Monday

June 27, 2016
Total # Posts: 4,384

**AP Physics B**

a. PE = mgh=1.5•9.8•0.5 = 7.35 J b. KE=PE = 7.35 J c. KE = mv²/2 v= sqrt(2•KE/m) = =sqrt(2•7.35/1.5)=3.13 m/s mv + 0 = 2mu u=mv/2m = v/2= 3.13/2 =1.565 m/s d. KE₂=2m•u²/2 = 2•1.5•1.565²/2 =3.68 J e. KE₂ = W(fr)=F(fr...
*August 9, 2013*

**physics**

v₀ = 0 v = ω R ω=v/R =15/0.3=50 rad/s ω =εt=> ε = ω/t= 50/8 =6.25 rad/s² 2πN = εt²/2 N= εt²/4π=6.25•8²/4π= 31.8 rev
*August 8, 2013*

**physics**

ω= ω₀- εt 2πn=2πn₀ - εt ε=2π(n₀-n)/t =2π(30-20)/2=31.4 rad/s² N= εt²/4π = 31.4•2²/4π =10 rev L=2πR•N =2π•0.05•10 =3.14 m
*August 8, 2013*

**Physics**

If the net force on Charge B is zero, => the electric field strength is zero at the point B (charge B location). vector E(A) +vector E(C) +vector E(D) = 0 => At the point B: Vectors E(A) and E(C) are direcred to the right => vector E(D) has to be directed to the left...
*August 8, 2013*

**physics**

If the net force on Charge B is zero the electric field strength is zero at this point. vector E(A) +vector E(C) +vector E(D) = 0 => At point B: Vectors E(A) and E(C) are direcred to the right => vector E(D) has to be directed to the left => charge D is positive. kq(A...
*August 8, 2013*

**science**

h=gt^2/2
*August 8, 2013*

**Physics**

ma=Fcosα –F(fr) a=0 0=Fcosα – μN= Fcosα – μmg (b) F = μmg/cosα = (c) W= Fcosα•d = (d) W(fr) = μmg•d =
*August 8, 2013*

**Physics**

mass of the bullet is m = 25 kg (???) mv²/2 =F(res)•d F(res) = mv²/2d d₁= mv₁²/2•F(res)
*August 8, 2013*

**Physical Science**

1) t=s/v= 28/4 = 7 hr 2) KE=PE=mgh =7.2 •9.8•1.2 = …J 3) Q=mcΔT ΔT= Q/mc = 30000/0.39•3.9 = …
*August 7, 2013*

**Physics**

N₁=N₀exp(-λt₁) N₁/N₀ = exp(-λt₁) ln(N₀/N₁) =λt₁ λ = ln(N₀/N₁)/t₁= =ln(2/1.6)/156=0.223/156 (day⁻¹) N₂=N₀exp(-λt₂) = =2•exp(-0.223•201/156)=2...
*August 7, 2013*

**Physics**

angular distance = 880+340 =1220 degr angular displacement =880-340 =540 degr
*August 6, 2013*

**Physics Angular Motion**

ma=T-mg mv²/R=T - mg T = mv²/R + mg
*August 4, 2013*

**Physics Angular Motion**

a= v²/R v=sqrt(aR)
*August 4, 2013*

**physical sciences**

water H₂O acetone CH₃-C(O)-CH₃ ethanol CH₃-CH₂-OH or C₂H₅OH methylated spirit CH₃ OH http://en.wikipedia.org/wiki/Water http://en.wikipedia.org/wiki/Acetone http://en.wikipedia.org/wiki/Ethanol http://en.wikipedia.org/wiki/Methanol
*August 4, 2013*

**Physics**

Upward motion of the weight h =v₀t-gt²/2 v= v₀ -gt v=0 => v₀ =gt => t = v₀ /g =9/9.8 =0.92 s h = v₀²/2g =9²/2•9.8 =4.13 m Let h₀ is the height of helicopter when the weight began its motion. Then for downward motion...
*August 4, 2013*

**physics**

h=gt²/2 t=sqrt(2h/g) v(y) = gt v(x) =15 m/s v= sqrt{v(x)²+v(y)²}
*August 3, 2013*

