Sunday

December 4, 2016
Total # Posts: 4,386

**Physics**

W= - p•ΔV = -0.8•(-5)•10⁻³•1.013•10⁵ = =405.2 J ΔU =Q+W=-380+405.2=252 J

*November 26, 2012*

**Physics**

(a) Ti=273+7.4=280.4 K Vi/Ti = Vf/Tf Vi/280.4 =8 Vi/Tf Tf=8•280.4 = 2243.2 K b) p•ΔV=νRΔT ν =1 mol p•ΔV= RΔT W = p•ΔV= RΔT= = 8.31•(2243.2-280.4)= =1.63•10⁴ J =16.3 kJ

*November 26, 2012*

**physics**

When the cart is 14 cm from the left end of the track, it has a displacement of x=14 -11 cm =3 cm= 0.03 m from the equilibrium position. The speed of the cart at this distance from equilibrium is v=sqrt{k(A²-x²)m}=…

*November 26, 2012*

**The Binding Energy of Electrons?**

Energy of photon: E = hc/λ h = Planck’s constant = 6.626•10⁻³⁴ J•s = 4.14•10⁻¹⁵ eV•s c = speed of light = 3•10⁸ m/s λ = wavelength = 0.954 nm = 0.954•10⁻⁹m E = 4.14•10⁻...

*November 26, 2012*

**physics**

A=4.2 cm = 0.045 m. v(max) =Aω=A•sqrt(k/m)=...

*November 26, 2012*

**physics**

T=2•π•sqrt(m/k)= 2•π•sqrt(0.223/10.20)= …

*November 26, 2012*

**Physics**

Flows to the Earth ΔS(E)=6000/295 =20.34 J/K Flows out of the Sun ΔS(S)=-6000/5650 =- 1.062 J/K ΔS= ΔS(E)+ ΔS(S)= 20.34 -1.062= 19.278 J/K

*November 26, 2012*

**PHYSICS - PLEASE HELP**

mv=MV M=mv/V=9.6•0.53/0.12=42.4 kg

*November 26, 2012*

**Physics**

Mv²/2 =4mcΔT ΔT = Mv²/8mc=1750•28²/8•8•448=47.9℃

*November 26, 2012*

**PHYSICS**

KE=mv²/2 ½KE=mcΔT ΔT= KE/2mc= mv²/4mc= = v²/4c=390²/4•130=292.5℃

*November 26, 2012*

**Physics Can you help me get started**

http://answers.yahoo.com/question/index?qid=20100629105308AAL8TBp

*November 26, 2012*

**Physics**

p=mv=2.3•10⁶•6•10⁶=13.8•10¹² kg•m/s. p₁= m₁v₁= 5.2•10⁵•2•10⁶=10.4•10¹¹kg•m/s. p₂=m₂v₂=8.4•10⁵•9•10 ⁵=75.6•10¹...

*November 26, 2012*

**PHYSICS**

For water, density is ρ =1000 kg/m ³, V =4•10¹¹ m³ => m=ρ•V=1000•4•10¹¹ =4•10¹⁴ kg, c =4183 J/kg•K, ΔT=11-9 =2 K. E=Q= m•c• (T₁-T)= m•c• ΔT =4•10¹...

*November 26, 2012*

**PHYSICS**

mgh=mv²/2 mv²/2=mcΔT => mgh=mcΔT gh=cΔT c =4183 J/kg•K, ΔT=gh/c=9.8•807/4183 =1.9° T1=T+ ΔT=16.7+1.9=18.6℃

*November 26, 2012*

**Physics**

Cu: m₁=0.1 kg, c₁=385 J/kg•K, T₁=460 K, Al: m₂=0.2 kg, c₂=930 J/kg•K, T₂=220 K, Water: m₃=0.5 kg, c₃=4183 J/kg•K, T₃280 K. T=? m₁c₁(T₁-T) = m₂c₂(T-T₂) +m₃c₃(T-T...

