Sunday

May 24, 2015

May 24, 2015

Total # Posts: 10,620

**math**

1/5x^(-1/2) = x^(1/2)/5.
*November 10, 2014*

**maths**

See previous post: Sun, 11-9-14, 11:00 AM.
*November 10, 2014*

**math**

Length = L Width = L/3 P = 2L + 2L/3 = 40 cm. Solve for L.
*November 10, 2014*

**physics**

Vo = 58.2km/h = 58200m/3600s. = 16.2 m/s a = u*g = 0.155 * -9.8 = -1.52 m/s^2. V^2 = Vo^2 + 2a*d = 0 d = -(Vo^2)/-2a = -(16.2^2)/-3.04 = 86.3 m.
*November 10, 2014*

**physics**

Tf = Tr = 5.8/2 = 2.9 s. Tr = Rise time. Tf = Fall time. h = 0.5g*Tf^2
*November 9, 2014*

**Math**

Incomplete.
*November 9, 2014*

**Check my physics answer**

When the arrow points toward the negative x axis.
*November 9, 2014*

**physics**

m*g = 75.95 N. = Wt. of the box. Fp = 75.95*sin29.7 = 37.63 N. = Force parallel to the ramp. a = (Fp-Fk)/m a = (37.63-5.88)/7.75 = 4.1 m/s^2 V^2 = Vo^2 + 2a*d = 0 + 8.2*3.11 = 25.5 V= 5.05 m/s = Xo = Initial hor. velocity. h = 0.5g*t^2 = 0.84 m. 4.9*t^2 = 0.84 t^2 = 0.171 Tf...
*November 9, 2014*

**Physics**

Vo = 22 m/s. u = 0.68 a = u*g = 0.68 * -9.8 = -6.66 m/s^2. V = Vo + a*t = 0 t = -Vo/a = -22/-6.66 = 3.30 s.
*November 9, 2014*

**physics**

m*g = 65 * 9.8 = 637 N. = Force of skier. Fp = 637*sin15 = 164.9 N. = Force parallel to the incline. Fn = 637*cos15 = 615.3 N. = Normal force = Force perpendicular to the incline. Fk = u*Fn = 0.25 * 615.3 = 153.8 N. = Force of kinetic friction. a = (Fap-Fp-Fk)/m. Fap = 500 N...
*November 9, 2014*

**physics 1**

Incomplete.
*November 9, 2014*

**Algebra 2**

+3x + 3y - z = -3 -2x - 3y + 2z = 3 x - 6y + 3z = 0 Eq1: 3x + 3y - z = -3 Eq2: -2x - 3y + 2z = 3 Sum: x + 0 + z = 0 z = -x Eq2: -2x - 3y + 2z = 3 Eq3: x - 6y + 3z = 0 Multiply Eq2 by 6 and Eq3 by (-3): -12x - 18y + 12z = 18 -3x + 18y - 9z = 0 Sum: -15x + 0 + 3z = 18 Replace z ...
*November 9, 2014*

**physical science**

a = (Fap-Ff)/M = (40-24)/8 = 2 m/s^2.
*November 9, 2014*

**maths**

(3/8+2/8) * 128 = (5/8) * 128 = 80 cards given away.
*November 9, 2014*

**Physics**

M*g = 178 N. Vo = 0.5 m/s. Ramp = 3m[24o] Fp = 178*sin24 = 72.4 N. = Force parallel to the ramp. M*g = 178 M = 178/g = 178/9.8 = 18.16 kg = mass of box. a = Fp/M = 72.4/18.16 = 3.99 m/s^2.
*November 9, 2014*

**Math**

Incomplete.
*November 9, 2014*

**Physics test review**

h = 30.1 m. Dx = 66.8 m. 1. h = Yo*t + 0.5g*t^2 = 30.1 0 + 4.9t^2 = 30.1 t^2 = 6.143 Tf = 2.48 s. = Fall time or time in motion. 2. Dx = Xo*Tf = 66.8 Xo * 2.48 = 66.8 Xo = 27 m/s. = Initial velocity. 3. X = Xo = 27 m/s = X component of velocity just before it hits gnd. 4. Y^2...
*November 8, 2014*