**physics**

L=10 log₁₀ (I/I₀) I₀= 10⁻¹² W/m² I/I₀ = 10^(L/10) I= I₀•10^(L/10) = =10⁻¹²•10^(130/10)=10 W/m² I₁ = I/19=10/19= 0.53 W/m² L=10 log₁₀ (I₁/I₀) = =10 log&#...
*August 2, 2013*

**physics**

L=10 log₁₀ (I/I₀) I₀= 10⁻¹² W/m² I/I₀ = 10^(L/10)= =10^(L/10)= I= I₀•10^(L/10) = =10⁻¹²•10^(95/10) =3.16•10⁻³W/m² E= I•A•t=I•πr²•t= =3.16•...
*August 2, 2013*

**physics**

Centripetal (centrifugal ) acceleration a= 2 m/s² a=v²/R v=sqrt(aR) =sqrt(2•200) =20 m/s Centripetal force is F=ma= 2500•2 = 5000 N http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html
*August 1, 2013*

**physics**

ΔL= L₁ - L₂ = =10 log₁₀(I₁/I₀) – 10log₁₀(I₂/I₀) = =10 log₁₀(I₁/I₂) ΔL/10 = log₁₀(I₁/I₂) I∼A² A₁/A₂=sqrt{ I₁/I }=sqrt{10^( &#...
*August 1, 2013*

**physics**

(a) L=10 log₁₀ (I₁/I₀) ΔL= L₁ - L₂ = =10 log₁₀(I₁/I₀) – 10log₁₀(I₂/I₀) = =10 log₁₀(I₁/I₂) ΔL/10 = log₁₀(I₁/I₂) I₁/I₂ =...
*August 1, 2013*

**physics**

(a) v=331.3 + 0.606•t =331.3 + 0.606•22 =344.6 m/s s=vt=344.6•0.8=275.7 m
*August 1, 2013*

**physics**

PE=KE mgh=mv^2/2 v=sqrt(2gh) KE= mv^2/2
*August 1, 2013*

**physics**

ΔL= αLΔtº Δtº = ΔL/αL= =0.1/11.8•10⁻⁶•1000=8.43º tº=0+8 =8℃
*August 1, 2013*

**Physics**

x₁=x₂=x F=F₁+F₂ F/x= F₁/x +F₂/x = F₁/x₁+F₂/x₂ => k=k₁+k₂= 35+55 = 90 N/m
*July 31, 2013*

**solid mechanics**

External diameter D=38 mm Inner diameter d=38 -2•2.5 = 33 mm E=2.05•10⁵ N/mm² A=π(38²-33²)/4 =278.8 mm² I = π(38⁴-33⁴)/64 = 44140 mm⁴ L=2300 mm P(cr) = π²•E•I/L² = = π²2.05•...
*July 31, 2013*

**Biomechanics**

h=v²/2g v=sqrt(2gh)=...
*July 31, 2013*

**physics**

T=2πsqrt(L/g) => L=T²g/4π²= =2.15²•9.8/4•π²=1.1475 m L₁=L+ΔL = L+αLΔt= =1.1475 + 19•10⁻⁶•1.1475•133 =1.1504 m T₁=2 πsqrt(L₁/g) = = 2 πsqrt(1.1504/9.8)=2.1527 s...
*July 30, 2013*

**physics**

ΔL=L₁-L = L+ΔL -L = αLΔt L= ΔL/ αΔt =0.195 m /12•10⁻⁶•50 =325 m
*July 30, 2013*

**physics**

λ=3.35•10⁵ J/kg r=2.26•10⁶J/kg Q₁=Q₂ λm₁=rm₂ m₁=rm₂/λ= 2.26•10⁶•2.32/3.35•10⁵=15.65 kg
*July 30, 2013*

**Physics**

Points: left end - A, Sam - B, center of mass - C , Joe - D , right end - E Torques about the point B: mg•BC –F(J) •BD =0 F(J) =mg•BC/BD= 405•2.9/4.8 = 244.7(N) Torques about the point D F(S) •BD - mg•CD = 0 F(S)= mg•CD/BD = 405•1.9...
*July 30, 2013*

