Sunday

November 23, 2014

November 23, 2014

Total # Posts: 4,334

**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*

**phisics**

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*

**Physics**

d•sinθ=2k•λ/2=k λ k=4 d•sinθ==4 λ
*October 31, 2012*

**Physics**

d•sinθ=2k•λ/2=k λ k=4 d sinθ = 4λ
*October 31, 2012*

**Physics**

θ=arcsin(1/n)=arcsin(1/1.501) =41.78°
*October 31, 2012*

**PHYSICS**

http://drcarlphysics.wikispaces.com/Week+5,+Group+3+-WA19,Q8
*October 31, 2012*

**physics**

d=sqrt(18²+25²)=... tanα=18/25, α= arctsn(18/25)
*October 31, 2012*

**physics**

The time of flare falling down t=8 s = > The flare fell from the height h=gt²/2 =9.8•8²/2=313.6 m since the time of accelerated helicopter motion is t1 =25 s, its acceleration was a=2h/t1²= 2•313.6/25²=1 m/s² The helicopter gets off by the...
*October 31, 2012*

**PHYSICS!**

W=mg m=W/g=528/9.8=53.88 kg
*October 31, 2012*

**Physics**

A. Focal length = half of radius of curvature = > f = 25/2=12.5 cm B. 1/o+1/i=1/f, where o is the object distance, I is the image distance , and f is the focal length, 1/I =1/f-1/o= 1/12.5- 1/40 = > I = 18.2 cm. C. Magnification = 12.5/(40-12.5) = 0.45 D. Height of image...
*October 31, 2012*

**Physics**

Its velocity is maximum.
*October 31, 2012*

**Physics**

32.04 r=sin⁻¹(sini/n)
*October 31, 2012*

**Physics**

The path of a light ray that crosses the boundary between two different media is irreversible
*October 31, 2012*

**Physics**

Light cannot be linearly polarized by: A transmission B reflection C diffraction D scattering D
*October 31, 2012*

**Physics**

t=sqrt(2h/g) v= gt=sqrt(2gh)
*October 31, 2012*

**physics**

http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion
*October 31, 2012*

**physics**

http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html
*October 31, 2012*

**physics ..PLEASE HELP ASSIGN DUE SOON :(**

m1•a=F-T-m1•g•sinα m2•a=T (m1+m2)a= F-T-m1•g•sinα+T a=(F-m1•g•sinα)/(m1+m2). T=m2•a=m2• (F-m1•g•sinα)/(m1+m2)=...
*October 31, 2012*

**physics**

E+PE1 =PE2 E=PE2-PE1=m•g•h2 -m•g•h1 = =m•g•s2•sinβ- m•g•s1•sinα = =m•g•( s2•sinβ -s1•sinα )= =3.5•9.8•(16•sin69.3 - 6•sin28.9) =402 J
*October 31, 2012*

**physics**

mg=Tcosα cosα=mg/T PE=KE PE= mgh=mgL(1-cosα) KE= mv²/2 mv²/2=mgL(1-cosα) v=sqrt{2gL(1-cos α)}= sqrt{2gL(1- mg/T)}=…
*October 31, 2012*

**physics**

Without air resistance KE=PE mv²/2=mgh h=v²/2g =27²/2•9.8 = 37.2 m (mgh-mgh1)/mgh = (h-h1)/h= (37.2-16.3)/37.2=0.56 56%
*October 31, 2012*

**Physics HELP!!**

ma=-mg+T T=m(a+g) = 7.37(3.62+9.8) =…
*October 31, 2012*

**Physics HELP!!**

(a) ma=F1+F2 =0 F2=F1 = 27.56 N (b)ma=F1+F2 1.11•(-9.3) = 27.56 +F2 F2= -37.883 N
*October 31, 2012*

**physics**

sinθ =R(cutting surface)/R(cutting tool) =4/40 =0.1 θ=sin⁻¹0.1 =5.7°
*October 31, 2012*

**Physics**

(a) M= F•W =46.6•1.26 = ... (b) M= F•W•sinα =46.6•1.26 •sin43°= (c) M=0
*October 30, 2012*

**Physics**

c•m1• (t-25) = c•m2• (75-t) 80t - 80•25 =150•75- 150t 230t= 13250 t=57.6°
*October 30, 2012*

**Physics**

This ball fell during t=sqrt2h/g) =sqrt(2•0.9/9.8)=0.43 s Horizontal motion (v(x)=const) => L=v(x) •t v(x)=L/t=2.6/0.43=6.1 m/s
*October 30, 2012*

**Physics**

ma=μmg a=μg v=v₀-at=v₀-μgt=...
*October 30, 2012*

**Physics**

PE=KE=KE(transl)+KE(rot) m•g•h=m•v²/2 +I•ω²/2 ω=v/R disc I=m•R²/2 => I•ω²/2= m•R²•v²/2•2• R²=m•v²/4 . m•g•h=m•v²/2 + m•v²/4=0.75 m&#...
*October 30, 2012*

