΔT₂ m (40.8 -30.9)=(190-m)(30.9-11.3) Solve for ‘m’
July 25, 2013

Physics
R=R1+R2+R3
July 25, 2013

physics
Q₁=mc₁ΔT₁ Q₂=mc₂ΔT₂ mc₁ΔT₁=mc₂ΔT₂ c₁ΔT₁=c₂ΔT₂ c₁(95-30)=c₂(292-30) c₂=c₁(95-30)/(292-30)= =840•65/262 = 208.4 J/kg•°C.
July 25, 2013

physics
mL= =m₁c ΔT m₁ = mL/c ΔT = =0.068•3.94•10⁵/4183•30.1= =0.213 kg
July 25, 2013

solid mechanics
E=205 kN/mm²= 205•10⁹ N/m². σ=E•ε(longitudinal) F/A = E•ε(longitudinal) ε=ε(longitudinal) = F/A•E= =40000/(20•10⁻³)²•205•10⁹=4.9•10⁻⁴. If a=0.2 m, b=c=0.002 m, &#...
July 25, 2013

solid mechanics
normal stress σ = F/A = 250000/0.1•0.1 = 2.5•10⁷ N/m² Tangential stress τ = F /(A/cosα) = Fcosα/A = =250000•0.5/0.1•0.1=1.25•10⁷ N/m²
July 25, 2013

Solid Mechanics
E=205 kN/mm²= 205•10⁹ N/m², μ = 0.3 σ=E•ε(longitudinal) F/A = E•ε(longitudinal) ε=ε(longitudinal) = F/A•E= =40000/(20•10⁻³)²•205•10⁹=4.9•10⁻⁴. If a=0.2 m, b=...
July 25, 2013

physics
η= 40% =>η =0.4 input heat Q₁=1.05•10¹⁴J waste heat Q₂=0.6 Q₁=6.3•10¹³ J Q₂/W=1.5 work W=0.4 Q₁=4.2•10¹³ J
July 24, 2013

physics
W=2.05•10⁸ J η= 4.00%=> η = 0,04 Q =W/0.04=5.125•10⁹ J Waste heat = Q - W = 5.125•10⁹-2.05•10⁸ = 4.92•10⁹ J N= 5.125•10⁹/6•10⁹ = 0.85
July 24, 2013

physics
Heat input Q=W+Q₁ = =8000+8000 = 16000 J η=W/Q =8000/16000=0.5 η =50%
July 24, 2013

physics
d=0.025 m Δx=0.31 m W=pΔV=p•πd²•Δx/4= =2.4•10⁵•3.14•0.025²•0.31/4 =36.5 J F=W/ Δx = 36.5/0.31 = 117.8 N
July 24, 2013

physics
0.2kcal/kg•℃=837.36 J/kg•℃ m = ρV=837.36• (a) Q=mcΔT=2.16•10⁴•837.36•10.5=1.9•10⁸ J (b) α = 2.5•10⁻⁶ /℃ ΔL = α•L•ΔT = 2.5•10⁻⁶•8.5&#...
July 24, 2013

Physics
mgH-mgh= mv²/2 v=sqrt{2g(H-h)} = …
July 24, 2013

Physics
s=(v²-v₀²)/2a
July 24, 2013

Dynamics
c.Velocity vectors before and after the collision.
July 22, 2013

Physics
λ=c/f => f=c/ λ = 3•10⁸/6•10⁻⁷=5•10¹⁴ Hz
July 22, 2013

Physics
ω=sqrt(k/m) = sqrt(20/2) = 3.16 rad/s x=Asinωt v=Aωcosωt t=0 => cosωt=1 A=v/ω =4/3.16=1.265 m x=Asinωt sinωt =x/A=1/1.265 =0.79 cos ωt =sqrt{1-sin²ωt} = 0.612 v₁=Aωcosωt =1.265•3.16•0.612 =2....
July 22, 2013

Dynamics
KE=PE+W(fr) m•v²/2 = m•g•h +F•s= =m•g•s•sinα +F•s. F=m(v²/2s - g•sinα)
July 22, 2013