**physics**

(a) 0.12•I•A = P, A=P/0.12•I = 150/0.12•700 = 1.79 m². (b) 2•365.25•(9•0.150)kW•h• 3 ¢/kW•h = =2958.5 ¢
*July 30, 2013*

**physics**

λ=v/f =s/t•f=85000/12•11.5 =.... (m)
*July 30, 2013*

**physics**

λ=v/f f=v/ λ =5.25/43 =0.122 s⁻¹=7.23 min⁻¹
*July 30, 2013*

**Physics**

A: p₁=3 atm =3.04•10⁵ Pa, V₁=1 L = 10⁻³ m³; B: p₂=6 atm = 6.08•10⁵ Pa, V₂ =5L=5•10⁻³m³; C: p₃= 4.5 atm = 4.56•10⁵ Pa, V₃=11L = 11•10⁻³m³. A->B: W&#...
*July 30, 2013*

**physical**

0=mgsinα - F(fr)= =mgsinα - μN= =mgsinα - μmgcosα. μ=sinα/cosα =tanα α =tan⁻¹0.58 =30.11º
*July 30, 2013*

**Physics**

p₁/T₁=p₂/T₂ T₁=273+20 =293 K, T₂=273+58=331 K p₂=p₁T₂/T₁= ...
*July 30, 2013*

**physics**

v(x) =v₀(x) =v₀cosα =const
*July 29, 2013*

**physics**

Q=Q₁+Q₂+Q₃= =mc(ice)(27-0)+mλ(ice) +mc(water) (35-0)= =5(2060•27 + 335000 + 4183•35) = …
*July 29, 2013*

**physics**

L=L₁+L₂=L₀₁+ΔL₁ +L₀₂ + ΔL ₂= = L₀₁+α₁•L₀₁•ΔT +L₀₂ + α₂•L₀₂•ΔT= =1+ 23.1•10⁻⁶(450-25) +1+ 12•10&#...
*July 29, 2013*

**physics**

T=F(buoyancy) –mg = = ρ(water)Vg - ρ(balsa)Vg= =Vg[ρ(water) - ρ(balsa)] = =0.25³•9.8(1000 – 160)=129 N
*July 29, 2013*

**physics**

p=p₀+ρgh= 1.013•10⁵ + 800•9.8•18 = =2.42•10⁵ Pa
*July 29, 2013*

**physics**

p= F₁/A₁= F₂/A₂ F₁/π R₁²= F₂/π R₂² F₁ = F₂ R₁²/R₂²= m₂g •R₁²/R₂²= =1500•9.8 •0.005²/0.5²=1.47 N
*July 29, 2013*

**physics**

Standard reference sound intensity I₀= 10⁻¹² W/m² I₁=10⁷W/m² L=10 log₁₀ (I₁/I₀) = =10 log₁₀(10⁷/10⁻¹²) = =190 dB
*July 29, 2013*

**physics**

N=L/λ =Lf/v =10•215/130=16.5 => 16
*July 29, 2013*

**physics**

f=f₀/{1 – (v/u)}=960/{1- (34/343)} = =1065 Hz
*July 29, 2013*

**physics**

T=2πsqrt(L/g), g=4π²L/T²=4π²•0.171/0.833² =...
*July 29, 2013*

**physics**

7.7•x = 53• 1 x=53/7.7 =6.88 m
*July 29, 2013*

**physics**

N=(n₂²-n₁²)/2a=(186²-226²)/(-2•6.4)= =1287.5 rev
*July 29, 2013*

**physics**

m₁v₁ - m₂v₂ =(m₁+m₂)u u= (m₁v₁ - m₂v₂)/(m₁+m₂)= =(1.8•10³•15-9•10²•15)/(1.8•10³+9•10²)= =13500/2700=5 m/s (eastbound)
*July 29, 2013*

**physics**

0=N-m₁g => N = m₁g 0=T-F(fr) => T=F(fr)=μN= μ•m₁g 0=T-m₂g => T= m₂g -------- μ•m₁g = m₂g μ = m₂g/m₁g= 50/100 =0.5
*July 29, 2013*