*November 26, 2012*

**optics**

http://www.optiboard.com/forums/showthread.php/18768-Need-Help-For-Ophthalmic-Question! and page 30 from http://www.optometry.co.uk/uploads/articles/949e0d09ac024c2edb6d55bad5da4d11_jalie20050225.pdf

*November 26, 2012*

**physics**

k = 1080 N/m, r = 0.0143 m, A = πr²= π•0.0143²=6.42•10⁻⁴ m² The pressure P = F/A = ρ•g•h, where water density is ρ = 1000 kg/m³ g = 9.8 m/s² , x = 0.0075 m For the elastic force of the spring F = kx. ...

*November 26, 2012*

**physics/optics**

Try this http://www.opticampus.com/files/introduction_to_ophthalmic_optics.pdf

*November 26, 2012*

**physics**

¹⁵₈O + ₋₁⁰e=¹⁵₇N

*November 26, 2012*

**Physics**

0=v-at a=v/t F=ma=mv/t (a) F₁ =m•v/t₁= =60.8•5.31/4.52•10⁻³=71426.5 N (b) F₂=m•v/t₂ = =60.8•5.31/0.209 = 1544.7 N. (c) mg=60.8•9.8=595.8 N ground force: F=F₁+mg=71426.5+595.8=72022.3 N F=F₂+mg=1544.7+...

*November 26, 2012*

**physics**

F =G•m1•m2/R² the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², R1=r/2 => F1=4F

*November 26, 2012*

**physics**

weight=buoyant force mg= F d₁•V•g =k•d₂•V•g k= d₁/d₂ =980/1025 =0.956

*November 26, 2012*

**physics**

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², PE₁=G•m₁•m₂/R₁ PE₂=G•m₁•m₂/R₂ E=ΔPE= PE₁-PE₂= =G•m₁•m₂/R₁-G•m₁&#...

*November 25, 2012*

**Physics**

p₁=8.9•10⁻²³ kg•m/s p₂=5.35•10⁻²³ kg•m/s Law of conservation of linear momentum 0=vector p₁+vector p₂+vector p, 0=vector p₁₂ + vector p, p₁₂=sqrt(p₁²+p₂²...

*November 25, 2012*

**Physics**

d•sinθ=k•λ N₀=1/d => λ=sinθ/k•N₀ sin30°=0.5, N₀=2•10⁵ m⁻¹ k=1: λ=sinθ/k•N₀=sin30°/1•2•10⁵=2.5•10⁻⁶ m invisible - Infra-red k=2 λ=sin...

*November 25, 2012*

**Physics**

(a) v=v₀-at a=( v-v₀)/t=(38.666-39.486)/0.231 =-3.55 m/s² F(dr)=ma=-0.209•(-3.55)=-0.7419 N (b) F(dr)=μN=μmg μ=F(dr)/mg=0.7419/0.209•9.8=0.362 (c) Work-energy theorem ΔKE=W(dr) 0-mv²/2=F•s. F= - mv²/2s= - 0.209•...

*November 25, 2012*

**physics**

Polarimetry, or polarization, determines sucrose concentration by passing polarized light through a solution. It determines the sugar content of solutions by measuring the angle of rotation of the plane of polarization of a light beam. The amount of rotation is proportional ...

*November 25, 2012*

**PHYSICS**

Polaroid filters are incorporated into sunglasses. Because the light reflecting off most surfaces will tend to be partly polarized with a predominance of the electric fields oscillating horizontally, that “glare” can be diminished by having the transmission axis of ...

*November 25, 2012*

**Science**

Correct answers. http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html#c3

*November 25, 2012*

**physics**

F(fr)=μN=μmg

*November 24, 2012*

**Physics**

Internal energy is U=3•n•R•T/2 => chage in internal energy is ΔU=3•n•R•ΔT/2. From the ideal gas law PV=nRT => n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) => ΔU=3•n•R•ΔT/2=3•p(V2-...

