Saturday

October 25, 2014

October 25, 2014

Total # Posts: 9,344

**Physics**

a. Tan A = Y/X = -100/38 = -2.63158 A = -69.2o = 69.2o S of E=20.8o E of S. The bird must head 20.8o W of S. b. d = V*t = 600 km 100t = 600 t = 6 h.
*February 19, 2014*

**Physics**

A. F = F1 + F2 = 81N[80o] + 90N[22o] F=(81*cos80+90*cos22)+(81*sin80+90*sin 22)i = 97.51 + 113.5i F = sqrt(97.51^+113.5^2) = 149.6 N. B. Tan A = Y/X = 113.5/97.51 = 1.16398 A = 49.3o = Direction.
*February 19, 2014*

**physics**

V^2 = Vo^2 + 2a*d a = (V*2-Vo^2)/2d a = (0.59^2-0)/0.86 = 0.405 m/s^2. Fn = m*a = 72 * 0.405 = 29.1 N.
*February 19, 2014*

**physics**

Y^2 = Yo^2 + 2g*h = 0 Yo^2 - 19.6*0.291 = 0 Yo^2 = 5.70 Yo = 2.39 m/s = Ver. component of initial velocity. Vo = Yo/sin A = 2.39/sin3.42 = 40.0 m/s. [3.42o]
*February 18, 2014*

**Physics**

Dx = Xo * Tf = 0.315 m. 0.582 * Tf = 0.315 Tf = 0.541 s. = Fall time or time in air Y = Yo + g*t = 0 + 9.8*0.541 = 5.30 m/s. = Ver. component of final velocity.
*February 18, 2014*

**Physics**

d = Vo*t + 0.5g*t^2 = 5.15 m. 0 + 4.9t^2 = 5.15 t^2 = 1.05 Tf = 1.025 s. = Fall time. Dx = Xo * Tf = 11.9 m. Xo * 1.025 = 11.9 Xo = 11.6 m/s = Required speed.
*February 18, 2014*

**physics**

Vo = 35.2 Ft/s * 1m/3.3Ft = 10.7 m/s.
*February 18, 2014*

**physics**

Y^2 = Yo^2 + 2g*h = 0 Yo^2 - 64*11.7 = 0 Yo^2 = 748.8 Yo = 27.4 Ft/s = Ver. component of the initial velocity. Vo = Yo/sinA = 27.4/sin51 = 35.2 Ft/s = Initial velocity = Speed at which the animal must leave gnd.
*February 18, 2014*

**Physics**

1a. 2.5m[30o] X = 2.5*cos30 = 2.17 m. Y = 2.5*sin30 = 1.25 m. P1(2.17,1.25). 3.8m[120o] X = 3.8*cos120 = -1.9 m. Y = 3.8*sin120 = 3.29 m. P2(-1.9,3.29). 1b. (2.17,1.25), (-1.9,3.29) d^2=(-1.9-2.17)^2 + (3.29-1.25)^2= 20.73 d = 4.55 m. 2. A + B = -14 29i + B = -14 B = -14 -29i...
*February 17, 2014*

**physics**

Fr = ((Vs+Vr)/(Vs-Vc))*Fc = 392 Hz. ((770+0)/(770-Vc))*349 = 392 ((770)/(770-Vc))*349 = 392 268,730/(770-Vc) = 392 301,840 - 392Vc = 268,730 -392Vc = -33,110 Vc = 84.5 mi/h = Velocity of the car. Vs = Speed of sound in air. Vr = Velocity of the receiver(hearer) of the sound.
*February 17, 2014*

**Physics**

L = V/F = 2*0.70 = 1.40 m. V/470 = 1.40 V = 1.40 * 470 = 658 m/s.
*February 17, 2014*

**Math**

The solution should satisfy both Eqs.
*February 17, 2014*

**Math**

Y = 15*sin(90-62) = 7.04 mi. X = 7 + 15*cos(90-62) = 20.2 mi. d = sqrt(20.2^2+7.04^2) = 21.4 mi.
*February 17, 2014*

**Math**

Tan13.25 = h/(X+156.25) h = (X+156.25)*Tan13.25 h = 0.23547x + 36.79 Tan18.17 = h/X h = 0.32820x h = 0.23547x + 36.79 = 0.32820x 0.23547x - 0.32820x = -36.79 -0.09273x = -36.79 X = 396.7 Ft. Tan18.17 = h/396.7 h = 396.7*Tan18.17 = 130.2 Ft = Ht. 0f tower.
*February 17, 2014*

