Friday

August 26, 2016
Total # Posts: 11,952

**physics**

See previous post: Wed, 8-12-15, 12:14 AM.
*August 12, 2015*

**Physics**

V1 = a*t = = 76.7 * 1.89 = 145 m/s. h1 = 0.5a*t^2 = 0.5*76.7*1.89^2 = 137 m. h2 = h1 + -V1^2/2g=137 + (-145)^2/-19.6 = 1210 m. Above gnd.
*August 12, 2015*

**Urgent math**

a. Let n = 3. 2*3^2 > (3+1)^2. 18 > 16. True. Let n = 4. 2*4^2 > (4+1)^2. 32 > 25. True. Check b, and c.
*August 11, 2015*

**physics**

Vo = 6.87m/s.[51.9o]. Yo = 6.87*sin51.9 = 5.41 m/s. = Vertical component of initial velocity. Y^2 = Yo^2 + 2g*h = 0. h = -Yo^2/2g = -6.87^2/-19.6 = 2.41 m.
*August 11, 2015*

**Math**

119-108 = 11db. db = 10*Log I2/I1 = 11. Log I2/I1 = 1.1. I2/I1 = 10^1.1 = 12.59 or 13.
*August 10, 2015*

**Algebra 2**

The answer is 1.
*August 10, 2015*

**engineering physics 1-science**

a = F/M = 68/32 = 2.125 m/s^2.
*August 9, 2015*

**physics**

See previous post: Sat, 8-8-15, 7:02 PM.
*August 8, 2015*

**physics**

M*g = 50.9 * 9.8 = 499 N. = Wt. of crate. = Normal(Fn). Fs = u*Fn = 0.526 * 499 = 264 N. = Force of static friction. Fk = 0.311 * 499 = 155.2 N. = Force of kinetic friction. a. (Fap-Fs) = M*a. Fap-264 = M*0 = 0. Fap = 264 N. = Fore applied. b. (Fap-Fk) = M*a. Fap-155.2 = M*0...
*August 8, 2015*

**physics**

Tan A = 1.31/1.81 = 0.72376. A = 35.9o Tan 35.9 = d/2.290. d1 = 2.290*Tan35.9 = 1.658 km downstream. sin35.9 = 1.31/V2. V2 = 1.31/sin35.9 = 2.23 m/s d2 = sqrt(2.29^2+1.66^2) = 2.83 km = Distance across with wind. d2 = V2*t = 2830 m. 2.23*t = 2830. t = 1269 s. = 21 Min. to cross.
*August 8, 2015*

**physics**

Vo = 19m/s[36o]. Xo = 19*Cos36 = 15.4 m/s. Yo = 19*sin36 = 11.2 m/s. Y^2 = Yo^2 + 2g*h = 0. h = -Yo^2/2g = -(11.2^2)/-19.6 = 6.4 m. Y^2 = Yo^2 + 2g*h = 0 + 19.6(6.4-2.7) = 72.52. Y = 8.5 m/s = Ver. component of final velocity. V = sqrt(Xo^2+Y^2) = sqrt(15.4^2+8.5^2)= 17.6 m/s.
*August 8, 2015*

**Physics**

D = sqrt(20^2+20^2) = 28.3 m. Tan Ar = 20/20 = 1. Ar = 45o A = 45 + 180 = 225o. CCW from +x-axis.
*August 8, 2015*

**Physics**

See previous post: Fri, 8-7-15, 10:42 AM.
*August 8, 2015*

**Physics**

Glad I could help.
*August 7, 2015*

**Physics**

d = V*t = 20 m = 0.02 km. 160 * t = 0.02. t = 1.25*10^-4 hr. = 0.45 s. To travel 20 meters. Rev = 1800rev/60s * 0.45s. = 13.5. Yes!!
*August 7, 2015*

**physics**

a. a = 10.5/3.18 = 3.3 m/s^2. b. d = 0.5a*t^2 = 0.5*3.3*3.18^2 = 16.7 m.
*August 7, 2015*

