Friday

October 21, 2016
Total # Posts: 4,384

**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*

**physics**

r= v²/a

*November 6, 2012*

**Physics B**

F =G•m1•m2/R² the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², R= 0.241 m F13= G•m1•m3/R² F23= G•m2•m3/R² F23-F13= =G •m3• (m2-m1)/R²= =6.67•10⁻¹¹•41.1&#...

*November 6, 2012*

**Physics**

T=2•π•r/v=2•π•sqrt(r³/GM) T1=T2

*November 6, 2012*

**physics**

T2=mg/sin52 =57/0.788=72.3 N T1 = T2cos52=72.3•0.616= 44.5 N

*November 6, 2012*

**physics**

F(fr) =μ(s) •m•g•cosα

*November 6, 2012*

**physics**

T=F(fr) =μ(s) •m•g

*November 6, 2012*

**physics**

0=v₀-at a=v₀/t s=at²/2

*November 6, 2012*

**physics**

v=at

*November 6, 2012*

**Physics B**

F =G•m1•m2/R² the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², R= 0.241 m F13= G•m1•m3/R² F23= G•m2•m3/R² F23-F13= =G •m3• (m2-m1)/R²= =6.67•10⁻¹¹•41.1&#...

*November 6, 2012*

**physics**

a=(v-v₀)/t=- 42/0.014=3000 m/s²

*November 6, 2012*

**Physics- PLEASE HELP**

http://www.mathematische-basteleien.de/ladder.htm Problem #7

*November 6, 2012*

**Physics- PLEASE HELP**

L•sinθ

*November 6, 2012*

**physics**

vector T +vector mg +vector F(el) =0 x: -T+F(el)cos37 =0 y: Fsin37 –mg =0 F(el) =qE q•E•cos37 = T q•E •sin37 =mg tan37 =mg/T T=mg/tan37= ... E=T/q•cos37=...

*November 6, 2012*

**physics**

(a) T= m1•g + F(el)= m1•g+ k•q1•q2/r²= =7.5•10⁻³•9.8 + 9•10⁹•32•10⁻⁹•58•10⁻⁹/0.02²=… (b) T1= m1•g + F(el)= m1•g+ k•q1•q2/r1²= T1=0.18 N. ...

*November 6, 2012*

**Physics**

T1=m1•a=1•7.885 = 7.885 N

*November 6, 2012*

**Physics**

For m1: m1•a=T1 (block m1=1 kg is on the surface ) For m2: m2•a=m2•g – T2 (m2=9 kg) For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R² a(m1+m2+ I/R²)=m2•g a= m2•g/(m1+m2+ I/R²) =9•9.8/{1+9+(0.3...

*November 6, 2012*

**Physics**

For m1: m1•a=T1 For m2: m2•a=m2•g – T2 For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R² a(m1+m2+ I/R²)=m2•g a= m2•g/(m1+m2+ I/R²) T1= m1•m2•g/(m1+m2+ I/R²) T2 = m2•g -m2•...

*November 6, 2012*

**Physics HELP!**

http://www.physics.umd.edu/~richman/phys141/hw8.pdf

*November 6, 2012*

**Physics HELP!**

The law of conservation of angular momentum I₁•ω₁ = I₂•ω₂ I₁ =mR²/2, ω₁ = 5.1 rad/s I₂ = mR²/2 +m₀•r². ω₂ = ? (mR²/2) • ω₁=(mR²/2 +m₀•r...

*November 6, 2012*

**Physics HELP!**

The law of conservation of angular momentum I₁•ω₁ = I₂•ω₂ I₁ =mR²/2, ω₁ = 5.1 rad/s I₂ = mR²/2 +m₀•R². ω₂ = ? (mR²/2) • ω₁=(mR²/2 +m₀•R...

*November 6, 2012*

**Physics- PLEASE HELP**

KE(tr) =m•v²/2, KE(rot) = I•ω²/2 =m•R²•v²/2•2• R²=m•v²/4. KE= KE(tr)+ KE(rot)= =m•v²/2 + m•v²/4= 3• m•v²/4.

*November 6, 2012*

**Physics**

M = I•ε I=M/ε

*November 6, 2012*

**Physics- Elena please help!**

Nnewton’s 2 Law for rotation M = I•ε, where angular acceleration ε=0.87 rad/s², the torque M = F•R , the moment of inertia I=mR²/2 , F•R= m•R²•ε/2, m=2•F/R•ε=2•180/0.45•0.87=919.54 kg I=mR&#...

