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February 1, 2015

February 1, 2015

Total # Posts: 4,351

**Physics**

A) W=∫F•dx=-∫kx•dx (limits ↓0↑x) =kx²/2 or ΔW=F•Δx =-kx •Δx, W=ΣΔW= Σ(-kx•Δx)= =- kx Σ{(0-x)/2}= kx²/2 B) W= kx²/2=80•0.35/2 =14 J
*January 3, 2014*

**physics**

h=gt²/2 => t=sqrt(2h/g) =sqrt(2•73/9.81) =3.86 s v(y) = gt = 9.81•3.86 =37.87 m/s v=sqrt{v(x)²+v(y)²} =sqrt{37.87²+40²}=55.08 m/s p=mv=Pv/g =177•55.08/9.81 =993.85 kg•m/s tan α =v(y)/v(x) =37.87/40 =0.947 α=arctan0.947...
*January 3, 2014*

**AP Calculus AB**

y´=3(cosx)²(-sinx) -3(sinx)²cosx= = - 3 sinx cosx(cosx+sinx)
*January 3, 2014*

**Need help World Geograpy**

3A 6C 7C 10B 12C 13D 17D 18B 20B
*January 3, 2014*

**Physics**

P=Fv=mav a=P/mv=4300/900•13.8=0.35 m/s²
*January 2, 2014*

**physics**

PE=KE mg(H+h) =mv²/2 v=sqrt{2g(H+h))
*January 2, 2014*

**Physics**

P=Fv=mav a=P/mv=4300/900•13.8=0.35 m/s²
*January 2, 2014*

**Physics**

Q =q•W•L•t= 1380•1000•1300•12•3600= =7.75•10¹³ J Q=cmΔT ΔT= Q/cm = Q/cρV=7.75•10¹³/4180•1000•1000•1300•300= =0.048 K
*January 2, 2014*

**Physics**

m₁=36 kg, c₁= 880 J/kg•K, m₂=4 kg, c₂=387 J/kg•K Q=m₁c₁ΔT + m₂c₂ΔT = = ΔT(m₁c₁ + m₂c₂). ΔT= Q/(m₁c₁ + m₂c₂)= =60000/(36•880+4•387) = 1.8 K
*January 2, 2014*

**Physics**

a=dv/dt= -b+2ct =-0.31+2.1•10⁻³•36= =-0.2344 m/s² P=Fv=ma•v = 1264•0.2344•18.36 = 5439.7 W
*January 2, 2014*

**physics**

Y=2•10¹¹ Pa ω=2πf =2π•2=4π rad/s. ma=T-mg m ω²R =T – mg T=m(ω²R+g) =m(16π²R +g)=… T/A =Y•ΔL/L ΔL=TL/AY =…
*January 2, 2014*

**physics**

V=53•10⁶ ft³=1.5•10⁶ m³ h=45720≈46000 m The density of the air ρ=2•10⁻³ kg/m³ The buoyant force is F=mg= ρVg=2•10⁻³•1.5•10⁶•9.8 =29400≈ 30000 N Answer (d) M=800 kg m...
*January 1, 2014*

**Physics**

W(horse) =W(fr) =μNs= μmgs m=W/ μ gs= 8000/0.25•9.8•62=52.7 kg If the rope is horizontal and v=const => T= F(fr) = μmg=0.25•52.7•9.8 =129 N
*January 1, 2014*

**Physics 221**

E=Δφ/Δx PE->W(work of electric field) -> KE W=e•Δφ=e•E•Δx KE= mv²/2 mv²/2= e•E•Δx v=sqrt{2•e•E•Δx/m}= =sqrt{2•1.6•10⁻¹⁹•1000•0.001/9.1•10&#...
*December 31, 2013*

**Physics**

mv²/2 = kx²/2 +μmgs ==== W(spring) = kx²/2=... E(fr) = μmgs=... v =sqrt{kx²/m +2μ gs}=...
*December 31, 2013*

**Physics - please help me..**

W(block) =kx²/2=252•0.29²/2=10.6 J W(spring) = - 10.6 J mv²/2 = kx²/2 v=x•sqrt(m/k)= =0.29•sqrt(0.395/252) =0.0115 m/s
*December 31, 2013*

