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August 20, 2014

# Posts by Elena

Total # Posts: 4,331

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
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

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
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

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

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

(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

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

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

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

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

Physics help
ma=F-F(fr) a=[F-F(fr)]/m s=at²/2
October 26, 2012

Physics
ω₀=2πn₀=2π/33.2=0.19 rad/s centripetal acceleration is a(cen) = ω₀²R=0.19²•42.2=1.52 m/s² tangential acceleration is a(tan) =ε•R =0.0542•42.2=0.12 m/s² acceleration is a=sqrt{ a(cen)²+a(tan)&#...
October 26, 2012

Physics
Man: φ =ω•t Dog: φ –π/2=ε•t²/2 ω•t= ε•t²/2 +π/2 ε•t² -2 ω•t+π=0 t={2±sqrt(4-2•3.14•0.31}/2•0.31 t1=3.74 s, t2 = 1.48 s ω(dog) = ε•t=0....
October 26, 2012

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