**physics**

1. V = Vo + g*t = 0 @ max ht. 10 - 10t = 0 Tr = 1 s. = Rise time or time to reach max. Ht. Tf = Tr = 1 s. = Time to fall back to top of bldg. d = Vo*t + 0.5g*t^2 d = 10*(3-2) + 5*(3-2)^2 = 15 m. Below top of blgd. 2. Tr + Tf = 4 s. Tr = Tf = 2 s. V = Vo + g*Tr = 0 @ max ht. Vo...
*February 17, 2014*

**physical science**

Wt. = 80kg * 1.625N/kg = 130.7 N. g on moon = 1.625 m/s^2.
*February 16, 2014*

**Math**

X = 6y + 4 In Eq2, replace x with 6y+4 3(6y+4)-18y = 4 18y + 12 - 18y = 4 12 = 4. NO solution!
*February 16, 2014*

**Math**

The point of intersection or solution should satisfy both Eqs.
*February 16, 2014*

**Math**

15Coupons * 5Candy bars/coupon=75 candy bars. Average = Sum/8 16/4 = 20/x. Solve for X.
*February 16, 2014*

**science**

In the above procedure, m*a means mass* acceleration. MA = 20/1.2 = 16.67=Mechanical advantage MA = Fo/Fin = 16.67 Fo/300 = 16.67 Fo = 16.67 * 300 = 5,000 N. = Output. m*g = 5000 m * 9.8 = 5000 m = 510.2 kg. = Combined mass.
*February 16, 2014*

**science**

sin A = 1.2m/20m = 0.06 A = 3.44o Fap-Fp = m*a = m*0 = 0 300 - Fp = 0 Fp = 300 N = Force parallel to the ramp. Fp = mg*sin A = 300 N. m * 9.8*sin3.44 = 300 0.588m = 300 m = 510.2 kg. = Combined mass.
*February 16, 2014*

**Physics**

calculate the work required to launch a satellite from the surface of the earth (radius6380km) to orbit 6000 km above the surface
*February 16, 2014*

**Foundations Math 12**

P = Po(1+r)n r = (5.2%/4)/100% = 0.013 = Quarterly % rate expressed as a decimal. n = 4comp/yr * 4yrs = 16 Compounding periods. P = 30,000(1.013)^16 = $36,886.92 ((36886.92-30000)/30000) * 100% = 23% Return rate.
*February 16, 2014*

**Foundations Math 12**

Glad I could help!
*February 16, 2014*

**Foundations Math 12**

P = Po*(1+r)^n r = (5.3%/12)/100% = 0.004417 = Monthly % rate expressed as a decimal. n = 12Comp/yr * T yrs = 12t Compounding periods. P = 150,000(1.004417)^12t = 200.000 (1.004417)^12t = 200,000/150,000=1.33333 12t*Log(1.004417) = Log1.33333 12t*1.914*10^-3 = 124.9377*10^-3 ...
*February 16, 2014*

**Foundations Math 12**

P = Po(1+r)^n Use same procedure as your 2-16-14, 4:08 AM post. Po = $46,000 P = $100,000
*February 16, 2014*

**Physics**

h = Vo*t + 0.5g*t^2 h = 0 + 4.9 * 2.4^2 = 28.22 m. = Ht. or elevation of gun above gnd.
*February 15, 2014*

**physics**

Y^2 = Yo^2 + 2g*h = 0 Yo^2 = -2g*h = 19.6*3.5 = 68.6 Yo = 8.28 m/s=Ver. component of initial velocity. Vo = Yo/sin A = 8.28/sin27 = 18.24 m/s. = Initial velocity.
*February 15, 2014*

**physic 1**

Wc = m*g = 1700kg * 9.8N/kg = 16,660 N. = Wt. of car. a. Fn=16,660*cos15 = 16.1 N.=Normal force. b. Fp = 16660*sin15 = 4312 N. = Force parallel to the road. Fp-Fs = 0 4312-Fs = 0 Fs = 4312 N. = Force of static friction.
*February 13, 2014*