**physics**

KE(rot)=Iω²/2 =MR²ω²/2•2 KE(tr)=0.25•KE(rot) KE(tr)=0.25•MR²ω²/4 mv²/2 = MR²ω²/16 v=(Rω/4)•sqrt{2M/m)=4.5 m/s C. 4.5 m/s
*December 30, 2013*

**physics**

M= 5.97•10²⁴ kg m=500 kg r = 6.378•10⁶ m. R= 2• 0⁶ km= 2• 10⁹ m GmM/r-GmM/R = mv²/2 GM(1/r -1/R) = v²/2 v=sqrt{2GM(r-R)/rR}= =sqrt {2•6.67•10⁻¹¹• 5.97•10²⁴(2• 10&#...
*December 27, 2013*

**science**

During 1 period El→Mag→El→Mag→El => T= 4•2 μs= 8 μs=8•10⁻⁶ s f=1/T=1/8•10⁻⁶=1.25•10⁵ Hz
*December 26, 2013*

**physics**

v=sqrt{2•6.67•10⁻¹¹•1.5•1.98•10³º/10⁴}= =1.99•10⁸ m/s ≈2•10⁸ m/s
*December 25, 2013*

**physics**

C. M=1.5•1.98•10³º kg G =6.67•10⁻¹¹ N•m²/kg² mv²/2=GmM/R v=sqrt{2GM/R}=2•10⁸ m/s
*December 25, 2013*

**Physics**

PE ->KE KE ->PE(spring) => PE= PE(spring) mgh=kx²/2 x=sqrt(2mgh/k)=… PE(spring) -> KE->PE => the same ‘h’ (due to the absence of friction)
*December 25, 2013*

**Physics**

ApWeight = mg±mv²/R
*December 24, 2013*

**Physics**

mv²/R=mg v=sqrt(gR)
*December 24, 2013*

**Physics**

a=g•tanθ T= sqrt{(ma)²+(mg)²}= =m•sqrt{a²+g²}
*December 24, 2013*

**Physics**

1. PE(B)=KE+PE(A) mgh= KE+mg2R KE= mgh- mg2R= =mg(h-2R) 2. mgh= mv²/2+mg2R v²= 2g(h-2R) a= v²/R= 2g(h-2R)/R 3. mgh= mv²/2+mg2R h= v²/2g +2R
*December 24, 2013*

**physics**

The total energy at the top point is mv²/2+mg2R. It is equal to the energy at lower point. Its height is h=R-Rcosθ = R(1-cosθ) => the total energy is mv₁²/2 +mgR(1-cosθ). Use the law of conservation of energy. mv²/2+mg2R = mv₁²...
*December 24, 2013*

**physics**

E=h²n²/8ma² Read http://www.umich.edu/~chem461/QMChap3.pdf
*December 23, 2013*

**physics**

V=8000cm³=8•10⁻³m³ p=2 atm= 2.03•10⁵Pa T=273 K ν=pV/RT = 2.03•10⁵•8•10⁻³/273•8.31 = = 0.716 moles V=const p₁/T₁=p/T p₁=T₁p/T=373•2.03•10⁵/273 =2.77•10&#...
*December 23, 2013*

**physics**

Δp•Δx≥h Δx=d=10⁻⁵ Δp•d ≥h Δp≥h/d=6.63•10⁻³⁴/10⁻⁵=6.63•10⁻²⁹ b. change in p greater or equal to 10^-29 kg m/s mΔv•Δx≥h Δx=d = m =...
*December 23, 2013*

**physic**

a= v²/R=420²/430 =410.2 m/s² F=ma=1750•410.2 =717907 N
*December 22, 2013*

**physics**

α=25º, β =35º, m=80 kg, μ=0.7 Fcosβ - F(fr) - mgsin α = 0 N+Fsinβ – mg cos α = 0 N= mg cos α - Fsinβ F(fr) = μN = μ(mg cos α – Fsinβ) Fcosβ -μ(mg cos α – Fsinβ) - ...
*December 21, 2013*