**physics**

mv²/R =GmM/R² v² =GM/R M= v²R/G (the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²)
*July 29, 2013*

**physics**

F=ma=mω²R ω↑=> F↑
*July 29, 2013*

**physics**

mv₀²/2= mgh h= v₀²/2g mg(H+h) = mv²/2 mgH +mv₀²/2 = mv²/2 v=sqrt{ v₀² +gH} = = sqrt{1.7²+9.8•3.2} = 5.85 m/s
*July 29, 2013*

**physics**

The 1st ball h=v₀t₁+gt₁²/2 The 2nd ball x=v₀t₂ h = gt₂²/2 t₁<t₂
*July 29, 2013*

**physics**

s=vt=90000•0.5/3600 = 12.5 m
*July 29, 2013*

**Physics**

ℰ = -N•dΦ/dt = -N•d(B₀Scosωt)/dt = = NB₀Sωsinωt, ℰ max= NB₀Sω, Without the figure, I believe that S=πR² => ℰ max= NB₀πR²ω = =100•10⁻⁶•3.14•0.1&#...
*July 29, 2013*

**physics**

ma=mω²R, ω=sqrt(a/R) = =sqrt(g/R) = =sqrt(9.8/100)=0.313 rad/s
*July 29, 2013*

**physics**

T=2πsqrt(L/g), T₁=2πsqrt(L/g₁), (T/T₁)²=g₁/g, g₁ = g(T/T₁)² =…
*July 28, 2013*

**physics**

f=1/T =1/2πsqrt(L/g)=...
*July 28, 2013*

**physics**

m₁=10 kg, m₂=m₁+62=72 kg, f₁= ω₁/2π = sqrt(k/m₁)/2 π , f₂=ω₂/2π = sqrt(m₂/k)/2 π, f₂ =f₁•sqrt(m₁/ m₂)=3sqrt(10/72)=1.12 Hz.
*July 28, 2013*

**physics**

R=0.04 m t= 0.012 m The solution is correct.May be there is mistake in calculations (or in your answer)...
*July 28, 2013*

**physics**

Newton’s 2nd law vector(ma) = vector(mg ) +vector( N) +vector(Ffr) Projections: ma=F(fr) 0=N-mg mv²/r =μN N=mg mv²/r = kN = μmg μ=v²/gr. tanα =F(fr)/N = μmg/mg = μ = v²/gr. v=sqrt{gr•tanα} T=2πR/v= 2πR...
*July 28, 2013*

**physics**

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html η=1-Tc/Th =1- 253.15/323.15 =0.2166, η = W/Qh => Qh= W/ η =3.6•10⁷/0.2166 = 1.66•10⁸ J= =39648.4 kCal 10 kW•h•10 Â¢/kW•h= 100 Â¢ (1.66&#...
*July 28, 2013*

**physics**

T=2π•sqrt(m/k) T₁²=4π²m₁/k T₂²=4π²m₂/k T₁²/T₂²=m₁/m₂ m₂=m₁T₂²/T₁² Δm= m₂-m₁ = =m₁(T₂²/T₁²-1)= =0.46...
*July 28, 2013*

**physics**

T₀=2πsqrt(L/g) = =2πsqrt(0.81/9.8)= =1.8064 s. ΔL=α•L•ΔT=19•10⁻⁶•0.81•18 =2.77•10⁻⁴m L₁=L+ ΔL =0.81+2.77•10⁻⁴= =0.810277 T₁=2πsqrt(L₁/g)= =2π...
*July 28, 2013*

**physic**

PE=KE=KE(transl) +KE (rot), mgh=mv²/2 + Iω²/2 = =mv²/2 + mR²v²/4R² =3mv²/4, mgh=3mv²/4, gLsinθ=3v²/4, v=sqrt(4gLsinθ/3).
*July 28, 2013*

**physics**

x is the length of unloaded spring F=k•Δx F=mg => m₁g=k(0.19-x) m₂g=k(0.75-x) m₁/m₂=(0.19-x)/(0.75-x) x=[0.19m₂ - 0.75m₁]/(m₂ -m₁) = =[0.19•2.8 – 0.75•0.3]/(2.8-0.3) = 0.1228 m k= m₁g/(0.19-x) = ...
*July 28, 2013*