**word problem math b**

Posts = 4 + (2*L/10) + (2*W/10)
*November 7, 2014*

**Math**

16pi/5 = 16*180/5 = 576o = 1.6 Rev. A = 0.6Rev * 360o/Rev = 216o, Q3. In Q3, A - Ar = 180o 216 - Ar = 180 Ar = 36o = Reference angle. Ar = (36o/360o) * 2pi = 2pi/10 Radians. Notes: In Q2, A + Ar = 180o In Q4, A + Ar = 360o
*November 7, 2014*

**math**

Tot. Funds=16*$1.00 + 5*5.00 + 5*10.00 + 3*20.00 + 4*0.05 + 20*0.10 + 16*0.25 + 66.00 + 53.75 Deposit = Tot. funds - 50.00
*November 7, 2014*

**physics**

Incomplete.
*November 7, 2014*

**physics**

False. The force of kinetic friction is smaller.
*November 7, 2014*

**CAL- derivative **

Find the derivative; y= x^3(5x-1)^4 So this is my answer y'=2x(5x-1)^3 + 10x + x(5x-1) Please correct me if my answer is wrong.
*November 7, 2014*

**Algebra**

10*Log(2.50*10^5/10^-16) = 10*Log(2.50*10^21) = 214 db.
*November 7, 2014*

**physics**

Vo = 1630 m/s[86o] Yo = 1630*sin86 = 1626 m/s. Y = Yo + g*Tr = = 0 @ max Ht. Tr = -Yo/g = -1626/-9.8 = 166 s. = 2.77 Min. Tf = Tr = 2.77 min. = Fall time. Tr + Tf = 2.77 + 2.77 = 5.54 Min. = Time in flight.
*November 6, 2014*

**physics**

d = 7.6 m. h = 0.5g*t^2 = 3.3 m. Solve for t. Xo = d/t m/s = Initial Hor. velocity
*November 6, 2014*

**science**

a = F/m = 85/5 = 17 m/s^2. V = Vo + a*t = 14 + 17*20 = 354 m/s.
*November 6, 2014*

**math**

Tan20 = h/d = 50/d d = 50/Tan20 = 137.4 m.
*November 6, 2014*

**science.**

The law of conservation of momentum says it will be the same.
*November 6, 2014*

**physics**

V^2 = Vo^2 + 2g*h Vo = 0 2g = 19.6 m/s^2 h = 35 m. Solve for V.
*November 6, 2014*

**Math, Please Help!**

X student tickets sold. (723-X) Non-student tickets sold. 3x + 5(723-X) = $2,815. Solve for X.
*November 6, 2014*

**Physics**

A. m*g = 19kg * 9.8N/kg = 186.2 N. = Force of the suitcase. u = a/g, a = u*g = 0.54 * 9.8 = 5.29 m/s^2 Fn = mg-F*sin41.3 = 186.2-0.66F = Normal force. F*Cos41.3-.54(186.2-0.66F) = 19*5.29 0.751F - 100.5 + 0.356F = 100.5 1.107F = 201 F = 181.6 N. = Force applied. Fn = 186.2 - 0...
*November 6, 2014*

**nqwiliso s.s.s.**

Fr = 30 + 25 + (-40) = 15 N.,Headed East
*November 6, 2014*

**Pre Algebra**

Sqrt(25) + (-9) = -4 = A Whole number.
*November 6, 2014*

**physics**

(1rev/0.4s) * 60s./min = 150 RPM.
*November 6, 2014*

**physics**

F = m*g = 1*9.8 = 9.8 N k = F/d = 9.8N/0.012m = 816.7 N/m.
*November 6, 2014*

**Math**

1. Log(9/300) = Log 9 - Log 300 2. Log2(4^2*3^4) = Log2(4^2) + Log2(3^4) = Log2(16) + Log2(81)
*November 6, 2014*

**Science**

F1+F2 = m*a = 0.750 * 0.450 = 0.3375 N. Fi-F2 = 0.750 * 0.240 = 0.18 N. Eq1: F1 + F2 = 0.3375 Eq2: F1 - F2 = 0.1800 Sum: 2F1 = 0.5175 F1 = 0.25875 N. In Eq1, replace F1 with 0.25875 0.25875 + F2 = 0.3375 F2 = 0.3375 - -0.25875 = 0.07875 N.
*November 6, 2014*