*November 24, 2012*

**Physics**

Internal energy is U=3•n•R•T/2 => chage in internal energy is ΔU=3•n•R•ΔT/2. From the ideal gas law PV=nRT => n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) => ΔU=3•n•R•ΔT/2=3•p(V2-...

*November 23, 2012*

**Physics**

1/f=1/di + 1/do 1/6=1/di + 1/8 Solve for ‘di’

*November 23, 2012*

**Physics**

x=kLλ/d k=3, L=2.51 m, d=5.35•10⁻⁶ m x₁= kLλ₁/d=3•2.51•640•10⁻⁹/5.35•10⁻⁶=0.9 m, x₂= kLλ₂/d=3•2.51•425•10⁻⁹/5.35•10⁻⁶=0.595 m, tan&#...

*November 23, 2012*

**Physics**

Internal energy is U=3•n•R•T/2 => chage in internal energy is ΔU=3•n•R•ΔT/2. From the ideal gas law PV=nRT => n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) => ΔU=3•n•R•ΔT/2=3•p(V2-...

*November 22, 2012*

**Physics**

Internal energy is U=3•n•R•T/2 => chage in internal energy is ΔU=3•n•R•ΔT/2. From the ideal gas law PV=nRT => n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) => ΔU=3•n•R•ΔT/2=3•p(V2-...

*November 22, 2012*

**Physics**

m1•v1=(m1+m2)v v1=(m1+m2)v/m1=(0.01+3.99)0.5/0.01=200 m/s

*November 21, 2012*

**Physics**

t= 2vₒ•sinα/g L=vₒ²•sin2α/g, h= vₒ²•sin²α/2g,

*November 21, 2012*

**Physics**

(a) PE= mgx=1.9•9.8•(1.13-3)= - … (b) PE =mgh=1.9•9.8•(3-1.13)= + … (c) ΔPE=0

*November 21, 2012*

**math**

23/24-15/24=(23-15)/24=8/24=1/3

*November 20, 2012*

**physics**

V1=4πR³/3=4π20³/3=33510 m³ T1-279K T2=305K V1/T1 =V2/T2 V2=V1•T2/T1

*November 20, 2012*

**physics**

Max displacement is the amplitude => A=26.6 cm

*November 20, 2012*

**Physics**

W=Fs=120•5 = 600 J W(fr)=100•5=500 J ΔKE=(120-100)•5=100J

*November 20, 2012*

**Physics**

mgh=4•9.8•1.5=... mgH=4•9.8•6.5 =...

*November 20, 2012*

**physics**

k =200 rad/m k=2π/λ => λ=2π/k=2π/200 =π/100=0.0314 m x= λ/2=0.0314/2=0.0157 m

*November 20, 2012*

**Physics**

F(x)= -F1+F2•cos α= =-23 +33.8•cos39°= 3.27 N F(y) = F2•sin α =33.8•sin39°= 21.27 N F(net) =sqrt{F(x)²+F(y)²}= sqrt{3.27²+21.27²}=21.52 N

*November 20, 2012*

**Physics**

-F1+F2•cos α=-23 +33.8•cos39°= 3.27 N

*November 20, 2012*

**university physics**

for Pt α =0.00392 (1/℃). R₁=R₀[1+α(T₁-T₀)] R₂= R₀[1+α(T₂-T₀)] R₁/R₂=[1+α(T₁-T₀)]/[1+α(T₂-T₀)] 200/253.8=(1- 0.00392 •20)/[1+ 0.00392• (T₂-20)] ...