**physics**

My mistake in the last line =1.22•10⁻¹¹ m=12.2 pm
*December 20, 2013*

**Physics 11 Forces Unit**

1. F =G•m₁•m₂/R² the gravitational constant G =6.67•10⁻¹¹ N•m²/kg², F =G•m₁•m₂/R²= =6.67•10⁻¹¹•(4.15•10⁸)²/60²=3191 N 2. g`=GM/R²= =6....
*December 19, 2013*

**Physics AP**

cos φ=0.25/1.2 =0.208 => φ=78° sinφ=0.98, tan φ=4.7. x: ma=Tcosφ y: 0=Tsinφ-mg a=v²/R, Tsinφ=mg Tcosφ= mv²/R Tsinφ/Tcosφ= mgR/ mv² tanφ= gR/v² v=sqrt{gR/tanφ}=sqrt(9.8•0.25/4.7) =0.72 m...
*December 19, 2013*

**physics**

KE=10⁴•1.6•10⁻¹⁹ =1.6•10 ⁻¹⁵J KE=mv²/2 =m²v²/2m =p²/2m => p=sqrt(2m•KE) λ=h/p =h/sqrt((2m•KE)= =6.63•10⁻³⁴/sqrt(2•9.1•10 ⁻³¹•1.6&#...
*December 19, 2013*

**physics**

a=v²/s=51²/75 =34.68 m/s²
*December 18, 2013*

**PHYSICS**

ma=F-μ mg a=F/m -μg
*December 18, 2013*

**physics**

a=μ(k)g s=v²/2a
*December 18, 2013*

**physical**

2) power
*December 18, 2013*

**physics**

↧W=m(g-a) ↥W=m(g+a)
*December 18, 2013*

**magnetis**

a. collapse (shorten)
*December 18, 2013*

**magnets**

Work=0
*December 18, 2013*

**PHYSICS URGENT HELP**

PE=qV V-potential
*December 18, 2013*

**magnets**

Prob. 9 http://www-physics.ucsd.edu/students/courses/winter2010/physics2b/hw/hw8.pdf
*December 18, 2013*

**electricity and magnets**

http://courses.washington.edu/phys431/hall_effect/hall_effect.pdf
*December 17, 2013*

**PHYSICS**

THETA =90
*December 17, 2013*

**physics**

The current in the loop and magnetic field direction are connected according to the right hand rule: curl fingers to the direction of the current in the loop, the thumb points in the direction of magnetic moment of the loop that coincides with the direction of B. => The ...
*December 17, 2013*

**magnetism**

Let x be the distance from the right end of the rails to the rod. Magnetic field of the long straight wire is B=μ₀I/2πr If the infinitesimal horizontal strip of length x and width dr, parallel to the wire and a distance r from it => area A = x dr and the ...
*December 17, 2013*

**manetism help!**

F=qvB mv²/R=qvB R=mv/qB T=2πR/v=2π mv/qBv= =2π m/qB T1=T2 (T doesn’t depend on v)
*December 17, 2013*

**Physical science**

When the balls are moving in opposite directions v₁= {-2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂) v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂)
*December 17, 2013*

**Physics multiple choice**

1- b) False 2 -c) 10 N [up] 3 - d) 400 N
*December 17, 2013*

**Physics**

Constant v = 2πR/T=2π•7.3/34 = 1.3 m/s Constant centripetal acceleration is a=v²/R =1.3²/7.3 = 0.23 m/s²
*December 17, 2013*

**PHYSICS PLS HELP**

a)theta=pi/2
*December 17, 2013*

**Physics**

At the top point v=0 => time of upward motion of the 1st stone is t=v/g =22.7/9.8=2.32 s. Δt =2.32 -2.14 =0.18 s. At t=2.32 s the 1st stone is at the top point h₁=v₀t-gt²/2 =27.7•2.32 – 9.8•2.32²/2 =37.9 m with v=0 and begins its ...
*December 17, 2013*

**Physics**

PE-> KE -> PE(springs) mgh =23 kx²/2 x=sqrt{2mgh/23k} = =sqrt{2•43•9.8•1.93/23•6000} =0.012 m
*December 16, 2013*

