Friday

August 26, 2016
Total # Posts: 4,384

**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*

**PHYSICS**

ΔQ = m•c•ΔT =1000•0.11•(95-20)=8250 cal
*July 3, 2013*

**gen physics**

a=v/t s=at²/2
*July 2, 2013*

**Physics**

PE=KE=KE(trans) +KE(rot) PE= mgh=mgLsinα KE= mv²/2 + Iω²/2= = 2 mv² /2 + 2mr²v²/5•2•r²= =0.5 mv² + 0.2 mv² =0.7 mv² mgLsinα= 0.7 mv² v =sqrt{gLsinα/0.7} = = sqrt{9.8•4•sin42⁰/0.7} =...
*July 2, 2013*

**phy**

Hook’s Law F₁=kx₁ F₂=kx₂ F₁/F₂=x₁/x₂ x₂=x₁F₂/F₁=0.4•100/80 =0.5 m
*July 2, 2013*

**phy**

242 – 22.8 =219.2 mm, 100⁰ /219.2 = 0.456 ⁰/mm, 22.8 + 0.456•20=31.92 mm.
*July 2, 2013*

**Physics**

Load = ultimate strength x Area/ Factor of safety= =630•10⁶•π•0.03²/4•5=8.9064•10⁴ N= =89.064 kN
*July 2, 2013*

**Physics**

cross sectional area has to be measured in cm² 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•5.3•10⁻&#...
*July 2, 2013*

**physics**

ave) = total distance/ total time =s/t t=s/v(ave) = 100/20 = 5 s s=v₀t+at²/2 ….(1) v= v₀+at a= (v -v₀)/t ….(2) Substitute (2) in (1) 2s= v₀t +vt v₀ =( 2s –vt)/t =( 200 -25•5)/5 = 15 m/s a=(v -v₀)/t =(25-15)/5 = 2 m/s...
*July 2, 2013*

**physics**

ave) = total distance/ total time =s/t t=s/v(ave) = 100/20 = 5 s s=v₀t+at²/2 ….(1) v= v₀+at a= (v -v₀)/t ….(2) Substitute (2) in (1) 2s= v₀t +vt v₀ =( 2s –vt)/t =( 200 -25•5)/5 = 15 m/s a=(v -v₀)/t =(25-15)/5 = 2 m/s...
*July 2, 2013*

**Sciences**

mg = Tsinα T=mg/sinα=0.875•9.8/sin10.5 =0.817 N mv²/R = Tcosα v=sqrt{TRcosα/m} = =sqrt{0.817•3.46cos10.5/0.875} = 1.78 m/s
*July 2, 2013*

**General physics**

a= v²/R. top ma= mg –N N=m(g-a) = =m (g- v²/R) = = 34.1{9.8 - (4.23²/9.15 )}= =267 N. bottom ma=N-mg N= m(g+a) = = m (g+ v²/R) = = 34.1{9.8 +(4.23²/9.15 )}= =401 N
*July 2, 2013*

**general physics**

a) KE=mv²/2= 13900•23.5²/2 = 3.83•10⁶ J. b) KE= PE₁+KE₁ KE₁ = KE-PE = mv²/2 -mgh₁ = =3.83•10⁶ - 13900•9.8•12.5= =3.83•10⁶- 1.7•10⁶ =2.13•10⁶ J. mv₁²/2 = KE&#...
*July 2, 2013*

**physics**

v(ave) = total distance/ total time =s/t t=s/v(ave) = 100/20 = 5 s s=v₀t+at²/2 ….(1) v= v₀+at a= (v -v₀)/t ….(2) Substitute (2) in (1) 2s= v₀t +vt v₀ =( 2s –vt)/t =( 200 -25•5)/5 = 15 m/s a=(v -v₀)/t =(25-15)/5 = 2 m...
*July 2, 2013*

**physics**

a= v²/R. top ma= mg –N N=m(g-a) = =m (g- v²/R) = = 34.1{9.8 - (4.23²/9.15 )}= =267 N. bottom ma=N-mg N= m(g+a) = = m (g+ v²/R) = = 34.1{9.8 +(4.23²/9.15 )}= =401 N
*July 1, 2013*