**physics**

(a) F=k•Δx F=mg => mg=k•Δx k= mg/Δx =165/0.015 = 11000 N/m (b) m₁a=N-m₁g N=m₁(g+a) k•Δx₁= m₁(g+a) Δx₁=m₁(g+a)/k= =94(9.8+1.25)/11000=0.094 m
*July 28, 2013*

**physics**

PE=KE m₁gh =m₁v₁²/2 v₁ =sqrt(2gh) = sqrt(2•0.03•0.05) = =0.055 m/s. m₁v₁ = m₁u₁ + m₂u₂ u₂=2m₁v₁/(m₁+m₂)=2•0.03•0.055/(0.03+0.02) = 0.066 m/s. 0= u₂ -at a= u...
*July 28, 2013*

**physics**

Mg•L=2mg•x 70•0.65 = 2•30x x=70•0.65/60=0.75 m
*July 28, 2013*

**physics**

(a) mv₁²/2=0.22•N₁•Q N₁=mv₁²/2•0.22•Q = =2•10⁵•55²/2•0.22•1.3•10⁸ =10.6 gal (b) 0.22•N₂•Q= mv₂²/2 -mv₁²/2 +mgh, N₂ = m(v₂² -v...
*July 28, 2013*

**physics**

average SPEED = distance/time taken =(32.5+78)/30•60=0.062 m/s displacement = sqrt(32.5²+78²)=84.5 average VELOCITY = displacement/time taken =84.5/30•60=0.047 m/s
*July 27, 2013*

**physics**

88 km/h =88000/3600= 24.44 m/s ------ average SPEED = distance/time taken =(32.5+78)/30•60=0.062 m/s displacement = sqrt(32.5²+78²)=84.5 average VELOCITY = displacement/time taken =84.5/30•60=0.047 m/s
*July 27, 2013*

**Physics**

You can find the solution in http://www.cabrillo.edu/~cfigueroa/11/11prob_sets/Solutions/11prob_sol_5%20.pdf
*July 27, 2013*

**MITx: 2.01x**

The above answers are correct because I got a green chek and not a red X.
*July 26, 2013*

**MITx: 2.01x**

Q2_1_1 TXC=-t_0*L Q2=1_2 a) 0<=x<L ===>> (2*t_0*L)/(pi*G_0*R4) b)L<=x<=3L =>>(2*t_0*(2*L-))/pi*G_0*R^4) c) x=2*L
*July 26, 2013*

**Physics Help ASAP**

I can’t understand the direction of Observer motion (towards or from the car?). If the car and observer move towards each other, their relative speed is 25-2 =23 m/s, then f=f₀/[1-v/c)] = =520/[1-23/334] = 558.5 Hz If they move in the same direction v =25+2 = 27 m/s...
*July 25, 2013*

**Physics Help ASAP**

http://en.wikipedia.org/wiki/Doppler_effect http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html
*July 25, 2013*

**physics**

ΔL = αLΔT α =12 •10⁻⁶ °C⁻¹ ΔL = 12 •10⁻⁶•65•( -37-22) = - 0.046 m
*July 25, 2013*

**physics**

ΔL = αLΔT ΔL = 12•10⁻⁶•3•2.4•33 = 2.85•10⁻³ m ΔL/2 = 2.85•10⁻³ /2 = 1.425•10⁻³ m
*July 25, 2013*

**physics**

The volume of liquid that spills over the can is the difference between the increase in the volume of the liquid and that of the aluminum can ΔV=β₁VΔT - β ₂VΔT β₁=(ΔV/VΔT) + β ₂ = =(6.1•10⁻⁶/...
*July 25, 2013*

**general physics**

Q₁=m₁cΔT₁ Q₂=m₂cΔT₂ m₁cΔT₁=m₂cΔT₂ m₁ΔT₁=m₂ΔT₂ m₁ΔT₁=m₂ΔT₂ m (40.8 -30.9)=(190-m)(30.9-11.3) Solve for ‘m’
*July 25, 2013*