**physics**

See previous post: Thu, 11-7-14, 6:30 AM.
*November 6, 2014*

**physics**

See previous post: Thu, 11-7-14, 6:30 AM.
*November 6, 2014*

**physics**

d = V*T = 3.50 km T = 3.50/V = 3.50/25 = 0.14 h. T = 3.50/40 = 0.0875 h. 0.14 - 0.0875 = 0.0525 h = 3.15 Min. gained.
*November 6, 2014*

**phisics**

Fap = mg*sin45 = 36 N. m = 36/g*sin45 = 5.2 kg Fap = Force applied.
*November 6, 2014*

**algebra**

3.65V^2 + 0.0875V^2/0.09 = 62.7 3.65V^2 + 0.9722V^2 = 62.7 4.622V^2 = 62.7 V^2 = 13.56 V = 3.68
*November 5, 2014*

**physics**

Fd = (Vs+Vd)/(Vs-Va)*Fa = 1050 Hz. (340+0)/(340-Va) * 980 = 1050 Divide both sides by 980: 340/(340-Va) = 1.0714 364.29-1.0714Va = 340 -1.0714Va = 340-364.29 = -24.29 Va = 22.7 m/s. = Velocity of the ambulance. Vs = Velocity of sound. Vd = Velocity of the detector.
*November 5, 2014*

**algebra**

V^2 = Vo^2 + 2g*h = 0 + 64*203 = 12,992 V = 114 Ft/s.
*November 5, 2014*

**Algebra**

V = 8*sqrt(h) = 8*sqrt(203) = 114 Ft/s.
*November 5, 2014*

**physics**

V^2 = Vo^2 + 2g*h = 0 + 19.6*(10-5) = 98 V = 9.90 m/s @ 5 m above the surface. V^2 = 0 + 19.6*10) = 196 V = 14 m. Just before striking the surface.
*November 5, 2014*

**science**

Incomplete.
*November 5, 2014*

**physics**

Vo = 10 m/s. V = 5 m/s. d = 150 m. a = ? a. V^2 = Vo^2 + 2a*d a = (V^2-Vo^2)/2d = b. d = (V^2-Vo^2)/2a. V = 0, Vo = 5 m/s.
*November 5, 2014*

**Physics**

Incomplete.
*November 4, 2014*

**Algebra**

F(x) = Y = -x^2 + 6x + 6 = 0 Use Quadratic Formula. (X, Y) = (-2, 0), (8, 0).
*November 4, 2014*

**Precalculus**

Do you mean e^2x - 1 = e^4 or e^(2x-1) = e^4. I will assume the 2nd Eq. (2x-1)*Ln e = 4*Ln e (2x-1)*1 = 4*1 2x-1 = 4 2x = 5 X = 5/2 = 2.5
*November 4, 2014*

**physics**

a. a = (V-Vo)/t. V = 510 m/s Vo = 0 t = 2.37 s. b. a = (V-Vo)/t V = 0 Vo = 510 m/s. t = 3.15 s. Note: Acceleration when braking will be negative.
*November 4, 2014*

**physics**

V^2 = Vo^2 + 2g*h. Vo = 0 @ top of hill. V^2 = 2g*h. Solve for V.
*November 4, 2014*

**physics**

V^2 = Vo^2 + 2g*h = 0 + 19.6*2000 = 3920 V = 62.6 m/s. KE = 0.5m*V^2 = 50*62.6^2 = 195,938 J.
*November 4, 2014*

**physics**

m*g = 180kg * 9.8N/kg = 1764 N. PE = mg*h = 8000 J @ top of hill. h = 8000/mg = 8000/1764 = 4.54 m.
*November 4, 2014*

**Math**

(40/360) * 2p1 = 2pi/9 Radians.
*November 4, 2014*

**Science**

a = (Fap-Ff)/m = (25-6)/1.5 =
*November 4, 2014*

**Physics**

A vector quantity shows magnitude and direction: 50 m/s @ 45o. Magnitude = 50 m/s. Direction = 45 Degrees.
*November 4, 2014*

