Sunday
August 30, 2015

Posts by henry


Total # Posts: 10,930

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

Physics
M*g = 60 * 9.8 = 588 N. = Wt. of the person. Eq1: T1*Cos9.4 + T2*Cos35.3 = 0. T1*Cos9.4 = -T2*Cos35.3. T1 = -0.827T2. Eq2: T1*sin9.4 + T2*sin35.3 =-588[270o]. 0.1633T1 + 0.578T2 = 0 + 588 Replace T1 with -0.827T2: 0.1633*(-0.827T2) + 0.578T2 = 588 -0.135T2 + 0.578T2 = 588 0....
July 7, 2015

physics 20
See Related Questions: Tue, 2-7-12, 11:59 AM.
July 6, 2015

Physics
Yo = 15*sin30 = 7.5 m/s. = Ver. component of initial velocity. Y^2 = Yo^2 + 2g*h = 0. h = -Yo^2/2g = -(7.5)^2/-19.6 = 2.87 m. above the roof. V^2 = Vo^2 + 2g*(h+40) = 0 +19.6*42.87 = 840.25 V = 29 m/s.
July 4, 2015

physics
Glad I could help.
July 3, 2015

physics
1. Work = 0.5M*V^2 = 0.5*1500*20^2 = 300,000 J. 2. hp * 746W/hp = 1492 Watts=1492J/s. P = F*V = Mg * V = 1492 J/s. 200*9.8 * V = 1492 1960V = 1492 V = 0.76 m/s.
July 3, 2015

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