**physics**

Vo = 5 + i7.35 = 8.89m/s[55.8o] Dx = Vo^2*sin(2A)/g Dx = 8.89^2*sin(111.6)/9.8 = 7.5 m.
*February 13, 2014*

**physics**

P = 1/7 s. F = 1/P = 7 cycles/s = 7 Hz.
*February 13, 2014*

**Math**

P = 35,000 + 0.03*17*35,000 = $52,850
*February 13, 2014*

**Physics**

False. Any wire carrying current has an electric field around it.
*February 13, 2014*

**Algebra**

See related Questions: Tue,2-11-14,6:23 PM.
*February 12, 2014*

**algebra**

Dana's speed = r mi/h Chuck's speed = (r+5) mi/h r*t = 132 mi. t = 132/r = Dana's time. (r+%)*t = 152 t = 152/(r+5) = Chuck's time. Dana's time = Chuck's time: 132/r = 152/(r+5). Solve for r.
*February 12, 2014*

**physics**

1c. Not true. 4. 90km/h = 90000m/3600s = 25 m/s. 5. 100N[170o] + 100N[50o] X = 100*cos170 + 100*cos50 = -34.2 N. Y = 100*sin170 + 100*sin50 = 93.97 N. Tan Ar = Y/X = 93.97/-34.2 = -2.74764 Ar = -70o = Reference angle. A = -70 + 180 = 110o Fr = Y/sin110 = 93.97/sin110 = 100[...
*February 12, 2014*

**Physics**

Wm = m*g = 30kg * 9.8N/kg = 294 N. = Wt. of the mass. Fp = 294*sin35 = 168.6 N. = Force parallel to the incline. Fn = 294*cos35 = 240.8 N. = Normal force or force perpendicular to the incline. Fs = u*Fn = 0.22 * 240.8 = 52.98 N. = Force of static friction. Fap-Fp-Fs = 0 Fap = ...
*February 12, 2014*

**Algebra**

Job A: X hours/wk. Job B: Y hours/wk. 10x + 7.5y => $300 job B:
*February 11, 2014*

**Algebra**

a. C = 8x + 75. b. C = 8x + 75 = 18,955 8x = 18,955 - 75 = 18,880 X = 2360. c.
*February 11, 2014*

**physics**

d = (Vs-Vw)t = 6000 m. (10-Vw)1080s = 6000 10800-1080Vw = 6000 1080Vw = 10,800-6000 = 4800 Vw = 4.444 m/s.
*February 11, 2014*

**physic**

Fn = 3580 - 660 - 1470 = 1450 N. = Net force. a = Fn/m = 1450 / 4110 = 0.362 m/s^2.
*February 11, 2014*

**Physics**

a = sqrt(750^2+850^2) = 1134 m/s^2. F = m*a = 0.39 * 1134 = 442.1 N.
*February 11, 2014*

**physic**

Incomplete.
*February 11, 2014*

**physics**

See previous post.
*February 11, 2014*

**physics**

1. V^2 = Vo^2 + 2g*h h = (V^2-Vo)/2g = (0-6^2)/-19.6=1.84 m. 2. Zero. 3. Calculated in #1.
*February 11, 2014*

**physics**

V = 25km/h = 25000m/3600s = 6.94 m/s. KE = 0.5m*V^2 = 10^4 J. 0.5m*6.94^2 = 10000 24.08m = 10000 m = 415.3 kg.
*February 11, 2014*

**Math**

P = Po(1+r)^n. r = (5.6%/4) / 100% = 0.014 = Quarterly % rate expressed as a decimal. n = 4comp/yr. * 4yrs = 16 Compounding periods. Plug the above values into the given Eq and solve for P.
*February 11, 2014*

**physics**

h = Vo*t + 0.5g*t^2 = 1.75 m 0 + 4.9*t^2 = 1.75 Solve for t.
*February 11, 2014*

**Physics**

To get help, You will have to explain how the resistors are connected and identify Vx.
*February 11, 2014*

**Physics**

a. Fp = 518*sin25 = 218.9 N. = Force parallel to the incline. m*g = 518 N. m = 518/g = 518/9.8 = 52.86 kg. a = F/m = 218.9/52.86 = 4.14 m/s^2. b. V = Vo + a*t = 0 + 4.14*6=24.85 m/s.
*February 11, 2014*