**physics**

M*g = 1000 * 9.8 = 9800 N. = Wt. of car. a = 25/10 = 2.5 m/s^2. d = 0.5a*t^2 = 1.25*10^2 = 125 m. P = F * d/t = 9800 * 125/10 = 122,500J/s = 122,500 Watts.
*August 7, 2015*

**calculus**

h = Vo*t + 0.5g*t^2 = -15*8 + 16*8^2 = -120 + 1024 = 904 Ft.
*August 6, 2015*

**trigonometry**

Tan(75/2) = &an37.5o = 0.76733.
*August 6, 2015*

**physics**

Gun A: V = Vo + g*Tr = 0. Tr = -Vo/g = -44.7/-9.8 = 4.56 s. = Rise time. h = Vo*Tr + 0.5g*Tr^2 = 44.7*4.56 - 4.9*4.56^2 = 102 m. Tf = Tr = 4.56 s. = Time to fall back to top of cliff. T = Tr+Tf = 4.56 + 4.56 = 9.12 s. = Time to rise and fall back to top of cliff. The time ...
*August 6, 2015*

**physics**

d = V*t = 2350 m. 25.5*t = 2350. t = 92.2 s. To reach the next exit. d = 0.5a*t^2 = 2350 m. 0.5a*92.2^2 = 2350. 4250.42a = 2350. a = 0.553 m/s^2.
*August 6, 2015*

**Calculus**

Can someone show how to solve this problem Find the equation of the tangent line to the curve given by, x=3e^t y=5e^-t at the point where t=0 Thanks!
*August 5, 2015*

**Physical Science**

Fr = 10[0o] + 8[90o] + 6[135o] + 4[270o]= Resultant force. X = 10 + 6*Cos135 = 5.76 N. Y = 8 + 6*sin135 - 4 = 8.24 N. Tan A = Y/X. A = Fr = Y/sinA =
*August 5, 2015*

**mth 156**

P = Po*(1+r)^n. P = $55,000. r = 5%/4/100% = 0.0125. n = 7yrs. * 4Comp./yr. = 28 Compounding periods. Solve for Po.
*August 5, 2015*

**Calculus**

Can someone show how to solve this problem Find the equation of the line normal to the graph of y = x^2 + 1 that passes through the point (1, 2). Thanks!
*August 5, 2015*

**physics**

230*10*CosA = 1410. CosA = 0.6130 = Power factor. A = 52.2o. = Phase angle. I^2*R = 1410W. 10^2 * R = 1410. R1 = 14.1 Ohms. = Resistance of coil. Tan52.2 = Xl/R = Xl/14.1 Xl = 14.1*Tan52.2 = 18.2 Ohms=Reactance of coil. Z1 = R1/CosA = 14.1/Cos52.2 = 23 Ohms. Parallel: Zp = 230...
*August 4, 2015*

**math**

3^200/3^50 = 3^(200-50) = 3^150.
*August 3, 2015*

**math**

(1,0), (0,1). Slope = (1-0)/(0-1) = 1/-1 = -1.
*August 3, 2015*

**physics**

Dx = Xo*Tf = 25 m. 4m/s * Tf = 25. Tf = 6.25 s. = Fall time. h = 0.5g*Tf^2.
*August 3, 2015*

**Algebra**

6 3/4 + x = 8. 6 + 3/4 +x = 8. Multiply both sides by 4: 24 + 3 + 4x = 32. 4x = 32-27 = 5. X = 5/4 = 1 1/4 = 1.25.
*August 2, 2015*

**math**

See Related Questions: Fri, 6-27-14, 2:15 AM.
*August 2, 2015*

**Real Estate**

See previous post: Sat, 8-1-15, 5:01 PM.
*August 2, 2015*

**Real Estate**

Ps-0.06Ps = 75,000 + 10,000. 0.94Ps = 85,000. Ps = $90,426 = Selling price.
*August 2, 2015*

**SLCS**

F1 + F2 = 602 N. F1 = (1.5/0.5)F2 = 3F2. 3F2 + F2 = 602. F2 = 150.5 N. = Paul's force. 3F2 = 3 * 150.5 = 451.5 N. = Justin's force.
*August 2, 2015*

