Monday

May 4, 2015

May 4, 2015

Total # Posts: 4,364

**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*

**Physics**

F=Δp/Δt=mΔv/Δt= =m[v₂-(-v₁)]/Δt= =m(v₂+v₁)/Δt= =1.2•(6+2)/ Δt=… Check the given data (time of contact!!!!)and substitute it
*July 13, 2013*

**physic**

W = F • d = Fx•rx + Fy•ry = (12 N) (13 m) + (−1 0 N)(+11 m) = 156 -110=46 J
*July 13, 2013*

**physic**

mv²/R=GmM/R ² v= sqrt (GM/R) =… T=2πR/v= …
*July 13, 2013*

**College Physics**

ω₁= √(g/L₁) ω₂=√(g/L₂) 1.8 ω₁=ω₂ 1.8√(g/L₁)=√(g/L₂) 1.8²g/L₁=g/L₂ L₂=L₁/3.24=10/3.24 = =3.086 ≈3.19 m = 309 cm
*July 13, 2013*

**College Physics**

v= 7 moles, T₁=300 K V₁=1000cm³= 10⁻³ m³ V₂=0.316•10⁻³ m³ W= {(vRT₁)/(γ-1)} •[1-(V₁/V₂)^(γ-1)]= γ=(i+2)/I =(3+2)/3 = 5/3 γ-1=5/3 -1 =2/3= 0.667 W=- {(vRT₁)/(γ-1...
*July 13, 2013*

**Physics**

15 V
*July 13, 2013*

**phy**

F₁=G•m•M/R₁² F₂ =G•m•M/R₂² F₁/F₂=R₂²/R₁² F₂=F₁•R₁²/R₂²= …
*July 13, 2013*

**physic**

F₁=G•m•M/R₁² F₂ =G•m•M/R₂² F₁/F₂=R₂²/R₁² F₂=F₁•R₁²/R₂²= …
*July 13, 2013*

**Physics Please Answer**

The charge of He ion = e The aceleration of ion in the electric field ma=eE a=eE/m KE=mv²/2= m(at)²/2= = (eEt)²/2m ℇ=KE ℇ = hc/λ hc/λ=(eEt)²/2m t={sqrt(2mhc/λ)}/eE= ={sqrt(2•6.65•10⁻²⁷•6.63•10...
*July 13, 2013*

**physic**

mv²/R=GmM/R ² v= sqrt (GM/R) v₁/v₂= sqrt(R₂/R₁) =sqrt(1/4.5)=0.47
*July 13, 2013*

**physic**

See answer to a similar question below under related questions
*July 13, 2013*

**Physics**

0 = m₁v-m₂u u= m₁v/m₂=175•2.75/2400= ...
*July 12, 2013*

**Physics**

F=Δp/Δt=mΔv/Δt= =0.063•45/0.03=94.5 N
*July 12, 2013*

**physics**

KE=PE=mgh
*July 11, 2013*

**Physics**

PE=KE mgh=mv²/2 v=sqrt(2gh)
*July 11, 2013*

**Physics**

PE=KE mgh=mv²/2 v=sqrt(2gh)
*July 11, 2013*

**physics**

m₁v-m₂u=0 u= m₁v/m₂= =1150•26.7/2030=15.13 m/s
*July 11, 2013*

**gen physics**

(a) KE₁ =m₁v²/2 =85.6•4.58²/2=897.8 =898 J. I also can’t obtain the suggested answer (b) and, as a result, the answer (c) (b) According to the law of conservation of linear momentum m₁v =(m₁+m₂)u u= m₁v/(m₁+m₂...
*July 11, 2013*

**Physics**

KE(ini) =mv₁²/2 =8•4²/2=…
*July 11, 2013*

**Physics**

KE(ini) =mv₁2 =8•4²/2=… KE(fin)= mv₂²/2 =8•12²/2=…. W= KE(fin)- KE(ini)= …
*July 11, 2013*

**Physics**

(a) N=q/e=3.2•10⁹/1.6•10⁻¹⁹= 2•10²⁸ (b) The mass of one lead atom is m₀=207.2 amu =207.2•1.67•10⁻²⁷=3.46•10⁻²⁵kg The number of atoms in 8 g of lead N₀=m/m₀=0.008...
*July 10, 2013*

**physics**

F=k•q₁•q₂/r² k =9•10⁹ N•m²/C² q₂ =F•r²/k •q₁
*July 10, 2013*

**physics**

KE=PE mv²/ 2 =kx²/2 k=m(v/x) ²
*July 10, 2013*

**physics**

Pt =E =mv²/2 v=sqrt(2Pt/m) = =sqrt{2•2000•60/1400)= =13.09 m/s
*July 10, 2013*

**Physics**

m₁v=(m₁+m₂)u u= m₁v/(m₁+m₂)= =17000•23/(17000+5100)=17.7 m/s
*July 10, 2013*

**physics**

ma=Fcosα-F(fr) ma=0 F(fr) = μmg Fcosα= μmg μ=Fcosα/mg = =20•cos30/5•9.8=0.35
*July 10, 2013*