**physics**

Correction: a = (0.75^2-0)/3.2 = 0.1758 m/s^2. b. Fp-Fk = m*a 10.77-Fk = 2.6*0.1758 = 0.4570 Fk = 10.77 - 0.4570 = 10.31 N u = Fk/Fn = 10.31/23.09 = 0.4466
*February 11, 2014*

**physics**

a. Wb = M*g = 2.6kg * 9.8N/kg=25.48 N. = Weight of block. Fp = 25.48*sin25 = 10.77 N. = Force parallel to the inclined plane. Fn = 25.48*cos25 = 23.09 N. = Normal Force = Force perpendicular to the incline. a=V^2-Vo^2)/2d a = (0.75^2-0)/2.6 = 0.2163 m/s^2. b. Fp-Fk = m*a 10.77...
*February 11, 2014*

**Physics**

F = mg*sin30 = 180 * sin30 = 90 N.
*February 11, 2014*

**Physics**

Use same procedure as previous problem.
*February 11, 2014*

**physics**

F1+F2+F3+F4+F5+F6 = 0 i57+40-i77-40+50*cos60+i50*sin60+F6=0 The student can finish this problem by using the same procedure as the previous problem.
*February 11, 2014*

**Physics**

F1 + F2 + F3 = 0 33N[215o] + 45N[315] + F3 = 0 33*cos215+i33*sin215+45*cos315+i45*sin315+F3 = 0 23.33 - i18.93 + 31.82 - i31.82 + F3=0 55.15 - i50.75 + F3 = 0 F3=-55.15 + i50.75=sqr(-55.15^2+50.75^2)= 74.95 N. = The equilibrant. Tan Ar = Y/X = 50.75/-55.15=-0.92022 Ar = -42....
*February 11, 2014*

**math**

(8,-5), (-6,2) m = (2-(-5))/(-6-8) = 7/-14 = -1/2. Y = mx + b = -5 (-1/2)*8 + b = -5 b = -5 + 4 = -1 Y = (-1/2)x -1 Y = -x/2 - 1 Multiply both sides by 2: 2y = -x - 2 X + 2Y = -2
*February 10, 2014*

**Physics**

1. V=62mi/h*1600m/mi*1h/3600s=27.56m/s a = (V-0)/t = (27.56-0)/2.5 = 11.02m/s^2 2. V=90mi/h * 1600m/mi * 1h/3600s=40m/s a = (V-0)/t = 11.02 m/s^2. Solve for t. 3. V = (180/90) * 40m/s = 80 m/s. a = (V-0)/t = 11,02 m/s^2. Solve for t. 4. Yes, with constant acceleration.
*February 9, 2014*

**algebra**

I = (240/540) * 12
*February 9, 2014*

**physical science**

a. Fap-Ff = m*a = m*0 Fap - Ff = 0 The velocity is constant. Therefore, the acceleration is zero. Fap = Force applied. Ff = Force of friction. b. Fap - Ff = 0 100 - Ff = 0 Ff = 100 N.
*February 9, 2014*

**Algebra**

Pitch = (Y/X) * 100% = (13/13.8) * 100%
*February 9, 2014*

**math**

cos52 = X/L = 100/L L = 100/cos52
*February 9, 2014*

**physics**

V = Vo + g*Tr = 0 10.5 - 9.8*Tr = 0 9.8t = 10.5 Tr = 1.07 s. = Rise time. hmax = ho + (V^2-Vo^2)/2g = 76 + (0-10.5^2)/-19.6 = 81.63 m. Above gnd. hmax = Vo*t + 0.5g*t^2 = 81.63 m. 0 + 9.8*t^2 = 81.63 t^2 = 8.33 Tf = 2.89 s. = Fall time. Tr+Tf = 1.07 + 2.89 = 3.96 s. To reach ...
*February 9, 2014*

**Physics**

h = Vo*t + 0.5g*t^2 = 115.7 m Vo*5.6 - 4.9*5.6^2 = 115.7 Vo*5.6 = 115.7+153.7 = 269.4 Vo = 48.1 m/s.
*February 9, 2014*

**Physics**

a. V = Vo + g*t = 0 + 9.8*2.04 = 20 m/s b. d = 0.5g*t^2 = 4.9*2.04^2 = 20.39 m. Downward.
*February 9, 2014*