*November 6, 2012*

**Physics- Elena please help!**

x=(m1•x1+m2•x2+m3•x3+m4•x4)/(m1+m2+m3+m4) = 0+0+3•4+5•x4/ 22 = 0 y=( m1•y1+m2•y2+m3•y3+m4•y4)/(m1+m2+m3+m4) = 0+3•5+0+5•y4/22 =0 12+5•x4 = 0 => x4 = - 2.4 m 15+5•y4 =0 => y4= - 3 m

*November 6, 2012*

**physical science**

ΔPE=KE mgΔh=mv²/2 v=sqrt(2gΔh) (a) Δh=5 m, (b) Δh=7.5 m.

*November 6, 2012*

**Physics**

You must know the directions of the forces

*November 6, 2012*

**help one question**

moonquakes. http://www.wcskids.net/grissom/staff_websites/fleming/Textbook%20pdfs/Space%20text%20pdfs/Chapter%207%20-%20Earth,%20Moon,%20Sun%20System.pdf

*November 6, 2012*

**physics**

Electric flux through the sphere with the point charge in its center is Φ = q/ε₀, the area of this sphere is A = 4πr². If the area of the spherical segment is A1=2πr(r-h) Then, from the proportion: q/ε₀ ….. 4πr². Φ1...

*November 6, 2012*

**science**

V=mass/density

*November 6, 2012*

**Physics... URGENT !!!!**

mg=66700 N, m=66700/9.8 kg, s= 418 m, v ₀=66.5 mph =29.7 m/s v=28.9 mph=12.9 m/s, h=13.1 m KE1-KE2= W(dr)+ PE m•v₀²/2 - m•v²/2= F(dr) •s +m•g•h Solve for “F(dr)”

*November 6, 2012*

**physics**

F= m•v²/R

*November 6, 2012*

**Physics**

(a) ω=2π/24•3600 rad/s v= ω•R=... a=ω²R=… (b) r=R•cos42° v1= ω•r=... a1=ω²r=…

*November 6, 2012*

**Physics**

θ=29°=0.5 rad The angular displacement θ=ω•t=(v/R) •t => t=θ•R/v OA=v•t=v• θ•R/v= θ•R=0.5•3.6=1.8 m

*November 6, 2012*

**physics**

ω=ε•t ε=ω/t=100/10=10 rad/s² M=I•ε=(m•R²/2) • ε= 1•0.2²•10/2=0.5 N•m

*November 6, 2012*

**physics**

ω=ε•t ε=ω/t M=I•ε=(m•R²/2) • ε=…

*November 6, 2012*

**physics**

mg=kx k=mg/x=760/0.5•10⁻³=... mgh=mv²/2 mv²/2=kx²/2. mgh=kx²/2. x=sqrt(2mgh/k)

*November 5, 2012*

**physics**

http://www.jiskha.com/display.cgi?id=1351542188

*November 5, 2012*

**Physics**

N1=500, U1 = 120 V, U2 = 24V N2=N1•U2/U1= 500•24/120 =100

*November 5, 2012*

**physics**

mg=55.5•9.8 =...

*November 5, 2012*

**Physics**

vector m•a=vector m•g +vector F(spring) a=0 => 0=vector m•g +vector F(spring) x: 0 = m•g•sinα –k•x, x=m•g•sinα/k=6.1•9.8•sin 39°/126 =0.3 m

*November 5, 2012*

**physics**

R=4.3 m, a(τ)=8.8 m/s². a(τ)=a(c) =v²/R => v=sqrt{a(c) •R}= sqrt{a(τ) •R}= =sqrt{8.8•4.3}=6.15 m/s. v=a(τ) •t, t= v/a(τ)=6.15/8.8 =0.699 s= 0.7 s. s= a(τ) •t²/2 =8.8•0.7²/2=2.156 m.

*November 5, 2012*

**physics**

If polarized light of intensity I₀ is incident on a polarizing sheet, the intensity transmitted by the sheet depends on the angle between the polarization direction of the light and the transmission axis of the sheet, φ, via I= I₀•cos²φ. After ...

*November 5, 2012*

**physics HELP**

F=ma F=F(fr)=> ma=μmg μ=a/g

*November 5, 2012*

**Physics**

V(pool) =π•D²•h/4 V= π•d²•vt/4 π•D²•h/4= π•d²•vvt/4 t=D²vh/d²•v

*November 5, 2012*

**Physics**

Centripetal acceleration a=ω²•R=T/m R=T/m•ω² The law of conservation of angular momentum I• ω²=I₀•ω₀² ω=(I₀/I)•ω₀²=(R₀²/R²)•ω₀ R³=(m&#...

*November 5, 2012*

**Physics**

m=7,45 g, m1=1220 g, m2=1603 g, v=348 m/s, v1 = 0.648 m/s The law of conservation of linear momentum for the 1st collision m•v+0=m1•v1+m•u u=(m•v-m1•v1)/m = v –(m1/m)v1= 348 – (1220/7.45) •0.648 = 241.9 m/s, The law of conservation of ...

