Monday

September 1, 2014

September 1, 2014

Total # Posts: 8,968

**Physics**

At top of hill: KE + PE = mg*h 0 + PE = mg*h PE = mg*h At bottom of hill: KE + PE = mg*h KE + 0 = mg*h KE = mg*h = 0.5m*V^2 0.5m*V^2 = mg*h V^2 = 2g*h V = Sqrt(2g*h). V = Sqrt(19.6*h). Sqrt means Square root.
*December 3, 2013*

**physics**

See previous post.
*December 3, 2013*

**physics**

Fc = m*g = 29kg * 9.8N/kg = 284.2 N. = Force of crate. Fap-Fs = m*a 90-u*Fc = m*0 90 - u*284.2 = 0 u*284.2 = 90 us = 0.317
*December 3, 2013*

**math**

T = (2/5) * 2 = 4/5 or 0.8 h.
*December 3, 2013*

**Physics**

a. a=F/m = 3.34*10^7/2.77*10^6 = 12.06 m/s^2. b. t = 2.5min * 60s/min = 150 s. h = 0.5a(2t-1) h = 6.03*(300-1) = 1803 m.
*December 3, 2013*

**Maths**

100% - 40.25% - 30.76% = 28.99 % Greater than 5.5 Ft. People = 0.2899 * 120M=34.79 M Greater than 5.5 Ft.
*December 3, 2013*

**science**

P = 40.2hp * 746W/hp = 29,989 Watts. = 29,989 Joules/s. T = 5.9*10*5J/(2.9989*10^4J/s=19.67 s.
*December 3, 2013*

**Physics**

B. L = V/F = 343/239.7 1.43 When flying away.
*December 2, 2013*

**Physics**

A. L = V/F = 343/258 = 1.33 m. B. F = ((Vs+Vr)/(Vs-Vg))*Fo = 258 Hz ((343+0)/(343-12.6))*Fo = 258 (343/330.4) * Fo = 258 1.0381Fo = 258 Fo = 248.5 Hz=Initial Freq. of the goose F = ((343-0)/(343+12.6))*248.5 F = (343/355.6) * 248.5 = 239.7 Hz When flying away.
*December 2, 2013*

**Physics**

A. L = V*T =V*(1/F)=V/F L = 343/4.4*10^6 = 7.95*10^-5m. B. L = 1580/4.4*10^6 = 359*10^-6=3.59*10^-4 m.
*December 2, 2013*

**math **

what property is described in this equation: 24x5=(20x5)+(4x5)
*December 2, 2013*

**Physics**

Fs = m*g = 87.1kg * 9.8N/kg = 853.6 N. = Force of snowboarder. Fp = 853.6*sin36.7 = 510.1 N. = Force parallel to hill. Fn = 853.6*cos36.7 = 684.4 N = Normal = Force perpendicular to hill. Fk = u*Fn = 0.106 * 684.4 = 72.55 N. = Force of kinetic friction. a=(Fp-Fk)/m=(510.1-72....
*December 2, 2013*

**Algebra**

x^2 + y^2 + 8x + 2y + 8 = 0 x^2+8x+(8/2)^2+y^2+2y+(2/2)^2=-8+(8/2)^2 +(2/2)^2. x^2+8x+16 + y^2+2y+1 = -8+16+1 Eq: (x+4)^2 + (y+1)^2 = 9 C(-4,-1). r2 = 9 r = +-3
*December 2, 2013*

**Geometry**

See previous post: Mon,12-2-13,7:39 AM.
*December 2, 2013*

**Geometry**

See previous post:Mon,12-2-13,7:39 AM.
*December 2, 2013*

**Geometry**

The diagonals form 2 30o-60o rt. triangles. h = 24*sin30 = 12 In. W = 24*cos30 = 20.8 In. P = 2*(W+h) = 2*(20.8+12) = 65.6 In.
*December 2, 2013*

**physics**

v = Vmax*sinWt v = 100*sin(1000t) Vrms = 0.707 * 100 = 70.7 Volts. W = 2pi*F = 1000 6.28*F = 1000 F = 159 Hz. Xl = W*L = 1000 * 0.5 = 500 Ohms. Xc=1/W*C=1/(1000*4.6*10^-6)=-217.4 Ohms tan A = (Xl+Xc)/R tanA = (500+(-217.4))/400 = 0.7065 A = 35.2o Z = R/cos A = 400/cos35.2 = ...
*December 2, 2013*