**phys**

V = Vo + a*t = 12 + 4*3 = 24 m/s = Final velocity. d=(V^2-Vo^2)/2a = (24^2-12^2)/8 = 54 m.
*February 9, 2014*

**Physics**

41km/h[0o], 28min. 41km/h[37o], 15min. 41km/h[180o], 64min a. d1=41km/h[0o] * (28/60)h=19.13km[0o] d2=41km/h[37o] * (15/60)h=10.25km[37o]= 8.19 + i6.17 d3=41km/h[180o] * (64/60)h=43.73km[180o] V=(d1+d2+d3)/(t1+t2+t3)=(19.13+8.19+i6.17-43.73)/(28+15+64) = (-16.41 + i6.17)/107=...
*February 9, 2014*

**Physics**

a. d = 407km[0o] + 871km[270o] X = 407 + 871*cos270 = 407 km Y = 871*sin270 = -871 km. D = sqrt(407^2 + (-871^2)) = 961.4 km. b. Tan A = Y/X = -871/407 = -2.14005 A = -64.95o = 64.95o S of E. = Direction. c. V=D/T = 961.4/(1.9+0.7333)=2.63 km/h. d. 64.95o S of E.
*February 9, 2014*

**physic**

See previous post.
*February 9, 2014*

**physic**

X = 4 km Y = -6 km R = sqrt(4^2 + (-6)^2)
*February 9, 2014*

**Algebra 1**

P = 300 - 0.3*300
*February 9, 2014*

**Physics**

17.6m[43o] a. X = 17.6*cos43 b. Y = 17.6*sin43
*February 9, 2014*

**Physics**

10.2m[48o] + B = 15m[15o] 10.2*cos48+i10.2*sin48+B=15^cos22+i15*sin22. 6.83 + i7.58 + B = 13.91 + i5.62 B=13.91-6.83 + i5.62 - i7.58=7.08-i1.96 B = sqrt(7.08^2 + (-1.96)^2) = 7.35 m. b. Tan A = Y/X = -1.96/7.08 = -0.27684 A = -15.47o S of E = 344.5o, CCW.
*February 9, 2014*

**Physics**

Note: Xt = 7-(-3.82) = 10.82 km.
*February 9, 2014*

**Physics**

a. 22km[260o] + 34[90o]. X = 22*cos260 = -3.82 km. Xt = 7.0 (-3.820 = 10.82 km = Total hor distance to B. Y = 22*sin260 + 34*sin90 = 12.33 km. D = sqrt((11.82)^2 + 12.33^2)= 17.1 km b. Tan A = Y/Xt = 12.33/10.82 = 1.13956 A = 48.7o
*February 9, 2014*

**Physics**

See previous post.
*February 9, 2014*

**Physics**

(Vb+Vr)2 = 26 km Eq1: 2Vb + 2Vr = 26 (Vb-Vr)6 = 26 km Eq2: 6Vb - 6Vr = 26 Add Eq1 and Eq2: 2Vb + 2Vr = 26 6Vb - 6Vr = 26 Multiply Eq1 by 3: 6Vb + 6Vr = 78 6Vb - 6Vr = 26 Sum: 12Vb = 104 Vb = 8.6667 km/h = Velocity of the boat. In Eq1, replace Vb with 8.6667: 2*8.6667 + 2Vr = ...
*February 9, 2014*

**Physics**

Vo = 0 V = 34.72 m/s. a = 5.6 m/s^2 A. V = Vo + a*t = 34.72 0 + 5.6t = 34.72 t = 6.2 s. Length = Vo*t + 0.5a*t^2 Length = 0 + 2.8*6.2^2 = 107.6 m. B. V^2 = Vo^2 + 2a*L = 34.72^2 = 1205.5 0 + 2a*75 = 1205.5 150a = 1205.5 a = 8.04 m/s^2. C. V = Vo + a*t = 34.72 m/s. 0 + 8.04t = ...
*February 8, 2014*

**Algebra 2**

(68-48sqrt2)^(1/4) = (68-67.882)^(1/4) = (0.118)^(1/4) = 0.5861
*February 8, 2014*

**Wave**

a. Period
*February 8, 2014*

**physics**

d. Reduced by a factor of 2: P = V^2/2R.
*February 7, 2014*

**physics**

1. False. 2. False. 3. False. 4. True. 5. True. 6. False. Note: The resistivity 0f both wires is the same, because they are made of the same material(copper).
*February 7, 2014*