**Physics**

a) KE=mv²/2= 13900•23.5²/2 = 3.83•10⁶ J. b) KE= PE₁+KE₁ KE₁ = KE-PE = mv²/2 -mgh₁ = =3.83•10⁶ - 13900•9.8•12.5= =3.83•10⁶- 1.7•10⁶ =2.13•10⁶ J. mv₁²/2 = KE&#...
*July 1, 2013*

**general physics**

a) the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², r=1.74•10⁶ m M=7.35•10²² kg ma=F F =G•m•M/R² ma=mv²/R G•m•M/R² =mv²/R v=sqrt(GM/R) T=2πR/v = 2π•sqrt{R&#...
*July 1, 2013*

**general 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. r=R+h =6.378•10⁶ +0.327•10⁶ =... ma=F F =G•m•M/r²...
*July 1, 2013*

**physics**

L=vₒ²•sin2α/g, vₒ² = Lg/sin2α F=mvₒ²/R=m Lg/Rsin2α=2259.69 N
*July 1, 2013*

**physics**

Weight =W= mg m=W/g =491 / 9.8 =50.1 kg F = mv²/ r=173.6 =50.1•4.2²/5.09 = 173.6 N since his partner is in the straight-down position, the man also has to apply force to keep his partner from falling down due to gravity, so you have to add the partner's ...
*July 1, 2013*

**physics**

W= mg=> m=W/g = 507 /9.8 = 51.7 kg. ma=T-mg a= T/m - g = =590/51.7 - 9.8 = 1.6 m/s² h=at²/2 t= sqrt(2h/a) = =sqrt(2•30.1/1.6) =6.13 s.
*July 1, 2013*

**physics**

http://answers.yahoo.com/question/index?qid=20090818145339AAdPF06
*July 1, 2013*

**Physics**

F=12 N
*July 1, 2013*

**physics**

ω=2π/T= 2π/24•3600= 7.27•10⁻⁵rad/s 1) equator v= ω•R=7.27•10⁻⁵•6.38•10⁶ = … m/s a=ω²•R= … m/s² 2) at a latitude of 76.0 ° north of the equator r=Rcos76.6⁰=6....
*July 1, 2013*

**physics**

v=at= 0.1g•3.15•10⁷=3.087•10⁷ m/s, s=at²/2=0.1g•(3.15•10⁷)²/2=4.86•10¹⁴m, s₁=d-s=4•10¹⁶-4.86•10¹⁴=3.95•10¹⁶ m, t₁=s₁/v = 3.95•10¹&#...
*July 1, 2013*

**Physics**

[Load] =N (not Pa)
*July 1, 2013*

**physics**

t=2πR/v
*June 30, 2013*

**physics**

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², F =G•m₁•m₂/R² m₁=m m₂=5.65-m F =G•m •(5.65-m)/R² Solve for m
*June 30, 2013*

**Phyiscs**

the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², Earth’s mass is M = 5.98•10²⁴kg, Earth’s radius is R = 6.38•10⁶ m. T=L/v F = m₁•m₂/R= G•m•M/(2.07R) ² F=ma=m•v²...
*June 30, 2013*

**Physics**

φ=ω₀t-εt²/2 ω=ω₀-εt ω=0 ω₀=10π rad/s ε=1.2 rad/s² φ=2πN 0=ω₀-εt => t = ω₀/ε 2πN = ω₀t-εt²/2= =ω₀²/ε - ε...
*June 30, 2013*

**Physics**

T =>he first law states that if the net force (the vector sum of all forces acting on an object) is zero, then the velocity of the object is constant => 1. One force acts on the object.
*June 30, 2013*