**Physics**

T=2π/ω
*October 30, 2012*

**college physics**

L=vₒ²•sin2α/g, solve for 'v'
*October 30, 2012*

**Physics- Elena please help!**

ma=F(grav) mv²/R =GmM/R² G •ρ•4πR³/3• m/R² =m(v²/R) R=sqrt{3v²/4πρG}=... M = ρ•4πR³/3= …. T=2π/ω=2πR/v=....
*October 29, 2012*

**Physics- Elena please help!**

(a) ω₃=ε•t₃ =0.65•3 =1.95 rad/s (b) ma=F(fr) centripetal acceleration a=v²/R =ω²R F(fr) =μmg m ω²R= μmg ω=sqrt(μg/R)=sqrt(0.42•9.8/0.18)=4.78 rad/s (c) ω =ε•t t= ω/ε=4.78...
*October 29, 2012*

**Physics- Elena please help!**

A full- length recording lasts for 74 min, 33 s , t=74 min33 sec =4473 s ω1=v/R1 =1.22/0.021 = 58.1 rad/s ω2=v/R2 =1.22/0.061 = 20 rad/s ε= (ω2- ω1)/t =(20-58.1)/ 4473=-0.0085 rad/s² φ=εt²/2=0.0085•4473²/2=85032.8 rad L=v...
*October 29, 2012*

**Physics- Elena please help!**

t=74 min33 sec =4473 sá ω1=v/R1 =1.22/0.021 = 58.1 rad/s ω2=v/R2 =1.22/0.061 = 20 rad/s ε= (ω2- ω1)/t =(20-58.1)/ 4473=-0.0085 rad/s²
*October 29, 2012*

**Physics, newtons laws**

v₀=sqrt{v₀² (x) +v₀²(y)}= =sqrt{86.6²+50²} = =100 m/s. tan α= v₀(y)/v₀ (x)=50/86.6 α= tan ⁻¹(50/86.6) =30° t= 2vₒ•sinα/g
*October 29, 2012*

**Physics, newtons laws**

x: ma=Tsinα y: mg=Tcosα ma/mg =Tsinα/ Tcosα a/g =tanα v²/R•g =tanα α=tan⁻¹(v²/R•g) = …
*October 29, 2012*

**physics**

If the carton is moved at uniform velocity ma=0 F-F(fr) –mgsinα = 0 F= F(fr) +mgsinα =μmgcosα+ mgsinα = =mg (μ•cosα + sinα)
*October 29, 2012*

**Physics- Elena please help!!!!**

n₀=3500 rev/min =3500/60 rev/s =58.33 rev/s N= 41 rev, (a) ω=2π•n₀=2π•58.33=366.5 rad/s (b) φ =2π•N =2π•41 =257.6 rad (c) φ= ω²/2•ε, ε= ω²/2•φ=58.33²/2•257...
*October 29, 2012*

**physics**

If the carton is moved at uniform velocity ma=0 F-F(fr) –mgsinα = 0 F= F(fr) +mgsinα =μmgcosα+ mgsinα = =mg (μ•cosα + sinα)
*October 29, 2012*

**Physics- Elena please help!**

PE(spring) =PE + W(fr) k•x²/2 =m•g•(2R) +μ•m•g•s. x= sqrt{2(m•g• 2R +μ•m•g•s)/k} = = sqrt{2m•g• (2R +μ•s)/k}= =sqrt{2•0.5•9.8(3.2 + 0.3•2.2)/84.8}= =0.668 m
*October 29, 2012*

**physics**

(a)W(gravity) = mgh (b) mgh =mv²/2 v=sqrt(2gh) (c) the same (d) mgh = W(fr) + mv1²/2 v1 <v
*October 29, 2012*

**Physics**

mgh =mv²/2 v=sqrt(2gh) = sqrt(2•9.8•5.6) = 10.5 m/s Correct further calculations...
*October 29, 2012*

**Physics**

(a) The law of conservation of energy: PE=KE mgh =mv²/2 v=sqrt(2gh) = sqrt(2•9.8•5) = 9.9 m/s m1==> v1 =+9.9 m/s m2 ==> v2 = - 9.8 m/s (b) the law of conservation of linear momentum: After elastic collision u1 = [(m1 - m2)/(m1 + m2)]•v1 + [ 2 •m2...
*October 29, 2012*

**Physics**

i1 =10A, i2 =1A sde a=10cm=0.1 m, distance between the straight cureent and the side 12 of the loop is b=10 cm =0.1 m Assume that the directions of the currents are following : i1↑I ⃞ ↓i2 Points 1,2,3, and 4 are located clockwise starting from the left bottom...
*October 29, 2012*

**Physics**

red and green
*October 29, 2012*

**Physics**

light reflected from the surface of a dry asphalt roadway
*October 29, 2012*

**physics**

The guage pressure at the height h = ρgh the atmospheric pressure = 1.013•10⁵ Pa ρgh•760/1.013•10⁵ =1030• 9.8•0.61•760/1.013•10⁵=46.2 mmHg
*October 28, 2012*

**phisics**

(a) ΔKE=ΔPE KE(B)-KE( A)=PE(A)-PE(B) mv²/2 -0 =mgh(A)-mgh(B) v=sqrt{2g[h(A)-h(B)]} (b) KE1/KE2 = {m1•gh(A)-mgh(B)}/ {m2•gh(A)-mgh(B)} = =m1/m2
*October 28, 2012*

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