**physics**

mv²/R =μmg v=sqrt(μgR) = =sqrt(0.45•9.8•0.12)=0.73 m/s T=2πR/v = 2π•0.12/0.73 = 1.03 s.
*July 10, 2013*

**science**

4πR²3/(4πR³) = 3/R = 3/3960=0.00076
*July 10, 2013*

**physics**

m₁g+m₂g =(m₁+m₂)g =380+175=555 (m₁+m₂) =555/g=56.6 kg m₁a =T m₂a=-T+m₂g m₂a = - m₁a +m₂g a=(m₂g)/(m₁+m₂) = =175/56.6=3.1 m/s²
*July 10, 2013*

**physics[**

(a) a=5.0 m/s² (b) a depends on power of the engine which doesn’t depend on the mass of fuel
*July 9, 2013*

**physics**

ma=mg-F(res) F(res) = m(g-a) = 78(9.8-7.4) = 187.2 N
*July 9, 2013*

**physics**

F=m₁ a₁ => m₁ =F/a₁ . If m₂=m₁/2 a₂=F/m₂=F/(m₁/2) =2F/m₁ = 2a₁
*July 9, 2013*

**Physics**

s₁ = 100 ya=91.44 m v=s₁/t₁ t₂=s₂/v= s₂•t₁/s₁= =100•9/91.44 =9.84 s
*July 9, 2013*

**Physics**

λ=v/f => f = v/ λ, v = sqrt(TL/m ), λ=2L, f = sqrt(TL/m )/ 2L= =sqrt(573•0.803/0.363•10⁻³)/2•0.803= =701 Hz
*July 9, 2013*

**Physics**

b. A Foccault pendulum (a large mass hanging from a tall structure using on a long wire) T=const during the long period of time
*July 9, 2013*

**Physics**

All explanations on the base of formula for period os spring pendulum T=2πsqrt(m/k) (a) m↓=>T↓ i. Shorter (b) iii. Unchanged (c) k↑↓ => T ii. Longer (d) iii. Unchanged
*July 9, 2013*

**Physics**

m₁=3.21 kg m₂=15.3 kg m₂g=T m₁v²/R=T m₁v²/R= m₂g v=sqrt{ m₂gR/ m₁}= =sqrt{15.3•9.8•0.855/3.21} = 1.8 m/s
*July 9, 2013*

**physics**

ax²+bx+c=0, 14x²-19 x +61 =0, x= [19±sqrt{19²+4•14•61}]/2•14 = =(19±61.5)/28. x₁=2.88 m, x₂=-1.52 m
*July 9, 2013*

**physics**

ma=mv²/2s =600•120²/2•400 =10800 P=W/t=[F+F(fr)]s/t = {10800+1200}•400/7.3 = 6.58•10⁵W =894 hp
*July 9, 2013*

**Physics**

E=E₁+E₂=k|q₁|/(d/2)²+ k|q₂|/(d/2)²= k/(d/2)²{ |q₁|+ q₂}= =9•10⁹(7.34+4.49) •10⁻⁶/0.1575= =6.76•10⁵V/m (directed to the negative charge)
*July 9, 2013*

**physics**

http://engineering.myindialist.com/2009/difference-between-a-heat-engine-refrigerator-and-heat-pump/#.UdxFaztM9u0
*July 9, 2013*

**physics**

KE₂-KE₁=W(fr) mv₂²/2 -mv₁²/2 = F(fr)d cosα = =μ•m•g•d•cos180º v₂²/2 - v₁²/2 =μ•g•d• (-1) v₂²/2= v₁²/2- μ•g•d v₂= sqrt(v&#...
*July 8, 2013*

**physics**

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², Earth’s mass is M₁ = 5.97•10²⁴kg, Earth’s radius is R₁ = 6.38•10⁶ m. Moon’s mass is M₂= 7.35•10² kg Moon’s radius ...
*July 8, 2013*

**physics**

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², Earth’s mass is M = 5.97•10²⁴kg, Earth’s radius is R = 6.378•10⁶ m. E(g) = F/m = G•m •M/m(R+h)²= = G•M/(R+h)²= =6.67•10&#...
*July 8, 2013*

**Phyics**

See answer in Related Questions
*July 8, 2013*

**physics**

m(ice)=m₁=0.046 kg L=3.33•10⁵J/kg m(water) =m₂=0.988 kg c=4186 J/kg•℃ m₁L+m₁c(t-0) =m₂c(90-t) Solve for t
*July 8, 2013*