*November 20, 2012*

**Physics**

mv²/2=mgh h= v²/2g

*November 20, 2012*

**PHYSICS**

c = e > a > d > b

*November 20, 2012*

**physics**

M/m=1

*November 20, 2012*

**Physics**

The law of conservation of linear momentum for inelastic collision m1•v1 - m2•v2=(m1+m2)•u u= {m1•v1 - m2•v2}/(m1+m2)= ={0.241•1.53 -0.321•2.49)/(0.241+0.321)= =-0.766 m/s. (in the direction of the second pizza motion)

*November 19, 2012*

**Physics**

Cosine Law u=sqrt(V²+v²-2•V•v•cos135°)= =sqrt(200²+90²-2•200•90•cos135°)= =sqrt(40000+8100 +25455.8)=271 km/h s=u•t=271•3=813 km Sine Law 271/sin135°=90/sinβ sinβ=90•sin135°/271=0.235 &#...

*November 19, 2012*

**Physics**

(a) Centripetal (or normal) acceleration is a(c)= m•v²/R=… (b) Tangential acceleration a(τ)= 1 m/s² a=sqrt{a(c)² +a(τ)²}=... (c)tanα= a(c)/a(τ)= ..

*November 19, 2012*

**physics**

ω=2π/T=2•3.14/7•10⁻⁵=… x(max) =A. v(max)=Aω => A= v(max)/ω=…

*November 19, 2012*

**Physics**

The mass of gravel is m2=441•2.67 = 1177.47 kg The law of conservation of linear momentum m1•v1= (m1+m2)v2 v2= m1•v1/(m1+m2) = =2150•2.29/(2150+1177.47)=1.48 m/s

*November 19, 2012*

**Physics**

m1•v1 =(m1+m2)v v= m1•v1/m1+m2)

*November 19, 2012*

**Physics**

h₁=40 m, h₂=5m, v₁=15 m/s PE1+KE1 =PE2+KE2 mgh₁+mv₁²/2= mgh₂+ mv₂²/2 Solve for v₂

*November 19, 2012*

**Physics**

(a) Centripetal (or normal) acceleration is a(c)= m•v²/R=… (b) Tangential acceleration a(τ)= 1 m/s² a=sqrt{a(c)² +a(τ)²}=... (c)tanα= a(c)/a(τ)= ..

*November 19, 2012*

**Physics**

Find the solution in Related Questions

*November 19, 2012*

**Physics**

Δp=FΔt Δp= mv₂-(-mv₁)= m(v₂+v₁), m(v₂+v₁)=FΔt F= m(v₂+v₁)/Δt= =0.148•(10.1+8.95)/10.1•10⁻³=132.6 N

*November 19, 2012*

**physics**

λ=v/f f= v/λ

*November 19, 2012*

**physics 209**

s=at²/2 => a=2s/t²=2•0.4/2²=0.2 m/s² The equation of the load motion ma=mg-T => T=m(g-a) The equation of the pulley motion Iε=M-M(fr) M(fr) = M-Iε = TR-Ia/R= =m(g-a)R – Ia/R = ...

*November 19, 2012*

**physics, university**

m=0.05 kg, ρ =11•10⁻⁸ Ω/m d=7860 kg/m³ R=1.5Ω R= ρ•L/A d=m/V=m/A•L R/d= ρ•L²•A/A•m = ρ•L²/m. L=sqrt{m•R/ρ•}=… A= ρ•L/R = … A=πD²/4 D=sqrt(4A...

*November 19, 2012*

**physics, university**

J=nev, where J is the current density, n is free electron density, e=1.6•10⁻¹⁹C , drift velocity v= 2.98•10⁻⁴ m/s J=U/RA, where U=10 V, R=0.2 Ω, A= πr² n•e•v= U/R•π•r² n= U/R•π•r...