**net**

Ps-0.07Ps = 6,8000 + 45,000. Solve for Ps, the selling price.
*August 1, 2015*

**physics(kinematics)**

90km/h = 90,000m/3600s = 25 m/s. a = (V-Vo)/t = (25-0)/8 = 3.125 m/s^2. d1 = 0.5a*t^2 = 0.5*3.125*8^2 = 100 m. = distance traveled while accelerating. t = 42-18 = 24 s. Driving time. A. d = 100 + 25m/s * (24-8)s. = 500 m. B. d = Vm*t = 500 m. V*42 = 500. V = 11.9 m/s. = 43 km/h.
*July 31, 2015*

**physics**

Dm = Vm*T. Dl = Dm + 25. Vl*T = Vm*T + 25. 5*T = 3*T + 25. 5T-3T = 25. 2T = 25. T = 12.5 s. To catch up. D = Vl*T = 5m/s * 12.5s. = 62.5 m. To catch up.
*July 31, 2015*

**math**

The above data was accidently posted; Please disregard it.
*July 30, 2015*

**math**

1cm = 4 mi. 1cm^2 = 16mi^2. 280cm^2 = 4480mi^2. 4mi/cm * sqrt(4480mi^2)
*July 30, 2015*

**maths**

3. X^x = 100^x. x*Logx = x*Log100 Divide both sides by x: Log X = Log100 = 2. 10^2 = X. X = 100.
*July 30, 2015*

**Math**

See previous post: Thu, 7-30-15, 3:28 AM
*July 30, 2015*

**Physics**

M*g = 200*9.8 = 1960 N. = Wt. of the body. = Normal(Fn). Fk = u*Fn = 50 N. u * 1960 = 50. u = 0.0255. a = u*g = 0.0255 * (-9.8) = -0.25 m/s^2. V = Vo + a*t = 0. 15 -0.25t = 0. 0.25t = 15. t = 60 s. d = Vo*t + 0.5a*t^2. d = 15*t - 0.125*60^2 = 450 m.
*July 30, 2015*

**Math**

Fp = 150*sin24 = 61 Lbs. = Force parallel to the incline = Force required
*July 30, 2015*

**math**

c. P+999 = 2003 + 999 = 3,002. and cannot be a prime number.
*July 29, 2015*

**physics**

Glad I could help.
*July 29, 2015*

**science**

R = E/I = 20/2 = 10 Ohms. Xl = E/I = 140/2 = 70 Ohms @ 40 Hz.= The inductive reactance. Xl == 2pi*F*L = 70 Ohms. 6.28*40*L = 70. L = 0.279 Henrys. = The inductance. a. Xl = 2pi*F*L = 6.28*50*0.279 = 87.5 Ohms. I = E/Xl = 230/87.5 = 2.63 Amps. b. Tan A = Xl/R = 87.5/10 = 8.75. ...
*July 29, 2015*

**Calculus**

Vpw = Vp + Vw = 700[67o] + 60[0o] = 700*Cos67 + 700*sin67 + 60 = 273.5 + 644.4i + 60 = 333.5 + 644.4i = 725.6km/h[62.64o]. Vpw = Vp + Vw = 700[67o]. Vp + 60 = 700[67]. Vp = 273.5 + 644.4i - 60=213.5 + 644.4i = 679km/h[71.7o] N. of E. = 18.3o E. of N.
*July 29, 2015*

**physics**

D = Xo * t = 800 Ft. Xo * 5 = 800 Xo = 160 Ft/s. Tr + Tf = 5 s. Tr = Tf. Tr+Tr = 5. Tr = 2.5 s. = Rise time. Y = Yo + g*Tr = 0. Yo = -g*Tr = -(-32)*2.5 = 80 Ft/s. 1. Tan A = Yo/Xo = 80/160 = 0.50. A = 26.6o. 2. Vo = Xo + Yoi = 160 + 80i = 179Ft/s[26.6o]. 3. h = Yo*Tr + 0.5g*Tr...
*July 28, 2015*