**Physics**

f. Force acting down and constant This force is gravity - mg
*June 30, 2013*

**Physics**

F=ma=mv/t
*June 30, 2013*

**Physics**

e. Force acting down and constant (mg - gravity)
*June 30, 2013*

**PHYSICS 2**

KE=W=eU=1• 10⁴ eV
*June 30, 2013*

**physics**

1) ma=T-mg T=m(a+g) 2) a=0 => 0=T-mg T=mg
*June 30, 2013*

**physics**

mgsinα=μN=μmgcosα μ=tanα
*June 30, 2013*

**physics**

v=const => ΣF=0 F₁₂=sqrt(F₁²+F₂²) F₃=F₁₂ F₃(x)= -F₂ F₃(y)= -F₁ tan α= F₃(y)/ F₃(x)= =F₁/F₂ α is the angle that F₃ makes with +x direction (south of east)
*June 30, 2013*

**Physics**

a=v/t
*June 29, 2013*

**physics**

a=v²/2s= 5.5²/2•0.45 = … t=v/a=….
*June 29, 2013*

**Physics (Stress & Strain)**

Factor of safety = ultimate strength/allowable strength Allowable strength= ultimate strength/ Factor of safety= =550•10⁶/3=183.33•10⁶ Allowable strength=Load/Area => Load= Allowable strength x Area = = Allowable strength x (π•d²/4) =183...
*June 28, 2013*

**Physics (Stress & Strain)**

Allowable strength=Load/Area, Factor of safety = =ultimate strength/allowable strength= = ultimate strength x Area/ Load= =550•10⁶•7•10⁻⁴/79000=4.873
*June 28, 2013*

**physics**

vt/L=5.45•10³•111•10⁻³/91.4 =6.62
*June 28, 2013*

**physics**

http://www.god-and-science.com/physics/phy111/2009/eoc02/ch01_a.pdf Problem #27
*June 28, 2013*

**physics**

L=37 m v₁=5.95 m/s v₂=7.8 m/s L+ v₁t= v₂t t=L/ (v₂-v₁)= =37/(7.8-5.95) =20 s d= v₁t=5.95•20=119 m.
*June 27, 2013*

**physics**

N=5540 lines/cm=554000 lines/m d=1/N=1.8•10⁻⁶ m 1) m=2 tanφ₂= x₂/L=26/42.5 =0.61 φ₂=tan⁻¹0.61= 31.5⁰ sinφ₂=sin 31.5⁰=0.52 dsinφ =mλ dsinφ₂=2λ λ= dsinφ₂/m=1...
*June 27, 2013*

**Physics**

C=εε₀A/d ε=1 ε₀=8.85 •10⁻¹² F/m A=Cd/ ε₀= =3.78•10⁻¹²•0.303•10⁻³/8.85•10⁻¹²= =1.29•10⁻⁴ m²
*June 26, 2013*

**Physics (Stress & Strain)**

Shearing stress (tangential stress) τ=V/A, where V ia resultant shearing force which passes through the area A being sheared. V= τA=τ•πdh Shear force is equal to the punching force P P=τπdh= =670•10⁹•3.14•0.03•0.08...
*June 26, 2013*

**Physics (Stress & Strain)**

OK
*June 26, 2013*

**Physics (Stress & Strain)**

Shearing stress (tangential stress) τ=V/A, where V ia resultant shearing force which passes through the area A being sheared. V= τA=τ•πdh Shear force is equal to the punching force P P=τπdh= =670•10⁹•3.14•0.03•0.08...
*June 26, 2013*

**astonomy**

A=4πr²
*June 25, 2013*

**physics**

a=v²/2s =1.9²/2•7.8=0.23 m/s² ma=F(fr)=μN=μmg μ=a/g = 0.23/9.8=0.23
*June 25, 2013*

**Physics**

v₀⒳=v₀•cosα v₀⒴=v₀•sinα L=vₒ²•sin2α/g,
*June 25, 2013*

**Please help how do I calculate this**

Hot water lost Δ H₁=CmΔT₁ =4.180• 25•(51.5-35) =1724 J Cold water got Δ H₂=CmΔT₂ =4.180• 25•( 35-25) =1045 J Δ H = Δ H₁-Δ H₂=1724-1045=679 J heat capacity of the calorimeter (Calorimeter...
*June 25, 2013*

**Physics**

F=ma=mv²/2s
*June 25, 2013*

**Physics**

F=ma=mv/t
*June 25, 2013*