**Physics**

R=R1+R2+R3
*July 25, 2013*

**physics**

Q₁=mc₁ΔT₁ Q₂=mc₂ΔT₂ mc₁ΔT₁=mc₂ΔT₂ c₁ΔT₁=c₂ΔT₂ c₁(95-30)=c₂(292-30) c₂=c₁(95-30)/(292-30)= =840•65/262 = 208.4 J/kg•°C.
*July 25, 2013*

**physics**

mL= =m₁c ΔT m₁ = mL/c ΔT = =0.068•3.94•10⁵/4183•30.1= =0.213 kg
*July 25, 2013*

**solid mechanics**

E=205 kN/mm²= 205•10⁹ N/m². σ=E•ε(longitudinal) F/A = E•ε(longitudinal) ε=ε(longitudinal) = F/A•E= =40000/(20•10⁻³)²•205•10⁹=4.9•10⁻⁴. If a=0.2 m, b=c=0.002 m, &#...
*July 25, 2013*

**solid mechanics**

normal stress σ = F/A = 250000/0.1•0.1 = 2.5•10⁷ N/m² Tangential stress τ = F /(A/cosα) = Fcosα/A = =250000•0.5/0.1•0.1=1.25•10⁷ N/m²
*July 25, 2013*

**Solid Mechanics**

E=205 kN/mm²= 205•10⁹ N/m², μ = 0.3 σ=E•ε(longitudinal) F/A = E•ε(longitudinal) ε=ε(longitudinal) = F/A•E= =40000/(20•10⁻³)²•205•10⁹=4.9•10⁻⁴. If a=0.2 m, b=...
*July 25, 2013*

**physics**

η= 40% =>η =0.4 input heat Q₁=1.05•10¹⁴J waste heat Q₂=0.6 Q₁=6.3•10¹³ J Q₂/W=1.5 work W=0.4 Q₁=4.2•10¹³ J
*July 24, 2013*

**physics**

W=2.05•10⁸ J η= 4.00%=> η = 0,04 Q =W/0.04=5.125•10⁹ J Waste heat = Q - W = 5.125•10⁹-2.05•10⁸ = 4.92•10⁹ J N= 5.125•10⁹/6•10⁹ = 0.85
*July 24, 2013*

**physics**

Heat input Q=W+Q₁ = =8000+8000 = 16000 J η=W/Q =8000/16000=0.5 η =50%
*July 24, 2013*

**physics**

d=0.025 m Δx=0.31 m W=pΔV=p•πd²•Δx/4= =2.4•10⁵•3.14•0.025²•0.31/4 =36.5 J F=W/ Δx = 36.5/0.31 = 117.8 N
*July 24, 2013*

**physics**

0.2kcal/kg•℃=837.36 J/kg•℃ m = ρV=837.36• (a) Q=mcΔT=2.16•10⁴•837.36•10.5=1.9•10⁸ J (b) α = 2.5•10⁻⁶ /℃ ΔL = α•L•ΔT = 2.5•10⁻⁶•8.5&#...
*July 24, 2013*

**Physics**

mgH-mgh= mv²/2 v=sqrt{2g(H-h)} = …
*July 24, 2013*

**Physics**

s=(v²-v₀²)/2a
*July 24, 2013*

**Dynamics**

c.Velocity vectors before and after the collision.
*July 22, 2013*

**Physics**

λ=c/f => f=c/ λ = 3•10⁸/6•10⁻⁷=5•10¹⁴ Hz
*July 22, 2013*

**Physics**

ω=sqrt(k/m) = sqrt(20/2) = 3.16 rad/s x=Asinωt v=Aωcosωt t=0 => cosωt=1 A=v/ω =4/3.16=1.265 m x=Asinωt sinωt =x/A=1/1.265 =0.79 cos ωt =sqrt{1-sin²ωt} = 0.612 v₁=Aωcosωt =1.265•3.16•0.612 =2....
*July 22, 2013*

**Dynamics**

KE=PE+W(fr) m•v²/2 = m•g•h +F•s= =m•g•s•sinα +F•s. F=m(v²/2s - g•sinα)
*July 22, 2013*