BTech, Mechanical
For the 1st ball, the time of the upward motion v=v₀-gt v=0 0=v₀-gt t= v₀/g =18/9.8=1.83 s. The height of the 1st ball h= v₀t-gt²/2=18•1.83 - 9.8•(1.83)²/2= =16.53 m The time of the 1st ball downward motion before the 2nd ball begins...
July 22, 2013

Physics
Ice m₁=200 g = 0.2 kg c₁ = 2060 J/kg•K λ =335000 J/kg Water c₂=4183 J/kg•K m₂=? m₁c₁[0-(-5)] + m₁λ+ m₁c₂(21-0) = m₂c₂(65-21) m₂=m₁(5c₁+λ+21c₂)/44 c₂ = =0.2(...
July 22, 2013

dynamics
F=P/v=mgsinα v=P/mgsinα=200/80•9.8•0.06 = 4.25 m/s
July 22, 2013

dynamics
0,06P=Fv=mgv v=0.06P/mg
July 22, 2013

Physics
m=F/a Weight =mg
July 22, 2013

physics
Steel ρ₁ =7800 kg/m³ V₁=m₁/ρ₁=2.6/7800 = 3.3•10⁻⁴ m³ Gasoline ρ₂ = 750 kg/m³ m₂=ρ₂V₂=750•15•10⁻³ =11.25 kg ρ(ave) =( m₁+m₂)/(V₁+V&#...
July 22, 2013

physics
V =m /ρ =41/750 = 0.055 m³ =14.53 gallons V=L•W•D= > D=V/L•W = 0.055/0.4•0.9 =0.15 m
July 22, 2013

physics
a=F/m F1=F/4.5 => a1=a/4.5=12.5/4.5=2.78 m/s^2
July 21, 2013

earth
Solar eClipse http://en.wikipedia.org/wiki/Solar_eclipse Lunar eClipse https://en.wikipedia.org/wiki/Lunar_eclipse
July 21, 2013

Physics
Q= mc Δt mc(50-20) = Mc(90-50) m=M•40/30 = 800•40/30 = 1067 kg
July 21, 2013

physics
cM[t(ini) –t(fin)] = mr cMΔt = mr Δt= mr/cM= 2.5•10⁻³•2260•10³/4183•0.28 = 4.8℃ T(fin) = 90-4.8 = 85.2℃
July 21, 2013

physics
(a) m=ρ•V= ρ•L•W•H=917•160000•45000•250=1.65•10¹⁵ kg (b) Q=mL=1.65•10¹⁵•334000=5.51•10²º J (c) Q/Q(US) =5.51•10²º/8•10¹⁹=6.89 (d) Q/q= 5.51•10&#...
July 21, 2013

physics
At 37.0ºC the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg. Heat of Fusion and vaporization at 0℃ for water(ice) L(f)= 334 kJ/kg. m(cond) •L(v)=ML(f) M= m(cond) •L(v)/ L(f)= =7.4•10⁻³•2430000/334000 =5.38•10&#...
July 21, 2013

physics
For human body c=3470 J/kg. At 37.0ºC the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg m•c•Δt= m(cond) •L(v) m(cond) = m•c•Δt/L(v) = =60•3470•0.75/2430000 =6.43•10⁻² kg=64.3 g
July 21, 2013

physics
(a) Stefan-Boltzmann Law R=σT⁴ R=P/A =P/4πR² σT⁴=P/4πR² T =forthroot{ P/4σπR²} = =forthroot{3.8•10²⁶/4•5.67•10⁻⁸•π•(7•10⁸)²} =5744 K (b) P₀= P/4...
July 21, 2013

physics
F(x) = F•cosα
July 21, 2013

physics
(a) ω=sqrt(k/m) = sqrt(305/0.26) = 34.2 rad/s y=Asin ωt = > y=0.28sin34.2t (meters) (b) y=A (longest) A= Asin ωt sin ωt = 1 ωt= π t= π/ω=3.14/34.2= 0.092 s. T=2π/ ω = 2•3.14/34.2 = 0.183 s The shortest spring after T/2...
July 20, 2013