**Physics**

(140-700) = -560 rev/min -560rev/min * 1min/60s * 6.28rad/rev = -58.6 rad/s. = The change in velocity. 60s/560rev * 50rev = 5.36 s. a = (-58.6)/5.36 = -10.9 rad/s^2.
*July 28, 2015*

**math**

V = 9 * 16 * sqrt(25) = 720 Cubic units.
*July 27, 2015*

**Physics**

a = u*g = 0.55 * (-9.8) = -5.39 m/s^2. V^2 = Vo^2 + 2a*d. Vo^2 = V^2 - 2a*d. = 0 - 2(-5.39)*52 = 561. Vo = 23.7 m/s.
*July 26, 2015*

**maths**

7/56 = 10/N. 7N = 560. N = 80.
*July 26, 2015*

**physics**

Wt. = M*g.
*July 25, 2015*

**mathematics**

P = Po(1+r)^n. Po = $1,024. r = 0.07. n = 1comp./yr. * 3yrs. = 3 Compounding periods. Solve for P. I = P-Po.
*July 25, 2015*

**physics**

db = 10*Log I2/I1 = 10*Log I2/6 = 20. Log I2/6 = 2. I2/6 = 10^2 = 100. I2 = 600 W/cm^2.
*July 24, 2015*

**College algebra**

2001: t = 0. P(t) = 1500*(2^)^0 = 1500*(1) = 1500 Thousands. Year 2008: t = 2008-2001 = 7 yrs.
*July 24, 2015*

**College Algebra**

X, (X+1), (X+2). x + (x+1) + (x+2) = 966. 3x + 3 = 966. 3x = 963. X = 321. X+1 = 322. X+2 = 323.
*July 23, 2015*

**Unisa**

Suppose the arc subteding the central angle theta has length 20 pi metres and the radius is 24 metres, then what is theta measured in radians
*July 23, 2015*

**Physics**

At what angle are the ropes pulling?
*July 22, 2015*

**Math**

15 = 3*5. 18 = 3*3*2. 27 = 3*3*3. GCF = 3x^12.
*July 22, 2015*

**Physics**

M1*V1 + M2*V2 = M1*V + M2*V. 7900*5 + 1650*20[30o+180o] = 7900V+1650V 39500 + 33,000[210o] = 9550V 39,500 -28,579 - 16,500i = 9550V. 10,921 - 16,500i = 9550V. 19,787[-56.5o] = 9550V. V = 2.07m/s.[-56.5o] S. of E. = 2.07m/s[304o] CCW from +x-axis.
*July 22, 2015*

**math**

Scale Change = (48-44)in/4cm = 1in/cm.
*July 22, 2015*

**math**

P = Po*(1+r)^n. Po = $3,000. r = (3%/4)/100% = 0.0075 = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 5yrs. = 20 Compounding periods. Solve for P. I = P-Po.
*July 21, 2015*

**math**

See previous post: Fri, 7-17-15, 11:46 AM.
*July 20, 2015*

**Physics**

T1*Cos(180-35) + T2*Cos50 = -36*Cos270. -0.819T1 + 0.643T2 = 0. T1 = 0.785T2. T1*sin(180-35) + T2*sin50 = -36*sin270. 0.574T1 + 0.766T2 = 36. Replace T1 with 0.785T2: 0.574*0.785T2 + 0.766T2 = 36. 0.451T2 + 0.766T2 = 36. 1.22T2 = 36. T2 = 29.6 N. T1 = 0.785T2 = 0.785 * 29.6 = ...
*July 19, 2015*

**Calculus**

Can someone show how this question is solved. Consider the curve given by the equation 2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2 Find all points at which the tangent line to the curve is horizontal or vertical. Thanks!
*July 18, 2015*

**Math**

Length = L1. Width = L1/4. L2 = L1+10. W2 = L1/4 + 6. A2 = A1 + 128 Ft^2. A2 = (L1*L1/4) + 128. L2*W2 = (L1*L1/4) + 128. (L1+10)*(L1/4) + 6) = (L1*L1/4)+128. Multiply both sides by 4: (L1+10)*(L1)+24 = (L1*L1) + 512. L1^2+10L1+24 = L1^2 + 512. L1^2 - L1^2 + 10L1 + 24 = 512. ...
*July 17, 2015*