**physics**

h = Vo*t + 0.5g*t^2 = 20 37.8t _ 4.9t^2 = 20 -4.9t^2 + 37.8t - 20 = 0 Use Quadratic Formula and get: t = 0.571, and 7.14 s. So, when the ball is rising; it reaches 20 m in 0.571 s. When it is falling, it reaches 20 m in 7.14 s.
*December 1, 2013*

**physics**

See your previous post: Sun,12-1-13,10:34 PM.
*December 1, 2013*

**physics**

See previous post: Fri,11-29-13,8:15 PM.
*December 1, 2013*

**physics. again.**

V== Vo + g*t = 10 m/s. 37.8 + (-9.8)t = 10 -9.8t = 10-37.8 =-27.8 t = 2.84 s. h = Vo*t + 0.5g*t^2 h = 37.8*2.84 - 4.9*2.84^2 = 67.83 m.
*December 1, 2013*

**physics**

Fs = m*g = 22kg * 9.8N/kg = 215.6 N. = Force of the sled. Fp = 215.6*sin6 = 22.54 N. = Force parallel to the incline. Fn = 215.6*cos6 = 214.4 N. = Normal = Force perpendicular to the incline. Fk = u*Fn = 0.2 * 214.4 = 42.88 N. = Force of kinetic friction. V = 64km/h = 64000m/...
*December 1, 2013*

**5th grade math**

I'm glad I could help, and thanks for your response.
*December 1, 2013*

**5th grade math**

4th Grade: 9 Left-handed students. Total = 3 * 9 = 27 Students. 9/27 = 1/3 Are left-handed. 5th Grade: 7 Left-handed students. Total = 30-2 = 28 Sudents. 7/28 = 1/4 Are left-handed. 6th Grade: 6 Left-handed students. Total = 27+3 = 30 Students. 6/30 = 1/5 Are left-handed. The ...
*December 1, 2013*

**Physics**

r2 = r1 + a*(T-To)r1 = 4.42 cm. 4.40 + 1.3*10^-4(T-20)4.40 = 4.42 4.40 + 5.72*10^-4(T-20) = 4.42 5.72*10^-4(T-20) = 4.42-4.40 = 0.02 5.72*10^-4T - 114.4*10^-4 = 0.02 5.72*10^-4T = 0.02 + 0.01144 = 0.03144 T = 314.4*10^-4/5.72*10^-4 = 54.97o
*December 1, 2013*

**Math**

Same procedure as previous post: Sun,12-1-13,7:39 PM.
*December 1, 2013*

**Math**

Given: A(-8,3), B(-6,1). m = -1 A(-8,3), P(x,y). m== -1. m = (y-3)/(x-(-8)) = -1 m = (y-3)/(x+8) = -1 Cross multiply: y-3 = -1(x+8) y-3 = -x-8 Eq: x + y = -5 Or x + y +5 = 0 Replace y with 0: x + 0 = -5 X = -5 = X-intercept. Replace X with 0: 0 + y = -5 Y = -5 = y-int.
*December 1, 2013*

**Physics**

Wr = m*g = 0.5kg * 9.8N/kg = 4.9 N = Wt. of the rat. Fp = 4.9*sin34 = 2.740 N. = Force to the plane. Fn = 4.9*cos34 = 4.062 N. Normal = Force perpendicular to the plane. Fk = u*Fn = 0.2 * 4.062 = 0.8125 N. Fc-Fp-Fk = m*a 0.4-2.74-0.8125 = m*a m*a = -3.15 a = -3.15/m = -3.15/0...
*November 30, 2013*

**Physics**

Fex = 15000 N. ? m = 500kg a = Fex/m = 15000/500 = 30 m/s^2.
*November 30, 2013*

**physics**

Change in L = Lo * a*(T-To) = 65.1 * 10^-5 * (-31.2-15.8) = -0.030597 m.
*November 30, 2013*

**physics**

Fap*cos45-u*mg = m*a Fap*cos45-0.02*588 = m*0 = 0 Fap * cos45 - 11.76 = 0 Fap*cos45 = 11.76 N. = Hor. component of Force applied. Work = 11.76 * 1000. = 11,760 Joules.
*November 29, 2013*