**Physics**

F=ma E=F/e=ma/e E=Δφ/Δx => Δφ =E Δx= ma Δx /e= =9.1•10⁻³¹•10¹²•0.04/1.6•10⁻¹⁹=0.2275 V
*December 16, 2013*

**Physics Elena Help pls**

Let x be the distance from the right end of the rails to the rod. Magnetic field of the long straight wire is B=μ₀I/2πr If the infinitesimal horizontal strip of length x and width dr, parallel to the wire and a distance r from it => area A = x dr and the ...
*December 16, 2013*

**science plz help right now!!!!**

1 B 2 D 3 c 4 d 5 c 6 b 7 d 8 b
*December 16, 2013*

**Physics**

Distance between the current I₀ and I is b=sqrt{0.015²+0.1²} =0.101 m. According to Bio-Savart Law, the magnetic fields (B₁=B₂) created by the currents lying on the ground at the point where current I is located are B₁=B₂=μ₀I/...
*December 16, 2013*

**physics-important fast**

a=v²/2s F=ma= mv²/2s
*December 13, 2013*

**physics**

v=ωR t=s/v= s/ ωR=8600/9.1•0.45=2100 s=35 min
*December 13, 2013*

**Physics-urgent**

s=v(t+3) s=at²/2 v(t+3) =at²/2 at²-2vt-6v=0 t²-10t-30=0 t=12.42 s s= v(t+3)=20(12.42+3) = 308.4 m
*December 13, 2013*

**physics**

m₁v=(m₁+m₂)u u= m₁v/(m₁+m₂)= =1.3•4.28/(1.3+0.93)=
*December 12, 2013*

**physics**

m=F/a
*December 12, 2013*

**Physics**

a=(v₂-v₁)/t= [78 – (-23)]/27 =3.74 m/s²
*December 12, 2013*

**PHYSICS HELP!**

m₁v₁-m₂v₂=(m₁+m₂)u u =(m₁v₁-m₂v₂)/(m₁+m₂)= … ΔKE=m₁v₁²/2 + m₂v₂²/2 -(m₁+m₂)u²/2=..
*December 12, 2013*

**physics HELP!**

m₁=0.004 kg, v₁₀=0.26 m/s m₂=0.012 g, v₂= -0.13 m/s v₁=? m₁v₁₀ = m₁v₁ +m₂v₂ v₁= (m₁-m₂)v₁₀/(m₁+m₂) ΔKE = m₁v₁₀²/2- m₁v&#...
*December 12, 2013*

**physics**

F=Δp/Δt =m(v₂-v₁)/Δt v₂ =0 Δt= m v₂/F=(-0.2•25.2)/(- 387)= 0.013 s a= - v₁/Δt=-25.2/0.013 = - 1938.5 m/s² s=- v₁²/2a=-25.2²/2•(-1938.5)=0.16 m
*December 12, 2013*

**physics**

Velocity in the stretched string is v = sqrt(TL/m) For the 3rd harmonic λ=2L/3 λ=v/f => v= λf=2Lf/3 sqrt(TL/m)= 2Lf/3 TL/m=(2Lf/3)² m=9TL/4L²f²=9T/4Lf²= =9•100/4•1.8•79²=0.02 kg=20 g
*December 12, 2013*

**physics**

elastic
*December 12, 2013*

**physics**

h=gt²/2 => t=sqrt(2h/g) L=vt=v• sqrt(2h/g)
*December 12, 2013*

**physics**

h=gt²/2 => t=sqrt(2h/g) L=vt=v• sqrt(2h/g)
*December 12, 2013*

**Physics Classical Mechanics**

http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W09D3-2.pdf
*December 12, 2013*

**Science HELP!!!!**

3. radio waves
*December 10, 2013*

**physics**

At rest: circle A=πR² Motion with v=0.8c r=Rsqrt(1-β²) =R(1-0.8²) =0.6R Area of the ellipse A₁=πRr=π0.6R² =0.6A
*December 8, 2013*

**Physics Classical Mechanics**

http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W13D1-8.pdf
*December 8, 2013*