*November 19, 2012*

**Physics**

h₁=40 m, h₂=5m, v₁=15 m/s PE1+KE1 =PE2+KE2 mgh₁+mv ₁²= mgh₂+ mv₂²/2 Solve for v₂

*November 19, 2012*

**Physics**

m1•v1 =(m1+m2)v v= m1•v1/m1+m2)

*November 19, 2012*

**Physics**

P =mgh/t, t=20 min =1200 s

*November 19, 2012*

**physics**

λ=v/f=s/tf=2.5/1.7•3=0.49 m

*November 19, 2012*

**Physics**

m•a=F(fr) m•v²/R=μ•m•g v=sqrt(μ•R•g)

*November 19, 2012*

**Physics**

Flows to the Earth ΔS(E)=1000/290 =3.448 J/K Flows out of the Sun ΔS(S)=-1000/5700 =- 0.1754J/K ΔS= ΔS(E)+ ΔS(S)= 3.448-0.1754=3.2726 J/K

*November 19, 2012*

**Physics**

(3500-2500)•30/4100=7.32 pounds

*November 19, 2012*

**Physics**

Coefficient of performance = energy removed from fridge / ( total energy expelled – energy removed from fridge ) k=Q1/(Q-Q1) Q=Q1(k+1)/k=2.4•10⁴•5/4=3•10⁴ J

*November 19, 2012*

**Physics**

η =1 –T(cold)/T(hot) = = 1- (273+18)/(273+22) = =1-291/493 = 0.41

*November 19, 2012*

**Physics**

Thermal E~kT, where k is the Boltzmann constant, T is the temperature in Kelvins 5℃=278K Answ. 278 x 2=556 K

*November 19, 2012*

**Physics HELP!**

(a) Compare x = (7.1 cm) sin (2.1π t) with x=Asin(ωt) Then ω= 2.1π, => T=2π/ ω = = 2 π/2.1π=0.7 rad/s (b) f=1/T=1.42 Hz (c) A=7.1 cm (d) x=Asin(ωt) 2.6 = 7.1 sin (2.1π t) sin (2.1π t) =2.6/7.1=0.366 (2.1π t)=sin&#...

*November 19, 2012*

**Physics URGENT!!!**

k=F/Δx=8.4/0.0214 = 393 N/m ω= sqrt(393/1.3)=302.3 rad/s ω=2πf f= ω/2π=302.3/2•3.14 =48.14 Hz

*November 19, 2012*

**Physics**

b) False

*November 18, 2012*

**Physics**

Δ=2bn-λ/2 = 2k(λ/2) 2bn =(2k+1)λ/2 λ=4bn/(2k+1) For k=0, λ=4bn=4•1.25•10⁻⁷•1.38=6.9•10⁻⁷m= =6.9•10⁻⁵cm ============= λ=3mm=3•10⁻³m v=340 m/s f=v/ λ=340/3•...

*November 18, 2012*

**physics**

(a) At the point D ma=mg mv²/r=mg v=sqrt(g•r)=sqrt(9.8•2.87)=5.3 m/s (b) KE(C)=PE+KE= =mg•2r+mv²/2= =m(2gr+5.3²/2)=54.8 J. (c) KE(B)=KE(C) + W(fr)= = KE(C) + μ•mg•s= =54.8+0.29•0.78•9.8•4.08= =63.8 J (d) PE1=KE(B) mgh...

*November 18, 2012*

**Physics**

Problem#62 http://www.physics.umn.edu/classes/2012/fall/Phys%201401V.001/downloads/169171-Ch09_ISM.pdf

*November 17, 2012*

**physics**

(a) ma=mgsinα sinα=a/g=5/9.8 =0.51 α=30.7° (b) s=at²/2 t=sqrt(2s/a)=... (c) the same acceleration

*November 15, 2012*

**physics**

KE=mv²/2

*November 15, 2012*

**Physics. Please please help!**

Δφ=φ₁-φ₂=kQ/r₁-kQ/r₂= =kQ(r₂-r₁)/r₁•r₂= =9•10⁹•4.5•10⁻⁶•8•10⁻²/20•10⁻⁴=1.62•10⁶ V mv²/2= q•Δφ, v= sqrt...

