Monday

June 27, 2016
Total # Posts: 4,384

**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*

**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*

**Physics**

Man: φ=ω•t Dog: φ=ε•t²/2 ω•t= ε•t²/2 t=2 ω/ ε=2•0.5/0.31 =3.23 s ω(dog) = ε•t=0.31•3.23= 1 rad/s
*October 26, 2012*

**Physics- PLEASE HELP**

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.
*October 25, 2012*

**Physics**

-0.33
*October 25, 2012*

**Physics**

13.3 cm
*October 25, 2012*

**Physics**

angle of incidence = 30o angle of reflection = 60o angle between incident ray and reflected ray = 120o
*October 25, 2012*

**physics**

α=arctan(Fy/Fx)=arctan(60/40)=arctan 1.5=56.3°
*October 25, 2012*

**Physics**

y: mg=T+N x: F(fr)=T N=mg-T F(fr)= μN=μ(mg- T) = μmg-μT =T T(1+μ)= μmg T= μmg/(1+μ)=0.49•710/1.49=233.5 Newtons
*October 25, 2012*

**Physics**

2•T•sinφ=2•19.5•sin16º=10.75 N
*October 25, 2012*

**Physics!**

F(x) =F1(x) +F2(x) F(x)= 57•cos30°+57•cos60° = F(y) =F1(y) +F2(y) F(y) =57•sin30°+57•sin 60° = F=sqrt{F(x)²+F(y)²} tan θ=F(y)/F(x) F(x) =F1(x) +F2(x) +F3(x) =0 57•cos30°+57•cos60° +F3(x) =0 F(y) =F1(y) +F2(y...
*October 25, 2012*

**Physics**

Initial heights of the blocks: m1 -> H; m2 -> h Initial energy of the “two blocks” system is E1 = m1•g•H+m2•g•h When the blocks covered s=0.9 m the energy of ”two blocks+spring” system is E2= m1•g(H-s•sinα) +m1•...
*October 25, 2012*

**Physics**

0.73 Tesla
*October 25, 2012*

**Physics**

Electrons are emitted with more velocity.
*October 25, 2012*

**Physics**

The atomic number increases by one.
*October 25, 2012*

**Physics**

Both the number of emission of electrons and the maximum velocity will increase.
*October 25, 2012*

**Physics**

1.103 m
*October 25, 2012*

**Physics URGENT! PLEASE**

ΣF(x) =F1(x) +F2(x) +F3(x) =0 0+42•cos60°+F3(x) =0 F3(x) = -42•cos60°= -21 N ΣF(y) =F1(y) +F2(y) +F3(y) =0 36+42•sin60°+F3(y) =0 F3(y) =-36 -42•sin60°= -72.4 N F=sqrt{F(x)²+F(y)²} tan θ=F(y)/F(x)
*October 25, 2012*

**Physics**

F=mv²/R m=FR/ v²
*October 25, 2012*

**Physics**

(m1+m2)a=F m1+m2 =F/a m2=F/a – m1
*October 25, 2012*

**physics**

change of current
*October 25, 2012*

**physics**

h=g•t²/2 t=sqrt(2•h/g) S=v(x) •t
*October 25, 2012*

**physics**

KE=W(fr) KE = m•v²/2 W(fr) =F(fr) •s m•v²/2 =F(fr) •s F(fr)= m•v²/2•s
*October 25, 2012*

**physics**

m1=130/9.8 kg , m2=63/9.8 = m1•a=0=T-F(fr) m2•a=0= m2g-T => T=m2•g F(fr) =T =m2•g
*October 25, 2012*

**Physics**

v=2πR/t = … a= a(n) =mv²/R F=ma
*October 25, 2012*

**High School Physics**

L=vₒ²•sin2α/g, sin2α=L•g/ vₒ²=126•32.174/98²=0.42 1) 2α=25° , α=12.5° 2) 2α =25+90=115°,, α=57.5° t= 2vₒ•sinα/g=...
*October 25, 2012*

**Physics**

m₀=ρV=ρ•πD³/6 ρ =920 kg/m³, D=0.021 m
*October 24, 2012*

**Physics**

mgh =mv²/2 v²= 2gh) = 2•9.8•10 =196 m²/s² a=v²/2•s =196/2•2.7=36.3 m/s² F(fr) = ma=69•36.3 =2504.7 N
*October 24, 2012*