**physics help**

http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W13D1-8.pdf
*December 8, 2013*

**physics**

http://www.columbia.edu/~crg2133/Files/Physics/8_01/lotsOfProbs.pdf Prob/ #29
*December 6, 2013*

**physics**

α=12•10⁻⁶ K⁻¹ T₁=22° T₂=-28°, T₃= 48° ΔT₁ = T₃-T₂ = 48-(-25) = 73° ΔT₂ =48-22 = 26° ΔL₁=L αΔT₁=9•12•10⁻⁶•73=0.0079 m ...
*December 3, 2013*

**Physics**

for stell α=(11-13) •10⁻⁶ K⁻¹ take α=12•10⁻⁶ K⁻¹ ΔA=A₀2 αΔT ΔA•100%/A₀=2αΔT•100%= =2•12•10⁻⁶ 220•100%= 0.528%
*December 3, 2013*

**Physics**

F₁/A₁=F₂/A₂ F₁=F₂A₁/A₂=mg A₁/A₂= =3100 •9.8•3.7/210 = 535.3 N p=F₁/A₁=535.3/3.7•10⁻⁴=1.447•10⁶ Pa
*December 3, 2013*

**Physics**

ΔL =L αΔT =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm
*December 3, 2013*

**Physics**

p = ρgh = 1000•9.8•0.61= 5978 N/m²
*December 3, 2013*

**PHYSICS**

X(L)=2πfL X(C) =1/2πfC Z=sqrt{R²+ [X(L)-X(C)]²} ΔVmax = I₀•Z cosφ=R/Z
*December 2, 2013*

**Physics**

y=0.105 m is the correct answer
*December 2, 2013*

**physics**

Describe the configuration of the bent wire
*December 2, 2013*

**Physics**

R=mv/eB m =9.1•10 ⁻³¹kg e =1.6•10⁻¹⁹ C.
*December 2, 2013*

**magnetism and electricity**

Desribe the configuration of the bent wire
*December 2, 2013*

**physics**

μ₀I₁/2πy= μ₀I₂/2π(0.28-y) I₁(0.28-y)= I₂y 30(0.28-y)=50y 80y=8.4 y=8.4/80=0.105 m
*December 1, 2013*

**physics**

r=d/2 =0.5 mm =0.0005 m I(max) = B(cr) c r /2 = 0.01•3•10⁸•0.0005 /2 = 750 A
*December 1, 2013*

**Physics**

a=(F-F(fr))/(m1+m2)
*December 1, 2013*

**magnetism**

r=d/2 =0.5 mm =0.0005 m I(max) = B(cr) c r /2 = 0.01•3•10⁸•0.0005 /2 = 750 A
*December 1, 2013*

**magnetic field**

μ₀I₁/2πy= μ₀I₂/2π(0.28-y) I₁(0.28-y)= I₂y 30(0.28-y)=50y 80y=8.4 y=8.4/80=0.105 m
*December 1, 2013*

**physics**

E=Pt/η=4500•20/0.9=100000 J =0.1 MJ
*December 1, 2013*

**physics**

μ₀I₁/2πy= μ₀I₂/2π(0.28-y) I₁(0.28-y)= I₂y 30(0.28-y)=50y 80y=8.4 y=8.4/80=0.105 m
*November 30, 2013*

**physics-magnets**

b=a•cos45 =0.1 •0.707 = 0.0707 m B=2 (μ₀/4π)(I/b)∫cosφdφ {limits: (- π/4) (+π/2)} =2(μ₀/4π)(I/b)(1+0.707)= =2.1414 •(μ₀/4π) •2/0.0707=6.06•10⁻⁶T
*November 29, 2013*

**Physics -Solinoid**

d)
*November 29, 2013*

**physics**

circle: L=2πR R=L/2π B₀=μ₀I/2R= μ₀I2 π /2L= =μ₀πI/L square: L=4a=> a=L/4 The distance from the center of the square to the midpoint of the side of the square is b=a/2tan45=a/2 B⋄=4B₁=4B= =4(μ₀...
*November 29, 2013*

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