*November 14, 2012*

**physics**

n₀= 9.80 101 rev/min =….rev/s ε=2 rad/s² 2•π•n=2•π•n₀ - ε•t, since n=0, 2•π•n₀ - ε•t=0, t=2•π•n₀/ ε=… φ=2•π•n₀•t- &#...

*November 14, 2012*

**Physics**

W=(F,s)=F•s•cosα = =8740•1130•cos180º= =8740•1130•(-1)= = -9876200 J

*November 14, 2012*

**Physics HW Help, Please**

mg=76300 N, m=76300/9.8 kg, s= 418 m, v₀=66.5 mph =29.7 m/s v=28.9 mph=12.2 m/s, h=10.7 m KE1-KE2= W(dr)+ PE m•v₀²/2 - m•v²/2= F(dr) •s +m•g•h Solve for “F(dr)” Answer this Question

*November 12, 2012*

**Physics**

The amplitude at resonance (ω₁ = ω₀) ia Ar =F₀/b•ω₀, A/Ar=A/( F₀/b•ω₀)= { b•ω₀/m}/sqrt{( ω₁²- ω₀²)²+(b²•ω₁²/m²)}= =1/sqrt{(m&#...

*November 12, 2012*

**Physics**

hole d=0.09 m ball D=d+Δd=d+10⁻⁵m=0.09001 m t=27.3℃ ΔL=αLΔT D+ΔD = d+Δd D+ α1•D•ΔT =d + α2•d•ΔT D-d = α2•d•ΔT - α1•D•ΔT= ΔT(α2•d- &#...

*November 9, 2012*

**physics, help!!**

q = Ne=- (8.2•10⁹•1.6•10⁻ ¹⁹)= =- 1.36•10⁻⁹ C. q(sph)+q =4•10⁻⁹ - 1.36•10⁻⁹ = + 2.64•10⁻⁹ C

*November 9, 2012*

**Physics**

The drops begin to move with horizontal speed v(x)=2•π•n•R=2•π•N•R/t= =2•π•22•0.5/44=1.57 m/s The time for falling down is t=sqrt(2h/g) = sqrt(2•1.5/9.8)=0.55 s. L=v(x) •t =1.57•0.55 =0.87 m

*November 9, 2012*

**Physics**

Change in velocity vecorΔv =vecor v₂-vecorv₁

*November 9, 2012*

**Physics**

Change in velocity Δv⃗=v⃗₂-v⃗₁ |Δv|=sqrt(v²+v²-2v•v•cos60°)=sqrt(2v²(1-cos60°))= =sqrt{2•60(1-0.5)}=7.74 m/s The distance covered by the car is s= 2•π•R•60°/360°=2•&#...

*November 9, 2012*

**Physics**

F=ma=mv²/R = =9.1•10⁻³¹• (2.3•10⁶ )²/ 5.1•10⁻¹¹= =9.45•10⁻⁸ N

*November 9, 2012*

**Physics**

http://answers.yahoo.com/question/index?qid=20090929074248AA1Q7sZ

*November 8, 2012*

**physics URGENT! HELP**

v₀ =21.82 mi/h, H= 1.60 x 102 ft. Upward motion h=v₀²/2g Downward motion H+h=v²/2g. v=sqrt{2g(H+h)}

*November 7, 2012*

**physics**

v₀ =21.82 mi/h, H= 1.60 x 102 ft. Upward motion h=v₀²/2g Downward motion H+h=v²/2g. v=sqrt{2g(H+h)}

*November 7, 2012*

**Physics**

a1=ω²•R1 a2=ω²•R2 a1/a2= R1/R2 a2=a1•R2/R1

*November 7, 2012*

**physics**

T=2•π•r/v=2•π•sqrt(r³/GM) T²=4π²r³/GM, r=cuberoot{GMT²/4π²}=.... v=sqrt(GM/r)=.....

*November 7, 2012*