**physics**

KE=PE KE=mgh h=KE/mg =50/0.55•9.8 =9.28 m
*October 24, 2012*

**Physics- Projectiles**

h= vₒ²•sin²α/2g,
*October 24, 2012*

**physics college**

Forward force = T•cosα
*October 24, 2012*

**physics**

Units????
*October 24, 2012*

**Physics**

m1•g=N+T m2•g=T m1•g=N+ m2•g N=(m1-m2)g
*October 24, 2012*

**Physics**

PE=KE + W(fr) mgh =mv²/2 + F(fr)•s m•g•s•sinα =mv²/2 +μ•m•g•cosα•s Solve for μ
*October 24, 2012*

**Physics**

m1 =4.3•10^-3 kg, m2 = 22.6•10^-3 kg, h =1.4 m, L= 3.1 m., v1 =? m1 •v1 = (m1+m2) •v, v= m1 •v1 /(m1 + m2) . h =g•t²/2 => t = sqrt(2•h/g) L= v•t = v• sqrt(2•h/g), v =L/sqrt(2•h/g), m1 •v1 /(m1 + m2) = L/sqrt(2...
*October 24, 2012*

**Physics HELP!!**

F(x) =F1(x)+F2(x)+F3(x) =0+19.9 +0 =19.9 N F(y) =F1(y)+F2(y)+F3(y) =11.1+0-14=-3 N F=sqrt{F(x)^2 + F(y)^2} a=F/m tan alpha=F(y)/F(x) unknown angle =appha +90 degrees
*October 24, 2012*

**Physics HELP!!**

(a) ma=F1-F2 F2=F1-ma (b)F2=F1 (c) -ma=F1-F2 F2=F1+ma
*October 24, 2012*

**Physics**

You accidentally got the right answer. Correect solution is following: The Earth makes one revolution (2π) during 1 year => the angular displacement for 5 days is θ= 2π•5/365 = 0.086 rad. The angular velocity of the Earth is ω= θ/t = 0.086/5&#...
*October 24, 2012*

**Physics**

If your question is: If a rod is moving at a velocity equal to 1/2 the speed of light parallel to its length, what will a stationary observer observe about its length? The answer is L=L₀•sqrt{1-(v/c)²}=0.866L₀ The length of the rod will decrease.
*October 24, 2012*

**Physics**

F=ILBsinα sinα=1 B=F/IL = 2.2/2•1.5 =0.73 T
*October 24, 2012*

**Physics**

Electrons are emitted with more velocity. ε(energy of photon) = Work function + KE ε(blue)>ε(green) => KE(blue) >KE(green) => v(blue) > v(green)
*October 24, 2012*

**Physics**

The atomic number increases by one.
*October 24, 2012*

**Physics**

The atomic number increases by one.
*October 24, 2012*

**Physics**

Both the number of emission of electrons and the maximum velocity will increase
*October 24, 2012*

**physics**

(a) sin r = sini/n =sin 58°/1.6=0.53 r =32 ° α=180°-58°-32°=90° (b) δ= 58 -32 =26°
*October 24, 2012*

**Physics**

Q(t)=Q₀•exp(-t/RC) E~Q² => E=E₀•exp(-2•t/RC) 100-80=20% 0.2•E₀=E₀•exp(-2•t/R•C) ln0.2 =- 2•t/R•C t= - ln0.2• R•C/2 =- (-1.6) •60•90•10⁻⁶/2 =0.0043 s.
*October 24, 2012*

**texas southern university**

p1•V1=p2•V2 p2=p1•V1/V2
*October 24, 2012*

**texas southern university**

T=1/f
*October 24, 2012*

**physics/light reflection**

(a) sin r = sini/n =sin 58°/1.6=0.53 r =32 ° α=180°-58°-32°=90° (b) δ= 58 -32 =26°
*October 24, 2012*

**physics**

Find the solution in Related Questions
*October 24, 2012*

**physics**

m1=1.7 kg, m2 =2.4 kg, L=1.29 m. PE=KE m1•g•L=m1•v²/2 v=sqrt(2•g•L) The velocities of the bodies after elastic collision are u1=(m1-m2)v/(m1+m2), u2=2m1v1/(m1+m2) “-“ means the motion in opposite direction
*October 23, 2012*