**physics**

a. Xl = 2pi*F*l = 6.28*50*0.405=127.2 Ohms Xc = 1/(6.28*50*4.43*10-6) = -718.9 Ohms tan A = (Xl-Xc)/R tan A = (127.2-718.9)/400 = -1.47925 A = -55.94o Z=R/cosA = 400/cos(-55.94) = 714.2 Ohms[-55.94o]= The Impedance. Vmax = Imax * Z=0.25 * 714.2 =178.6 Volts. b. The negative ...
*November 29, 2013*

**physics**

Fn = 75N[40o] + 85N[340o] X = 75*cos40 + 85*cos340 = 137.3 N. Y = 75*sin40 + 85*sin340 = 19.14 N. tan A = Y/X = 19.14/137.3 = 0.13940 A = 7.94o Fn=X/cosA = 137.3/cos7.94=138.6 N[7.94o]
*November 29, 2013*

**PHYSICS**

a. Xc = 1/2pi*F*c Xc = 1/(6.28*60*4*10^-6) = 663.5 Ohms. I = V/Xc = 120/663.5 = 0.181A rms. b. Same procedure as "a".
*November 29, 2013*

**Physics**

We need to know the length of the hill. Length = 100 m? A = 30o m = 75 kg h = 100*sin30 = 50 m. = Ht. of hill. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*50 = 980 V = 31.3 m/s At bottom of hill. V^2 = Vo^2 + 2a*d a=(V^2-Vo^2)/2d=(31.3^2-0)/200=4.9 m/s^2 V = Vo + a*t = 31.3 0 + 4.9*t = ...
*November 28, 2013*

**math**

CORRECTION: Cost = 2375*0.85*0.80*0.80 = $1292. Discount = 2375 - 1292 = $1083.
*November 27, 2013*

**math**

Discount = 2375*0.15*0.20*0.20 = $14.25
*November 27, 2013*

**Physics**

V1 = 6N[180o] X1 = -6N Y1 = 0 V2 = 4.5N[45o] X2 = 4.5*cos45 = 3.18 N. Y2 = 4.5*sin45 = 3.18 N. X = X1 + X2 = -6 + 3.18 = -2.82 N. Y = Y1 + Y2 = 0 + 3.18 = 3.18 N. tan Ar = Y/X = 3.18/-2.82 = -1.12766 Ar = -48.43o = Reference angle. A = -48.43 + 180 = 131.6o R=Y/sinA = 3.18/...
*November 26, 2013*

**Finance - Annuities**

P = Po*(1+r)^n P = $2250/mo. * 48mo. = $108,000 r = (3%/12)/100% = 0.0025 = Monthly % rate expressed as a decimal. n = 1comp./mo * 48mo = 48 Compounding periods. a. P = (Po-12000)*(1.0025)^48 = 108,000 (Po-12000)*1.12733 = 108000 Po-12000 = 95,801.75 Po = $107,801.75 = Initial...
*November 26, 2013*

**math**

A*(1-0.0653) = 1,225,000 A*0.9347 = 1225000 A = 1,310,580.94
*November 26, 2013*

**physics**

F1 = m*g = 1.6kg * 9.8N/kg = 15.68 N F2 = (0.05m/1m) * 400N = 20 N. F1-Fs = m*a 15.68-Fs = m*0 = 0 Fs = 15.68 = Force of static friction. Fs = u*F2 = 15.68 u*20 = 15.68 u = 0.784 = Coefficient of static friction.
*November 26, 2013*

**Physic: Projectile Motion**

Range = Vo^2*sin(2A)/g = 15 m. Vo^2*sin(40)/9.8 = 15 Vo^2*0.0656 = 15 Vo^2 = 228.7 Vo = 15.12 m/s.
*November 26, 2013*

**Math**

2. A cube has 6 edges. As = 6 * 3^2 = 6 * 9 = 54 Sq. In.
*November 25, 2013*

**finance**

1. P = (Po*r*t)/(1-(1+r)^-t) r = (9%/12)/100% = 0.0075 = Monthly % rate expressed as a decimal. t = 12mo/yr * 5yrs. = 60 Months. Plug th above values into the givenEq and get: P = $12,455.01 I = P-Po 2. Monthly = P/t 3. P = Po + Po*r*t Po = $10,000 r = 0.0075 t = 60 Months ...
*November 25, 2013*