**physics**

F=Δp/Δt FΔt=Δp =p₂-(-p₁)=mv₁+mv₂ (since p₁ directed ↓and p₂ ↑) v₂ = {FΔt- mv₁}/m = ={300•0.01 - 1•2}/1 = 1 m/s
*July 8, 2013*

**physics**

0=mvcosα -Mu v=Mu/ mcosα= =2000•0.5/10•cos60º= 200 m/s
*July 8, 2013*

**math**

V=25•20•12 =... A= 2.5•0.75 = ,,,
*July 8, 2013*

**physics**

L=L₀•sqrt{1-(v/c)²}= =4.4•10⁶•sqrt{1-0.69²}=3.2•10⁶ m
*July 8, 2013*

**physics**

ΔKE= ΔPE 480-200=mgh h=(480-200)/27•9.8 = 1.06 m
*July 8, 2013*

**physics**

1. W(grav) = -mgh 2. W(es) = mgh
*July 8, 2013*

**physics**

W=(F,s)=F•s•cosα = = 2350•817•cos180º= = 2350•817•(-1)= = - 1919950 J
*July 8, 2013*

**math**

9 x 8 +22 =94
*July 8, 2013*

**Maths**

h=sqrt{H²-(a/2)²} = =sqrt{15²-6²} = 13.75 cm, V=ha²/3 =13.75•12²/3=660 cm³, S= 4•(ha/2) =2•15•12 =360 cm².
*July 8, 2013*

**Math**

http://www.mathcaptain.com/algebra/simultaneous-equations.html ax +by =a-b bx-ay=a+b ax+by +(b-a) = 0 bx –ay +(-a-b) = 0 x y 1 -------------------------------- b ↘ b-a ↘ a ↘ b -a ↗ -a-b ↗ b ↗ -a x/{-ab-b²+ab-a²} = y/{b²-...
*July 8, 2013*

**Math**

AB=sqrt{(x₂-x₁)²+(y₂-y₁)²} = =sqrt{(0-(-2))²+(5-(-2)) ²} = =sqrt(4+49)=7.3 BC = sqrt{(x₃-x₂ )²+(y₃-y₂)²} = =sqrt{(3-0)²+(-1-5) ²} = =sqrt(9+36)=6.7 CA = sqrt{(x₁-x₃ )²+(y&#...
*July 8, 2013*

**physics**

a=v/t = 28/20 = 1.4 m/s^2
*July 7, 2013*

**physics**

Work= W=ΔKE= mv²/2 Power= P=W /t = mv²/2t t= mv²/2P = 1000•50²/2•100000 = =12.5 s.
*July 7, 2013*

**physics**

L=vₒ²•sin2α/g, vₒ =sqrt{Lg/ sin2α}= =sqrt{8.59•9.8/sin46º}= =10.8 m/s
*July 7, 2013*

**physics**

... if the vertical distance between the lid of the garbage can andthe runner’s point of release is 0.50m h=gt²/2 t=sqrt(2h/g)= sqrt(2•0.5/9.8) = 0.32 s s=vt= 6.2•0.32 = 2 m
*July 7, 2013*

**Physics**

a) I=mR²/2 http://cnx.org/content/m14292/latest/ b) KE= Iω²/2 http://www.dummies.com/how-to/content/how-to-calculate-rotational-kinetic-energy.html
*July 7, 2013*

**physics**

A. 100 cm B. 200 cm C. 50 cm
*July 7, 2013*

**Physics**

F=(m+M) g +F(fr)= (1000+800)9,8 + 4000 = 5440 N P=Fv = 5440•3=16320 W
*July 7, 2013*

**physics**

mv²/2=mgh h= v²/2g
*July 6, 2013*

**Physics**

(1) I=mR² ω=2π/T KE=Iω²/2 (2) PE= KE(hoop)=KE(trans) +KE(rot) = =mv²/2 + Iω²/2= = mv²/2 + mR²v²/2 R²= = mv²/2+ mv²/2 = mv² PE= KE(block) = mu²/2 => mv²=mu²/2 v²=u²/2 s=v²...
*July 6, 2013*

**phy**

Q1 the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², Radius of the Earth is 6400 000 m =6.4•10⁶ m ===Your data is incorrect!!! mg=G•m•M/R² mg`= G•m•M/(R+h)² (R+h)²/ R² = g /g`=9.8/4.9 = 2 R...
*July 5, 2013*

**physics**

C=εε₀A/d Energy =CU²/2 = εε₀A U² /2d New energy = εε₀A (2U)² /2•2d = εε₀A U² /d= Doubled iniutial energy
*July 5, 2013*