Monday

December 22, 2014

December 22, 2014

Total # Posts: 4,338

**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*

**Thermodynamics**

Q= 460000 J m = 27.300kg c = 390 J/kg t₁ = 35℃ t₂ = Final temperature in ℃ t₂ = Q/mc +t₁ = 460000/27.3•390 + 35 =43.2 +35 = 78.2℃
*July 5, 2013*

**Thermodynamics**

ΔL = αLΔT = =12•10⁻⁶•920•420=4.637 m
*July 4, 2013*

**Physics (Thermodynamics)**

Q= 460000 J m = 27.300kg c = 390 J/kg t₁ = 35℃ t₂ = Final temperature in ℃ t₂ = Q/mc +t₁ = 460000/27.3•390 + 35 =43.2 +35 = 78.2℃
*July 4, 2013*

**physics**

s=at²/2 +> a=2s/t²= =2•910⁻³/3²=2•10⁻³ m/s². m₁a=T-F(fr) -m₁gsinα, 0=m₁gcosα –N, m₂a=m₂g –T. F(fr) =μN=μm₁gcosα . m₁a=T- μm₁gcosα...
*July 4, 2013*

**phy**

η={Q(h) –Q(c)}/ Q(h) = =(2000-750)/2000=0.625 => 62.5%. W= Q(h) –Q(c) =2000-750=1250 J. P=W/t=1250/0.5 =2500 W
*July 4, 2013*

**Phisics**

P=Fv
*July 3, 2013*

**physics**

km/h -> m/s, a=v²/2s
*July 3, 2013*

**physics**

km/h -> m/s km -> m a=v²/2s
*July 3, 2013*

**Physics**

km -> m a=v²/2s s=at²/2 t=sqrt(2s/a)
*July 3, 2013*

**Physics**

s=at²/2
*July 3, 2013*

**Physics**

s=v₀t+at²/2
*July 3, 2013*

**physical science**

Find solution in the Related Questions
*July 3, 2013*

**physics**

0.262KE=PE 0.262m₁v²/2 = m₂gh v=sqrt{2m₂gh/0.262m)= =sqrt{2•0.413•9.8•4.22/0.262•11.6} = =3.35 m/s
*July 3, 2013*

**Physics**

cross sectional area has to be measures in cm ² or m² => A= 5.300cm² = 5.3•10⁻⁴ m² = = 0.00053 m² ............... !!!! Factor of safety = ultimate strength/allowable strength= = ultimate strength x Area/ Load = =630•10⁶&#...
*July 3, 2013*

**Physics**

Factor of safety = Ultimate strength/Allowable strength …(1) Allowable strength =Load/Area ….(2) Substitute (2) in (1) => Factor of safety = =Ultimate strength x Area/ Load => Load = ultimate strength x Area/ Factor of safety= =630•10⁶•π&#...
*July 3, 2013*

**Physics**

630.000MPa = 630000kN ?????? Factor of safety = ultimate strength/allowable strength= = ultimate strength x Area/ Load Load = ultimate strength x Area/ Factor of safety= =630•10⁶•π•0.03²/4•5=8.9064•10⁴ N= =89.064 kN
*July 3, 2013*

**Physics**

Correct
*July 3, 2013*

**Physics**

cross sectional area of 10.000cm^2 => Area =10 cm²=10•10⁻⁴ m²=0.001 m² !!!!! Stress =Load/Area = = 5750/0.001 =5750000 Pa = = 5750 kPa Strain = ΔL/L = 2.74•10⁻³/1 = 2.74•10⁻³=0.00274
*July 3, 2013*

**Physics**

Correct
*July 3, 2013*

**Physics**

Correct
*July 3, 2013*

**phy**

Q1 p₁V₁/T₁=p₂V₂/T₂ p₁=p₂ V₁/T₁=V₂/T₂ T₁=27℃= 300K T₂=V₂T₁/V₂ =2 T₁= 600K =327 ℃ Q2 mgh=mcΔT ΔT= gh/c=10•210/4200 =0.5℃
*July 3, 2013*

**phy**

Q1 R is the resistance at 1730℃ P=U²/R = > R=U²/P = 12²/36= 4 Ohms ΔR/R₀=αΔ T (R-R₀)/R₀=αΔ T Δ T=T-T₀=1730-0 =1730 R- R₀=R₀•α•Δ T R₀=R/(1+ α•Δ T...
*July 3, 2013*

**phy**

Q1 σ=Eε F/A = E ΔL/L E= F•L/A• ΔL= 3000•0.5/0.00006•0.2•10⁻³ = 125000000000 N/m² Q2 α= ΔL/L•ΔT ΔL= α•L•ΔT= =0.00002 •0.1•70 =0.00014 m L₁=0.1+0.00014=0....
*July 3, 2013*

**phy**

Q1 ΔQ = mcΔT + mL = 2•4186•10 + 2.26•10⁶•2 = 4.6•10⁶ J initial volume V=m/ρ(water) = 2/1000 =2•10⁻³ m³ final volume = 3.3 m³ Change in volume ΔV=V(fin) – V(init) =3.3-2•10⁻³...
*July 3, 2013*

**phy**

Q1 ΔQ/Δt =- κ•A•ΔT/L = > ΔQ =- κ•A•ΔTt/L= =- 50.4•0.1•(-20) •30•60/0.02 = = 9072000 ≈9100000 J Q2 q= ΔQ/Δt =- κ•A•ΔT/L κ = - q•L/ A•ΔT = = - 4...
*July 3, 2013*

**phy**

dS= dQ/T = m•c•dT/T After integration ΔS=mc ln(T₂/T₁). The temperatures must be in Kelvins. For piece of metal ΔS₁=0.5•600•ln(293/573) =201.2 J/K. For the pool of water (T=const) dS=dQ/T = > ΔS₂ = ΔQ/T = m•...
*July 3, 2013*

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