Physics
(a) Stefan-Boltzmann Law R=σT⁴ R=P/A =P/4πR² σT⁴=P/4πR² T =forthroot{ P/4σπR²} = =forthroot{3.8•10² /4•5.67•10⁻⁸•π•(7•10⁸)²} =5744 K (b) P₀= P/4πR...
July 20, 2013

general physics
ρ= m/V= 6.3•10²⁸/1.6•10³=3.93 •10²⁵ kg/m³ m= ρV₁=3.93 •10²⁵•2 •10⁻⁷=7.875•10¹⁸ kg=17.4•10¹⁸ lb
July 20, 2013

general physics
p = 2 atm = 202650 Pa p₀ = mg/a•b = 160/0.4•0.24 =1667 Pa N = p/p₀ =202650/1667=121.6 ≈122
July 20, 2013

general physics
http://www.phy.ilstu.edu/~holland/phy108/Homework%2015%20solution.pdf http://answers.yahoo.com/question/index?qid=20120322205457AAh2i77
July 20, 2013

general physics
http://answers.yahoo.com/question/index?qid=20090419164616AAiRaHT
July 20, 2013

physics
(a) m=ρ•V= ρ•L•W•H=917•160000•45000•250=1.65•10¹⁵ kg (b) Q=mL=1.65•10¹⁵•334000=5.51•10²º J (c) Q/Q(US) =5.51•10²º/8•10¹⁹=6.89 (d) Q/q= 5.51•10&#...
July 20, 2013

Physics
Q=Mc ΔT =mL(v) m/M = c ΔT/L9v) = =4183•1.6/2430000 = 0.00275 or 0.275%
July 20, 2013

physics
1.25 kcal = 5233.5 J c(merc) = 139 J/kg•K c(steel) = 470 J/kg•K c(concrete) = 880 J/kg•K c(water) = 4183 J/kg•K ΔT=Q/mc =5233.5/1.25• 139 =30.1º => 50,1º ΔT=Q/mc =5233.5/1.25•470 =8.91º => 28,91º ΔT=Q/mc =...
July 19, 2013

Physics
PE₁=k•q₁•q₂/r₁ r₁=∞ =>PE₁ =0 PE₂=k•q₁•q₂/r₂ r₂= 7.07•10⁻¹² m ΔPE= PE₂ - PE₁=PE₂= = 9•10⁹•8e(-8e)/ 7.07•10⁻¹...
July 18, 2013

Physical Science
A. increasing its thickness R=ρL/A A↑=>R↓
July 17, 2013

physics
The image size of a distant object is simply its angular size multiplied by the lens focal length. 1/f = 1/f₁+ 1/f₂−d/f₁•f₂. f₁= 20 cm =0.2 m f₂= - 8 cm = - 0.08 m 1/f = 1/0.2 – 1/0.08 - 1/0.2•0.08= =5-12.5-62.5 = - 70 f...
July 17, 2013

physics
The image size of a distant object is simply its angular size multiplied by the lens focal length. 1/f = 1/f₁+ 1/f₂−d/f₁•f₂. f₁= 20 cm =0.2 m f₂= - 8 cm = - 0.08 m 1/f = 1/0.2 – 1/0.08 - 1/0.2•0.08= =5-12.5-62.5 = - 70 f...
July 17, 2013

Physics
v =v₀-at t=( v₀-v)/a = s= v₀t - at²/2 =…
July 16, 2013

Physics
Or it may be 60 s after the first 10 s v₃ = v₀+a(t₁+t₂) = 10+0.6•(10+60)=52 m/s
July 16, 2013

Physics
v₁=v₀+at₁ a= (v₁-v₀)/t₁ = (16-10)/10 = 0.6 m/s² v₂=v₀+at₂ = 10+0.6•60 = 46 m/s
July 16, 2013

velocity
http://hyperphysics.phy-astr.gsu.edu/hbase/vel2.html http://en.wikipedia.org/wiki/Velocity
July 15, 2013

physics
W=KE=mv²/2
July 15, 2013

Elementary mechanics
the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², M=1.9•10²⁷ kg !!!!!!!!!!!! R=7.15•10⁷ m !!!!!!!! v=sqrt(2GM/R)= =59539 m/s ≈59.5 km/s
July 15, 2013

physics
No. The plane mirror will give an image of the object that is on the opposite side. It creates an erect, virtual image of the same size as the object where the distance from the object to the mirror equals the distance from the image to the mirror.
July 15, 2013