**physics**

Fap-Fk = m*a 30-Fk = 2 * 10 = 20 Fk = 30-20 = 10 N. = Force of kinetic friction.
*November 25, 2013*

**geometry**

See Related Questions: Thu,5-25-11,11:28 AM.
*November 24, 2013*

**Science, Physics, math, calculus**

a. h = 0.5g*t^2 = 0.77 4.9t^2 = 0.77 t^2 = 0.1571 t = 0.396 s. = Fall time. Dx = Xo * t = 5.15 m. Xo * 0.396 = 5.15 Xo = 13 m/s = Break speed. b. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*0.77 = 15.09 V = 3.88 m/s
*November 24, 2013*

**Physics**

h = 0.5g*t^2 = 50 m. 4.9t^2 = 50 t^2 = 10.2 Tf = 3.19 s. = Fall time. d = V*Tf = 100m/s * 3.19s = 319 m.
*November 24, 2013*

**PHYSICS**

CORRECTION: T = 1.15 Milliseconds.
*November 24, 2013*

**PHYSICS**

CORRECTION: d. Vr = I*R = (0.8*0.857)*7 = 4.8 Volts Vi=6-4.8=1.2 Volts across the inductor. V/e^(T/0.714) = 1.2 6/e^(1.40T) = 1.2 e^1.40T = 6/1.2 = 5 1.40T*Ln e = Ln 5 1.40T = 1.609 T = 1.15 s. 7 =
*November 24, 2013*

**PHYSICS**

a. T.C.=L/R=5.0/7=0.714 Milliseconds b. T/TC = 0.25mS/0.714mS = 0.350 Vi = V/e^(T/TC) = 6/e^0.350 = 4.23 Volts after 250 uS. I=Vr/R = (V-Vi)/R = (6-4.23)/7= 0.253A. c. I=Vr/R = (V-Vi)/R = (6-0)/7 = 0.857A d. Vr=I*R = (0.8*0.857)*7 = 4.80 Volts. Vi = 6-4.8 = 1.2 Volts. = ...
*November 24, 2013*

**Math**

Incomplete.
*November 24, 2013*

**physics**

Fr = 975N[0o] + 745N[60o] + 1175N[90o] X=975 + 745*cos60 + 1175*cos90=1347.5 N Y = 745*sin60 + 1175*sin90 = 1820.2 N. tan A = Y/X = 1820.2/1347.5 = 1.31079 A = 53.5o Fr = Y/sin A = 1820.2/sin53.5=2264.3 N. [53.5o].
*November 24, 2013*

**physics**

Ht. = 50 m.? a. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*50 = 980 V = 31.3 m/s. b. V = Vo + g*t = 31.3 m/s. 0 + 9.8t = 31.3 t = 3.19 s. c. m = 2.0Lb * 0.454kg/Lb = 0.908 kg. KE = m*V^2/2 = 0.908*31.3^2/2 = 444.8 J.
*November 24, 2013*

**physics**

Ws = m*g = 77.7kg * 9.8N/kg = 761.5 N = Wt. of skier. Fp = 761.5*sin35.7 = 444.3 N. = Force parallel to slope. Fn = 761.5*cos35.7 = 618.4 = Normal = Force perpendicular to slope. Fk = u*Fn = 0.08 * 618.4 = 49.47 N. = Force of kinetic friction. Fp-Fk = m*a a=(Fp-Fk)/m=(444.3-49...
*November 22, 2013*

**science**

KE = 0.5m*V^2 = 2.5*4^2 = 40 J.
*November 22, 2013*

**Please help!- Physics**

d = 483[180o ] + 347[106o] + 347[106o] d = 483[180 + 694[106o] X = 483*cos180 + 694*cos106 = -674.3 km Y = 483*sin180 + 694*sin106 = 667.1 km tan Ar = Y/X = 667.1/-674.3 = -0.98935 Ar = -44.69o = Reference angle. A = 180-44.69 = 135.3o d = X/cosA = -674.3/cos135.3 = 948.7 km [...
*November 22, 2013*