**math**

8/x = 2. X = 4 = Scale factor. Area = 8/4 * 12/4 = 6 Ft^2.
*July 17, 2015*

**math**

Area = 7*12 * 5*12 = 5,040 Ft^2
*July 17, 2015*

**Intermediate algebra**

post it.
*July 15, 2015*

**Math**

P(6, 4), m = 5. Y = mx + b. 4 = 5*6 + b. b = -26. Y = 5x - 26.
*July 15, 2015*

**science,**

4400N./0.02m^2 = 220,000 N./m^2. F = 220,000N/m^2 * 0.04m^2 = 8800 N.
*July 15, 2015*

**physics**

h = Vo*t + 0.5g*t^2. h = 2*2.5 + 4.9*2.5^2 = 35.63 m.
*July 15, 2015*

**Geometry**

L = Length of a side. A = Apothem = Altitude. L/A = L/L*sin60 = 1/sin60 = 1.1547. So the length of a side is approximately 1.15 times the apothem.
*July 13, 2015*

**science**

Unbalanced force.
*July 13, 2015*

**Physics**

500[90o] + 400[90o+60o] + F3 = 0 500i -346.41+200i + F3 = 0. -346.41 + 700i + F3 = 0. F3 = 346.41 - 700i = 781N.[-63.7o] = 781N.[63.7o] S. of E.
*July 13, 2015*

**physics**

See previous post: Sun, 7-12-15, 10:24 PM.
*July 12, 2015*

**physics**

See previous post: Sun, 7-12-15, 10:24 PM.
*July 12, 2015*

**physics**

W=0.5M*V2^2-0.5M*V1^2 = 0.5M(V2^2-V1^2)= 450(9^2-36^2) = -546,750 J. mg*h = -546,750. 8820h = 546,750. h = 62. m. 2. W = 450(36^2-9^2) = -546,750 J. Note: The negative sign means the system is doing the work.
*July 12, 2015*

**Physics**

Ti*sin(180-65) + T2*sin65 = -Wp*sin270. 1.3*0.90631 + 1.3*90,631 = Wp. Wp = 2.36 N. = Wt. of picture.
*July 12, 2015*

**Physics**

1a. 1.3*Cos(180-65) = -T2*Cos65. -0.549 = -0.423T2. T2 = 1.30 N. The Tensions are equal because the angles are equal.
*July 12, 2015*

**Physics**

1. Are the two strings tied to the same point on the wall? 2. a = u*g = 0.45 * (-9.8)=-4.41 m/s^2. V^2 = Vo^2 + 2a*d. V = 0. Vo = 7.75 m/s. a = 4.41 m/s^2. Solve for d.
*July 12, 2015*

**Calculus**

Thank You very much, I understand it now
*July 12, 2015*

**Calculus**

y = g(x) = cos(x) Can someone show how to estimate g'(pi/2) using the limit definition of the derivative and different values of h. Thanks!
*July 12, 2015*

**physics**

Glad I could help.
*July 11, 2015*

**physics**

4a. Mass = 1*10^-6m^3 * 19,300kg/m^3 = 0.0193 Kg . 4b. Fb = 1*10^-6m^3 * 1000Kg/m^3 = 0.001 Kg. = 0.0098 N. 5. V = 0.25 * 0.5 * 1 = 0.125 m^3. a. Fb = 0.125m^3 * 1000Kg/m^3 = 125 kg. = 1225 N. b. M = 0.125m^3 * 8,600kg/m^3 = c. = Wb = M*g Newtons. 6. V*D = 13 N. V*810 = 13. ...
*July 11, 2015*

**physics**

Glad I could help.
*July 11, 2015*

**physics**

P = E*I = 120 * 5 = 600 Watts = 600 J/s. Energy = P*t = 600J/s * 55s = 33,000 J.
*July 11, 2015*

