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August 1, 2015

August 1, 2015

Total # Posts: 4,370

**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*

**physics**

a=4.7g =v²/R v=sqrt(4.7•9.8•15)=26.3 m/s
*November 7, 2012*

**physics**

φ=90°=π/2 ω= φ/t a= ω²R= φ²R/t²=π²•0.59/0.31²=60.6 m/s²
*November 7, 2012*

**physics**

I1=m1•r² I2=m2•r² I3=m3•r²
*November 7, 2012*

**Physics HELP!**

M= m•g•L•sinα=7.4•9.8•2.1•sin3°=7.97 N•m
*November 7, 2012*

**Physics**

Before collision PE=KE m•g•h =m•v²/2 v=sqrt(2•g•h)= v₁₀=v₂₀=sqrt(2•9.8•5.8)=10.7 m/s (b) After collision v₁= {-2m₂•v₂₀ +(m₁-m₂)•v₁₀}/(m₁+m₂...
*November 7, 2012*

**Physics**

When the objects are moving in opposite directions v₁= {-2m₂•v₂₀ +(m₁-m₂)•v₁₀}/(m₁+m₂) v₂={ 2m₁•v₁₀ - (m₂-m₁)•v₂₀}/(m₁+m₂)
*November 7, 2012*

**Physics**

On the Earth mg=GmM/R² In the spacecraft mg'=GmM/(5R)²= {GmM/R²}/25=mg/25 His weight is 50.0/25 =2.0 N
*November 7, 2012*

**Physics**

tan α=ma₁/mg = (mv₁²/r)/mg= v₁²/rg tan β=ma₂/mg = (mv₂²/r)/mg= v₂²/rg tan α/ tan β= v₁²/v₂² v₂ = v₁•sqrt(tan β/tan α)=18.5•sqrt(tan16/tan11)= =22....
*November 7, 2012*

**physics**

Q=c•m•ΔT For brass c= 385 J/kg•K ΔT= Q/c•m=760/385•1=1.97º
*November 7, 2012*

**physics**

Q=c•m•ΔT For brass c= 385 J/kg•K ΔT= Q/c•m=760/385•1=1.97º
*November 7, 2012*

**physics**

KE of the objects transforms into PE, and then PE→KE => v(start) =v(land)
*November 7, 2012*

**Physics**

F(fr) =2•μ•N=m•v²/R R= m•v²/2•μ•N=6•10⁻³•17²/2•0.73•1.8 = …
*November 7, 2012*

**Physics Help!**

β = ΔV/V•ΔT β =10•10⁻⁴(1/℃) ΔV=β•V•ΔT=10•10⁻⁴43.3•10⁻³•(34.5-14.5)=8.66•10⁻⁴ m³=0.866 liter.
*November 6, 2012*

**Physics Help!**

Linear expansion coefficient α=ΔL/L•ΔT For steel α =12•10⁻⁶ (1/℃) ΔL=α• L •ΔT= =12•10⁻⁶•514•(32.7+20.3)=0.327 m
*November 6, 2012*