**Math**

Incomplete.
*November 21, 2013*

**physics**

3.2 cm from equilibrium position: KE + PE = mg*d 0 + PE = mg*d/2 Halfway to the equilibrium position: KE + PE = mg*d mg*d/2 + mg*d/2 = mg*d KE = mg*d/2 = 0.5m*V^2 0.5m*V^2 = mg*d/2 = 3.92*0.032/2 0.2*V^2 = 0.06272 V^2 = 0.3136 V = 0.56 m/s.
*November 21, 2013*

**Physics**

CORRECTION: V = 81.6km/h[74.93o] D = 245 km[74.93o]
*November 20, 2013*

**Physics**

V = 100km/h[90o] + 30km/h[315o] X = 100*cos90 + 30*cos315 = 21.21 km/h Y = 100*sin90 + 30*sin315 = 78.79 km/h tan A = Y/X = 78.79/21.21 = 3.71461 A = 74.93o V=Y/sinA = 78.79/sin74.93=81.6 km/h [78.79] D=V*t = 81.6[78.79] * 3 = 245 km[78.8o]
*November 20, 2013*

**Math**

Parallel lines have equal slopes.
*November 20, 2013*

**Math**

Parallel lines have equal slopes.
*November 20, 2013*

**Math**

Parallel lines have equal slopes.
*November 20, 2013*

**Math**

2x - 3 = 4y Eq1: 2x - 4y = 3 m1 = -A/B = -2/-4 = 1/2 m2 = -(2/1) = -2 Y = mx + b Eq2: Y =-2x - 5/4.
*November 20, 2013*

**Math**

See previous post: Wed,11-20-13,6:18PM.
*November 20, 2013*

**Math**

Eq1: x + 2y = 10. m1 = -A/B = -1/2 The slope of the 2nd line = the negative reciprocal of m1: m2 = -(-2/1) = 2. Y = mx + b Eq2: Y = 2x - 10 or 2x - y = 10
*November 20, 2013*

**physics**

Energy = 120 * 5 *6h /1000=3.6 KWh Used Cost = 3.6kwh * 0.090/kwh = 0.324 or 32.4 Cents.
*November 20, 2013*

**Physics**

L = h/sin A = 2/sin 30 = 4 m. = Length of incline. KE = mg*h-Fk*L 0.5m*V^2 = mg*h-Fk*l 0.2*4.50^2 = 3.92*2 - Fk*4 4.05 = 7.84 - 4Fk 4Fk = 7.84-4.05 = 3.79 Fk = 0.948 N. = Force of kinetic friction.
*November 19, 2013*

**physics**

Fn = m * a =47 * 4.50
*November 19, 2013*

**Physics**

Yes. For a given initial velocity, the larger the angle the greater the vertical component of initial velocity. The increase in vertical velocity causes an increase in the max. ht.
*November 19, 2013*

**Physics - projectile**

V = Vo = 50 m/s.
*November 19, 2013*

**Physics**

b. h = Vo*t + 0.5g*t^2 = 3.05 8t + 4.9t^2 - 3.05 = 0 Use Quadratic Formula and get: t = 0.319 s. c. V^2 = Vo^2 + 2g*h V^2 = 8^2 + 19.6*3.05 = 123.78 V = 11.13 m/s.
*November 19, 2013*

**MATH**

A. Y = ho - 8X ho = The initial ht. B. When he got on the elevator, it was at the initial height(ho) and decreased at a rate of 8 feet per second for 12 seconds. C. Y = ho _ 8x = 240 Ft. ho - 8*12 = 240 ho = 240 + 96 = 336 Ft. above gnd. D. Y = 336 - 8x = 0 336 - 8x = 0 8x = ...
*November 19, 2013*

**math**

N = 0.57 * 11
*November 19, 2013*

**Physics**

d = 0.5g*t^2 = 75 4.9t^2 = 75 t^2 = 15.31 Tf = 3.912 s. = Fall time. d = V*Tf = 7-4 = 3 m. V * 3.912 = 3 V = 0.769 m/s.
*November 19, 2013*

**math**

Did your book give the total # of Robinsons? Something is missing.
*November 18, 2013*

**physical science**

P = F * d/t = 96 * 2.8/4.6 = 58.43 J/s = 58.43 Watts.
*November 18, 2013*

**college algebra**

4^(x-3) = 3^(2x) (x-3)*Log4 = 2x*Log 3 Divide both sides by Log 4: x-3 = 2x * 0.79248 = 1.5850x x - 1.5850x = 3 -0.5850x = 3 X = -5.1282
*November 18, 2013*