**Physics**

d = Vo*t + (V^2-Vo^2)/2a d = 20*0.5 + (0-20^2)/-10 = 50 m. Note: Vo*t is the distance traveled during the reaction time. The 2nd Eq was derived from V^2 = Vo^2 + 2a*d. Solve for d.
*February 7, 2014*

**Physics**

Vc = Vo + a*(T-To)Vo Vc=3.5*10^-4 + 69*10^-6*(80-8)3.5*10^-4= 3.56*10^-4 m^3 = Vol. of can at 80oC. Vl = Vc + Vs = 3.56*10^-4 + 5.8*10^-6 = 3.618*10^-4 m^3=Vol. of liquid at 80oC. 3.618*10^-4 - 3.5*10^-4 = 1.18*10^-5 m^3 = The increase in vol. of the liquid. a*(T-To)Vo = 1.18*...
*February 7, 2014*

**Physics**

a. d = Vo*t + 0.5a*t^2 = 11.5 m. 0 + 0.5a*5.4^2 = 11.5 14.58a = 11.5 a = 0.789 m/s^2 m = F/a = 73/0.789 = 92.55 kg. b. V=Vo + a*t = 0 + 0.789*5.4=4.26 m/s. d = Vo*t = 4.26 m/s * 4.20s = 17.89 m.
*February 7, 2014*

**Physics**

m = 2.5 kg h = 1.7 m. Work=F*h = mg*h = 2.5*9.8*1.7=41.65 J. No work is done when can is stationary.
*February 7, 2014*

**physics**

X = 4.3 m/s. A = 44o V = X/cosA = 4.3/cos44 = 5.98 m/s = Total velocity.
*February 7, 2014*

**math**

9pages/15min = 0.6 Pages/min. 198pages * 1min./0.6pages = 330 min. reading. (330min/30min) * 5min = 55 min. on break. 330 + 55 = 385 min = 6 hours-25 min consumed. 8:00 AM + 6:25 = 14:25=2:25 PM.= Finish time.
*February 6, 2014*

**Science**

Ratio = 200/20 = 10/1. 10 Loops on Primary and 1 loop on Secondary. p
*February 6, 2014*

**physics**

Series: Ct = 4.98*8.98/(4.98+8.98) = 3.20 uF. Parallel: Ct = 4.98 + 8.98 = 13.96 uF.
*February 6, 2014*

**math**

2/9/(3/5) = X/1 X = 2/9 * 5/3 = 10/27 gal.
*February 6, 2014*

**Physics**

x^2 + 2x - 35 = 0 -35 = -1*35 = -5*7. Choose the pair of factors whose sum = 2, the coefficient of X. (x-5)(x+7) = 0 x-5 = 0 X = 5. x+7 = 0 X = -7. Solution Set: X = -7, and 5.
*February 6, 2014*

**Physics**

See Related Questions: Mon,9-5-11,11:19 PM.
*February 6, 2014*

**trig**

h = 80*sin40
*February 5, 2014*

**Physics**

See Related Questions: Fri, 9-9-11,11:20 AM.
*February 5, 2014*

**Physics**

What is the question?
*February 5, 2014*

**Physics**

a. a = (V^2-Vo^2)/2d a = (30.9^2-21^2)/382 = 1.35 m/s^2. b. 21 + 1.35t = 30.9 1.35t = 9.9 t = 7.33 s.
*February 5, 2014*

**physics**

a. h1 = 0.5a*t^2 = 7.5*21^2 = 3308 m V = a*t = 15 * 21 = 315 m/s. hmax = h1 + (V^2-Vo^2)/2g = 3308 + (0-315^2)/-19.6 = 8371 m.
*February 5, 2014*

**Physics**

Vo = 17.9m/s[39o] Xo = 17.9*cos39 = 13.91 m/s. Yo = 17.9*sin39 = 11.26 m/s. Y = Yo + g*Tr = 0 @ max ht. 11.26 - 9.8Tr = 0 9.8Tr = 11.26 Tr = 1.15 s. = Rise time. Tf = Tr = 1.15 s. = Fall time. Tr+Tf = 1.15 + 1.15 = 2.30 s. = Time to hit gnd.
*February 5, 2014*

**physics**

Incomplete.
*February 5, 2014*

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