Monday

January 26, 2015

January 26, 2015

Total # Posts: 4,350

**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*

**PHYSICS H**

http://www.physicsclassroom.com/mmedia/newtlaws/efar.cfm
*October 23, 2012*

**Physics**

v=sqrt(2gh)
*October 23, 2012*

**Physics**

units???
*October 23, 2012*

**physicsssss**

ω=sqrt(k/m) = sqrt(160/0.885) =13.45 rad/s x=A•sinωt v=A•ω•cosωt v(max) = A•ω A= v(max)/ω =0.38/13.45 =0.028 m a= - A•ω²•sinωt a(max)= A•ω²=0.028•13.45²=5.065 m/s²
*October 23, 2012*

**physics**

k=mg/x=0.625•9.8/0.16 =38.3 N/m T=2•π•sqrt(m/k) m= k•T²/4π²
*October 23, 2012*

**Physics**

v(max) = A•ω ω = v(max)/A T=2π/ω k=mg/A
*October 23, 2012*

**Physics**

y(max1) =mλL/d λ=y(max1)•d/m•L tanα=y(max1)/L λ=tanα•d/m=tan2.5°•20•10⁻⁶/1=0.873•10⁻⁶ m tanα=L •m/d=….
*October 23, 2012*

**Physics**

Frequency is inversely proportional to wavelength. Wavelength is directly proportional to speed. λ=c/f
*October 23, 2012*

**Physics**

f=c/λ=3•10⁸/10⁻⁸ =3Hz
*October 23, 2012*

**Physics**

f=c/λ=3•10⁸/10⁸ =3Hz
*October 23, 2012*

**Physics**

λ=c/f=3•10⁸/6•10¹⁴=
*October 23, 2012*

**Physics**

ultra violet region
*October 23, 2012*

**Physics**

ℰ(max) =NBAω =30•0.04•0.25•100 = 30
*October 23, 2012*

**Physics**

L=L₀•sqrt{1-(v/c)²}=0.866L₀ The length of the rod will decrease.
*October 23, 2012*

**Physics**

Jenifer is right 1/F = 1/0.2 – 1/0.125 – 1/0.4 +1/0.1429 = =5-8-2.5+7 = +1.5 F= + 0.667 m
*October 23, 2012*

**Physics**

Lens 1- F'= +200mm=+0.2 m Lens 2- F'= -125mm = - 0.125 m Lens 3- F'= -400mm =- 0.4 m Lens 4- F'= +142.9mm= +0.1429 m 1/F = 1/0.2 – 1/0.125 – 1/0.4 +1/0.1429 = =5-8-2.5+7 = -3.5 F= - 0.286 m
*October 23, 2012*

**Physics**

C=Q/(T2-T1) c=Q/m(T2-T1)
*October 23, 2012*

**physics to be conceptualize**

In an isolated system where only conservative forces cause energy changes, the kinetic energy and potential energy of the system can change, but the mechanical energy of the system cannot change. The work done by a conservative force on a particle moving between two points ...
*October 23, 2012*

**Physics helppp**

An isolated system is one in which objects inside the system are not influenced by any forces from outside the system.
*October 23, 2012*

**Physics Please help !!**

ΔU = U(x) - U(0) = ∫(6x - 12)dx Therefore, with U(0) = 23 J, we obtain by integrating U = 23 + 12x - 3x ² F = 0 when 6x - 12 = 0 Thus x = 2 m Substituting into U = 23 + 12x² - 3x , U = 35 J Using the quadratic equation find the negative and positive values...
*October 23, 2012*

**physics**

55 µm =55•10⁻⁶ m=0.000055 m=5.5 •10⁻⁷ m
*October 22, 2012*

**physics**

10.1 nm =0.0000000101=1.01•10⁻⁸ m= =1.01•10⁻ ⁵mm=1.01•10⁻ ²µm
*October 22, 2012*

**physics**

2. x = v t
*October 22, 2012*

**Physics**

F=qvBsinα sinα=1 => F=qvB F(p)=e2vB F(e) =evB F(p)/F(e) =2 ...2:1
*October 22, 2012*

**Physics**

m1+m2+m3 = 92.5 kg. m1= 0.31 kg m2+m3 =92.19 kg v=0.55 m/s v1= - 17 m/s v2=? (m1+m2+m3) •v= - m1•v1+(m2+m3) •v2 v2= {(m1+m2+m3) •v +m1•v1}/(m2+m3)=…
*October 22, 2012*

**Physics**

3/sqrt2 = 2.1 A
*October 22, 2012*

**physics**

r(x)= 13.8•cos15.5◦ r(x)= 13.8•sin15.5◦
*October 22, 2012*

**Physics**

Change of the horizontal component of the stone’s linear momentum is Δp(x) = m•v(0x) – (-mv(1x)} = m{v(ox)+v(1x)} = m{v(0) •cos15 +v1•cos12} For the boat Δp=p2-p1 = M•u – 0 = M•u. Thje law of conservation of linear momentum &#...
*October 22, 2012*

**physics**

h=v²/2g
*October 22, 2012*

**PHYSICS**

m•g=k•x₁ k= m•g/x₁=0.116•9.8/0.035= 32.5 N/m When the spring rises 5.7 cm =0.057 m and the spring is compressed to x₂=0.022 m F= k•x₂ =ma a =k•x₂ /m=32.5•0.022/0.116 =6.16 m/s²
*October 22, 2012*

**Physics**

The rope it attached to the man’s hands, and the chair he is in. So, there are two tensions pulling upward on the “child-chair” system. m=(287+223)/g=510/g, T=265 N 2T=mg +ma ma= (2T-mg)/m = =g (2•265- 510)/510= =9.8(530-510)/510= =0.38 m/s² N=...
*October 22, 2012*

**physics help i still dont get it**

sinα =2,5/sqrt{2.5²+(43/2)²} =0.116 m•g=2•T•sinα T=m•g/2•sinα =52/2•0.116 =225 N
*October 22, 2012*

**physics help i still dont get it**

sinα =2,5/sqrt{2.5²+(43/2)²} m•g=2•T•sinα T=m•g/2•sinα
*October 22, 2012*

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