**science**

Magnitude = 210 N. Direction = 30o South of West = 210o CCW.
*November 18, 2013*

**science**

Magnitude = 200 N. Direction = 60o North of East = 60o CCW.
*November 18, 2013*

**science**

Magnitude = 125 N. Direction = North(Positive Y-axis).
*November 18, 2013*

**math**

Sally has X postcards. Marta has X postcards. X = 4(X-18) Solve for X.
*November 18, 2013*

**Math**

Y = 5 For all values of X. (0,5),(5,5),(2,5),(-3,5).
*November 18, 2013*

**science**

500+(65-47)sk/wk*105days*1wk/7days=770 Skunks.
*November 18, 2013*

**Phyics**

d1 = 0.5a*t^2 = 15 m. 0.5*t^2 = 15 t^2 = 30 T1 = 5.477 s. V = a*t = 1m/s^2 * 5.477s = 5.477 m/s. d2 = V * t = 5.477m/s * 15s = 82.2 m. D = d1 + d2 = 15 + 82.2 = 97.2 m. = Distance at which he was caught. T = T1 + T2 = 5.477 + 15 = 20.48 s To catch up.
*November 18, 2013*

**math**

(25%/15%/10% = (1-0.25)(1-0.15)(1-0.10)= 0.75 * 0.85 * 0.90 = 0.57375
*November 18, 2013*

**math**

a. X + 27.80*20 = 914 X = $358. b. C = 358 + 27.80x X = The number of years since 1981. c. 2000-1981 = 19 years. C = 358 + 17.80*19 = $528.20
*November 18, 2013*

**Check my answers please- Physics**

1. A force that makes the net force on an object 0. 2. It is equal in magnitude and opposite in direction. 3. An object in equilibrium must not be in motion. 7. 90 N.
*November 18, 2013*

**algebra2**

See previous post: Mon,11-18-13,11:59 AM.
*November 18, 2013*

**algebra2**

CD: $X Savings: $4X Mutual: $(15,000-5X) 0.02X + 0.01*4X + 0.1(15000-5x) = 210 0.06X + 1500 - 0.5X = 210 -0.44X = 210-1500 = -1290 X = 2931.82 4X = 4*2931.82 = $11,727.27 = Savings.
*November 18, 2013*

**math**

Bal. = 423.78 + 123.42 - 100 = 447.20 Int. = 447.20 *(0.15/12) * 1 = $5.59
*November 18, 2013*

**Physics**

Incomplete.
*November 18, 2013*

**Business math**

P = Po*(1+r)^n r = (0.75%/365)/100% = 2.055*10^-5 = Daily % rate expressed as a decimal. n = 1comp./day * 21days = 21 compounding periods. Plug the above values into the given Eq and get: $9,0003.88
*November 17, 2013*

**physic**

Vo=36mi/h * 1600m/mi * 1h/3600s=16 m/s. V = Vo + a*t = 0 16 + a*2.4 = 0 2.4a = -16 a = -6.67 m/s^2. F = m*a = 58 * (-6.67) = -386.7 N. The negative sign means that the force was in the opposite direction of the motion.
*November 17, 2013*

**Math**

P = 350hp * 746W/hp = 261,100 Watts.
*November 17, 2013*

**physics**

Wb = m*g = 7.6kg * 9.8N/kg = 74.48 N. = Wt. of block. Fs = 8 N ? Fs = us*Fv = 8 N. us*74.48 = 6 us = 0.081
*November 17, 2013*

**geometry**

See previous post: Sat, 11-16-13, 7:13 PM.
*November 17, 2013*

**GEOMETRY**

1. Triangle #1 a = 8 b = 15 c = 18 Triangle #2: d = 10 e = ? f = ?. a/d = b/e = c/f = 8/10 15/e = 8/10 e = 18.75 18/f = 8/10 f = 22.5 2. Scale Factor=5/4 = 15/12=25/20 = 1.25 3. x/r = y/p = z/n.
*November 16, 2013*

**physics**

Fn = m*a m = Fn/a = 50/25 = 2 kg
*November 16, 2013*

Pages: <<Prev | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | **13** | 14 | 15 | 16 | 17 | 18 | 19 | Next>>