Physics
V=A•L A=V/L R=ρL/A = ρL²/V L=sqrt{RV/ρ} = =sqrt{ 0.15•2•10⁻⁶/1.72•10⁻⁸} = 4.17 m
July 15, 2013

Physics Help
Using Problem 5 page 234 from http://books.google.com.ua/books?id=PQRo3XuHHvYC&pg=PA262&lpg=PA262&dq#v=onepage&q&f=false From steam table at 22.5 bar Hw= 936.48 kJ/kg He = 1865.2 kJ/kg - enthalpy of evaporation (Latent heat) Enthalpy of 1 kg of wet steam is H=Hw+ζ •...
July 14, 2013

Physics Help
Another table http://www.chem.mtu.edu/~tbco/cm3230/steamtables.pdf gives more precise values Hs =2801.7 kJ/kg Hw = 936.48 kJ/kg As a result ζ= (Ht-Hw)/ (Hs-Hw)= =(2279.434-936.48)/(2801.7-936.48) = =13.42.954/1865.22=0.7199 => ≈72%
July 14, 2013

Physics Help
(A)Ht=Hs•ζ+(1-ζ)Hw, where Ht = total (actual) enthalpy (kJ/kg), Hs = enthalpy of steam (kJ/kg), Hw = enthalpy of saturated water or condensate (kJ/kg) Using steam-pressure table http://enpub.fulton.asu.edu/ece340/pdf/steam_tables.PDF we can find the magnitudes ...
July 14, 2013

Physics
p=F/A= mg/πr²= =0.001/3.14•0.0002²=7958 Pa
July 14, 2013

Physics
ρ= m/V =0.240/89•10⁻⁶=2696 kg/m³
July 14, 2013

Physics
PE=KE=KE(transl) + KE(rot) I= mR² mgh = mv²/2 + I ω²/2= = mv²/2 + mR²v²/2R²= = mv²/2+ mv²/2 = mv² v=sqrt(gh)= …
July 14, 2013

Physics
(a) L=Iω= I•2πn=0.4•2π•6 = 15.08 J•s (b) L₁=L₂ I₁•2 πn₁ =I₂•2 πn₂ I₂=I₁•n₁/n₂= 0.4•6/1.25 = 1.92 kg•m² (c) 2 πn₂= 2 πn₁...
July 14, 2013

Physics
PE=KE=KE(transl) + KE(rot). mgh = mv²/2 + I ω²/2= = mv²/2 + I v²/2R². I= (2R²/v²)(mgh- mv²/2)= =mR²(2gh/v² -1) = = mR²(2•9.8•2/36 -1)= =mR²(1.09-1)=0.09 mR²
July 14, 2013

Physics
τ=FR= 180•0.28 = 50.4 N•m I=mR²/2 = 75•0.28²/2=2.94 N•m² τ=Iε => ε= τ/I = 50.4/2.94 = 17.14 rad/s² τ₁=FR-F(fr)r =180•0.28 - 20•0.015 = 50.4 – 0.3 = 50.1 N•m ε₁= τ...
July 14, 2013

Physics
v=500000/3600 =138.9 m/s ω=v/R=138.9/30=4.63 rad/s n= ω/2π=4.63/2 π=0.74 rev/s
July 14, 2013

Physics
F₁+mg=F₂ ….(1) 2•F₁+1•mg=1.5•F₂…(2) From (1) F₁ =F₂ -mg ….(3) Substitute (3) in (2) 2(F₂ -mg) +mg = 1.5•F₂ F₂=2mg=.... F₁=mg= ....
July 14, 2013

physics
KE=mv²/2
July 14, 2013

PHYSIC
KE=PE+W(fr) KE=mv²/2 PE= mgh W(fr) =F(fr) •s=µmgcosα•h/sinα mv²/2 =mgh+µmgcosα•h/sinα Solve for ‘v’
July 14, 2013

physics
s=s₀+vt= =0.2-0.5•20 = - 9.8 m
July 14, 2013

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