*November 5, 2012*

**physics**

What is direction of initial velocity? If v=5 m/s is downward velocity, PE+KE1=KE2 mgh+mv1²/2 =mv2²/2 v2 =sqrt(2gh+v1²)

*November 5, 2012*

**physics**

The question is.....???

*November 4, 2012*

**physics**

m=7,45 g, m1=1220 g, m2=1603 g, v=348 m/s, v1 = 0.648 m/s The law of conservation of linear momentum for the 1st collision m•v+0=m1•v1+m•u u=(m•v-m1•v1)/m = v –(m1/m)v1= 348 – (1220/7.45) •0.648 = 241.9 m/s, The law of conservation of ...

*November 4, 2012*

**College Physic**

If the inclined plane is not frictionless, you have to know the coefficient of friction. If it is frictionless, then PE=KE=KE(transl) +KE(rot) mgh= mv²/2 + Iω²/2. I=mR²/2, ω=v/R mgh= mv²/2 + mR² v²/4 R²= = 3mv²/4 v=sqrt(4gh/3)

*November 4, 2012*

**physics**

x3=7•cos30=6.06 m, y3=7/2 =3.5 m. x=(m1•x1+m2•x2+m3•x3)/(m1+m2+m3) = =0+0+69•6.06/104 =4.022 m. y=( m1•y1+m2•y2+m3•y3)/(m1+m2+m3) = =0+21•7+69•3.5/104 =3.74 m.

*November 4, 2012*

**physics**

W=W(fr)= μ•m•g•s

*November 4, 2012*

**Physics**

The torque is M=F•cosα•L. The ball is material point => its moment of inertia is I=m•L². The Newton’s 2 law for rotation M=I•ε, F•cosα•L= m•L²•ε, ε= F•cosα•L/ m•L²=F&#...

*November 4, 2012*

**physics**

20.7-2.25=18.45 cm Effort x Effort-arm = Load x Load-arm Load is 18.45 29.3/2.25=…

*November 4, 2012*

**Physics**

s=v₀•t+a•t²/2 Solve for v₀

*November 4, 2012*

**physics**

R=4.3 m, a(τ)=8.8 m/s² a(τ)=a(c) =v²/R => v=sqrt{a(c) •R}= sqrt{a(τ) •R}= =sqrt{8.8•4.3}=6.15 m/s v=a(τ) •t, t= v/a(τ)=6.15/8.8 =0.699 s= 0.7 s. s= a(τ) •t²/2 =8.8•0.7²/2=2.156 m.

*November 4, 2012*

**Physics Help, Please**

m•a=F(fr)=μ(s)•m•g a= μ(s)•g

*November 4, 2012*

**Physics**

N = mg+Fsinα = 14•9.8 +200•sin37

*November 4, 2012*

**physics**

The torques: m(A) •g•2 +(m/2) •g•1 = =(m/2) •g•1+m(B) •g•2+m(C) •g•x 2{m(A)-m(B)}=m(C) •x x=2(42-26)/21 = 1.52 (meters)

*November 4, 2012*

**physics**

m1•0+m2•v2=(m1+m2)v v=m2•v2/(m1+m2)

*November 4, 2012*

**physics**

T=2•π•sqrt(l/g) T= 2•π•sqrt(m/k) sqrt(l/g) =sqrt(m/k) L/g=m/k Solve for ‚m’.

*November 4, 2012*

**physics**

Let V be the volume of the duck. Weight of displaced water = weight of the duck 0.26•V•ρ•g= mg 0.26•V•1000•g =mg 260 V = m The density of the duck is ρ= m/V= 260 V/V= 260 kg/m³ V= m/V=2.6/260=0.01 m³

*November 4, 2012*

**physics**

v₁= (m₁-m₂)v₁₀/(m₁+m₂) v₂=2m₁v₁₀ /(m₁+m₂)

*November 4, 2012*

**Physics**

2) 8.8 m/s2 m•a=F(fr)=μ(s) •m•g a= μ(s) • g

*November 4, 2012*

**Physic**

ε= ω/t=32.4/0.861=37.63 rad/s² φ=ω²/2ε=32.4²/2•37.63=13.95 rad=799.3°

*November 4, 2012*

**Physics**

Centripetal acceleration a=ω²•R=T/m R=T/m•ω² The law of conservation of angular momentum I• ω²=I₀•ω₀² ω=(I₀/I)•ω₀²=(R₀²/R²)•ω₀ R³=(m&#...