**MATH**

You did not give the y-intercept nor the point. So I am going to solve a problem with that INFO given. Given: Y-intercept = 4 or(0,4), and P2(2,6). Find the Eq. of the line. P1(0,4), P2(2,6). Slope = (y2-y1)/x2-x1) = (6-4)/(2-0) = 1. Y = mx + b. Y = x + 4.
*July 10, 2015*

**Mathematics**

From what I understood in the problem, the required is the total travel time of car X from town A to town B. In the 180minutes, I think that the "80minutes later" is already included so no need to add it up.
*July 9, 2015*

**Mathematics**

Let x = speed of car X y = speed of car Y L = distance of AB "meet 80minutes later" L = 80x + 80y (distance = velocity * time) The time for car X to reach town B is: tx = ty + 36 But, tx = L/x and ty = L/y Substitute, tx = ty + 36 [80(x+y)]/x = [80(x+y)]/y + 36 ...
*July 9, 2015*

**Mathematics**

A car X left town A for town B at the same time that another car Y left town B for town A, each travelling at constant speed. They met 80 minutes later and car X arrived at town B 36 minutes after car Y reached town A. How long did it take car X to reach town B?
*July 9, 2015*

**Physics**

Work = The change in kinetic energy: Work = KE2-KE1 = 0.5*M*V2^2 - 0.5*M*V1^2 = 31*5^2 - 31*2^2 = 651 J. Work = (Fe-30)*d = 651. (Fe-30)*25 = 651. 25Fe - 750 = 651. 25Fe = 1401. Fe = 56 N. = Force exerted.
*July 9, 2015*

**physics**

M*g = 95 * 9.8 = 931 N. = Wt. of load. Fp = 931*sin50 = 713.2 N. = Force parallel to the incline. Fn = 931*Cos50 = 598.4 N. = Normal = Force perpendicular to the incline. Fk = u*Fn = 0.10 * 598.4 = 59.84 N. = Force of kinetic friction. a. Work = Fk*d = 59.84 * 30 = 1795 N. b. ...
*July 9, 2015*

**Physics**

Correction: b. D1 + D2 = 30+300 = 330 m. 0.5*1.17*t^2 + 0.5*1.1*t^2 = 330. 0.585t^2 + 0.55t^2 = 330. 1.135t^2 = 330. t^2 = 290.75. t = 17.1 s.
*July 9, 2015*

**Physics**

D1 = Distance traveled by train on left. D2 = Distance traveled by train on right a. D1 + D2 = 30 m. 0.5a1*t^2 + 0.5a2*t^2 = 30 0.5*1.17*t^2 + 0.5*1.1*t^2 = 30 0.585t^2 + 0.55t^2 = 30. 1.135t^2 = 30. t^2 = 26.43 t = 5.14s. D1 = 0.5*a1*t^2. a1 = 1.17 m/s^2. t = 5.14 s. Solve ...
*July 9, 2015*

**Algebra**

F(x) = Y = -8x^2 +3x +7 = 0 Use Quadratic Formula: X = (-B +/- sqrt(B^2-4AC))/2A X = (-3 +/- sqrt(9+224))/-16 X = (-3 +- sqrt233)/-16. X = (-3 +- 15.26)/-16 = -0.767 and 1.14.
*July 8, 2015*

**Calculus**

Can someone provide a function that satisfies all of the following: • p(−1) = 3 and lim x→−1 p(x) = 2 • p(0) = 1 and p'(0) = 0 • lim x→1 p(x) = p(1) and p'(1) does not exist Thanks!
*July 8, 2015*

**ITT**

PE = 20J. at top of incline. PE = 10J. halfway down the incline. KE + PE = 20 KE + 10 = 20 KE = 10J. KE = 0.5M*V^2 = 10 M = 1kg. Solve for V.
*July 8, 2015*

**Math**

Diagram: Draw a rectangle with the vert. sides representing the ht. of the Metro Bldg.(h1).and hor. sides are the distance between bldgs.(45m). Draw a diagonal from lower left to upper right and notice that the alternate interior angles are equal(56o).Extend the vert. side on ...
*July 7, 2015*