*November 4, 2012*

**physics**

h = v²/2g Distance= 2h+100(m) The speed is necessary

*November 4, 2012*

**Physics B**

F =G•m1•m2/R² the gravitational constant G =6.67•10^-11 N•m²/kg², F1= G•1.5•3.4/1.8² F1= G•1.5•4.5/4.2² F=sqrt(F1²+F2²) α=arctan(F2/F1) The angle with respect to the positive x axis is 180°+ &#...

*November 4, 2012*

**Physics**

ma=Fcosα-F(fr) F(fr) =μmg a= (Fcosα- μmg)/m

*November 3, 2012*

**college physics**

When the bodies are at rest, the tension in the ropess just equal the weight that is suspended beneath them. The lower body (the stick) has gravity down of m3•g=2•9.8=19.6 N and the tension (T3) in the lower string is T3 = 19.6 N. The rope between the climber and the...

*November 3, 2012*

**physics**

ρ(water) =1000 kg/m³ !!!!!!!!!!!!

*November 3, 2012*

**physics**

k = 1160 N/m r = 1.22 cm = 0.0122 m x = 0.750 cm =0.0075 m Change in the pressure, when the gauge is taken into water ΔP = F/ A= k•x /π•r² = =1160•0.0075/ π•(0.0122)²= ..... ρ(water) =100 kg/m³ ΔP = ρwater• ...

*November 3, 2012*

**physics**

The speeds of the objects after elastic collision are v₁= {-2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂) v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂)

*November 2, 2012*

**Physics**

3.going down and speeding up

*November 2, 2012*

**Physics**

x: 0 =F-m•g•sinα-F(fr) y: 0=N-m•g•cosα, 0 =F-m•g•sinα-F(fr)= F-m•g•sinα- μ•N= =F - m•g•sinα- μ•m•g•cosα. F = m•g(sinα- μ•cosα)= = 340•(...

*November 2, 2012*

**Physics**

I believe that t mass of the traffic light is 27.5 kg!!! We choose the coordinate system with positive torques clockwise. The sum of torques ΣM= 0 about the point A ΣM =-T•H+M•g•L•cosθ+m•g•(L/2)•cos θ = 0; -T(3.80 )+ 27.5...

*November 2, 2012*

**Physics**

n=30000 rpm=30000/60=500rev/s, ω=2πn=2•3.14•500= ... ME=10•10⁶ J =10⁷J KE=Iω²/2 =mR²ω²/2 R=sqrt(2•KE/m• ω²)=… D=2R

*November 2, 2012*

**Physics**

m(hole)= m, M(plate)=M We must treat the holes as the objects of negative mass. The moment of inertia is I=I(plate) – 3I(hole). The plate is disk => I(plate)=M•R²/2. The moment of inertia of the hole using the parallel axis theorem is I(hole)= m•r²/...

*November 2, 2012*

**Physics**

Weight is the force with which the object presses the stationary support. Weight may be less or greater than the force of gravity (mg) in dependence on the downward or upward motion, => Weight = N, but directed in the opposite direction vector (ma)= vector (mg)+vector( N) ...

*November 2, 2012*

**physics**

how much force the floor is pressing up on the crate with F(fr)=μN

*November 2, 2012*

**physics**

sin θ= n(water)/n(glass) n(glass)= n(water)/ sin θ=1.33/sin60.1=1.53

*November 2, 2012*

**physic**

in the wire - 1D (linear waves) in the air -3D(space wave)

*November 2, 2012*

**Physics**

I believe that the refractive index of water is n=1.3 Incidence angle α. sin α =8/sqrt(8²+6)=0.8 sinα/sinβ=n => sinβ= sinα/n =0.8/1.3=0.61 =>β= 38° x=h•tan β=2•tan38°=1.56 m

*November 2, 2012*

**physics**

.."as shown below" ??????????

*November 2, 2012*

**physic question**

kx²/2=mv²/2 v=x•sqrt(k/m) t=sqrt(2h/g) s=v•t= x•sqrt(k/m) • sqrt(2h/g) => s1= x1•sqrt(k/m) • sqrt(2h/g) s2= x2•sqrt(k/m) • sqrt(2h/g) s1/s2=x1/x2 x2=x1•s2/s1 =1.1 (2.15/(2.15-0.23) =1.23 cm

*November 2, 2012*

**physics**

n(relative)=n(glass)/n(air)=n(glass)= =n=1.523 For equilateral prism α=60° n(relative)=sin{½• (α+θ)}/sin(½•α) sin{½• (α+θ)}= n• sin(½•α)=1.583•sin30°=0.7915 ½• (α+θ...

*November 1, 2012*

**physics**

E=500•24•365•24•3600 = ...(J)

*October 31, 2012*

**physics**

Earth’s radius is R = 6.378•10⁶ m. L=Iω=mR²•2•π/24•3600=84•(6.378•10⁶)²•2• π/24•3600=….

*October 31, 2012*