**physics**

F1+F2+F3 = 72.7N[0o] + 79.9N[45o] + 103N[315o]. X = 72.7 + 79.9*Cos45 + 103*Cos315=202 N Y = 79.9*sin45 + 103*sin315 = -16.33 N. F^2 = X^2 + Y^2 = 202^2 + (-16.33) = 41,071 F = 203 N. = Resultant force. Tan A = Y/X = -16.33/202 = -0.08084 A = -4.62o = 4.62o S. of E. = Direction.
*November 3, 2014*

**science**

1. The DF increases: DF = PW/PRF. 2. The PRF increases: PRF = 1/PRP. 4. When WL decreases, the PRF increases. The increase in PRF decreases the DF. PRF = V/WL
*November 3, 2014*

**physics**

A = (27.1-9.1)*7.7*10^-4 * 5 = 0.0693 Gal.
*November 3, 2014*

**science**

incomplete.
*November 3, 2014*

**Physics**

D = 4.19*Cos10.4
*November 3, 2014*

**Physics**

D = 4.19*sin10.4 =
*November 3, 2014*

**physics**

a. m*g = 1.0*10^4 N. m = 1.0*10^4/g = 1.0*10^4/9.8 = 1020 kg. Fp = 1.0*10^4*sin34 = 5592 N. = Force parallel to the hill. a = Fp/m = 5592/1020 = 5.48 m/s^2. b. V^2 = Vo^2 + 2a*d = 0 + 10.96*28 = 307 V = 17.5 m/s.
*November 2, 2014*

**Physics**

Vo = 20m/s[50o] Xo = 20*cos50 = 12.9 m/s Yo = 20*sin50 = 15.3 m/s. V = Vo + g*Tr = 0 @ max Ht. Tr = -Yo/g = -15.3/-9.8 = 1.56 s. = Rise time. h = Yo*Tr + 0.5g*Tr^2 h max = 15.3*1.56 - 4.9*1.56^2=11.9 m. h = 0.5g*t^2 = 11.9-3.04 = 8.90 m. 4.9t^2 = 8.90 t^2 = 1.82 Tf = 1.35 s...
*November 2, 2014*

**physics**

Vo = 50m/s[30o] Xo = 50*Cos(30) = 43.3 m/s. Yo = 50*sin(30) = 25 m/s. h = Yo*t + 0.5g*t = 59 m. 25t + 4.9t^2 = 50 4.9t^2 + 25t - 50 = 0 Use Quadratic Formula. Tf = 1.54 s. = Fall time. Dx = Xo * Tf = 43.3m/s * 1.54s = 66.7 m. From the cliff. Y = Yo + g*t = 25 + 9.8*1.54 = 40.1...
*November 2, 2014*

**Physics**

a. Fr = 328N[35o] + 464N[13o] X = 328*Cos35 + 464*Cos13 = 721 N. Y = 328*sin35 + 464*sin13 = 293 N. Fr^2 = X^2 + Y^2 = 721^2 + 293^2 = 605,690. Fr = 778 N. = Resultant force. b. a = F/m = 778/2950 = 0.264
*November 2, 2014*

**Physics**

Maximum power transfer occurs when the resistance of the load is equal to the internal resistance of the source: R = r.
*November 2, 2014*

**physics**

Vo = 0 V = 6,000rad/min = 6,000rad/60s = 100 rad/s. a = (V-Vo)/t = (100-0)/4 = 25 rad/s^2
*November 2, 2014*

**Physics**

a. D = Vo*t = 190rad/s * 13s = 2470 rad b. D = 300rad/s * 13s = 3900 Rad. c. a = (V-Vo)/t = (300-190)/13 = 8.46 rad/s^2. D = Vo*t + 0.5a*t^2 D = 190*13 + 4.23*13^2 = 3185 Rad.
*November 1, 2014*

**physics**

Multiply your answer by 100 to convert to cm.
*November 1, 2014*

**physics**

d = 0.5g^t^2. g = 9.8 m/s^2, t = 0.29 s. Solve for d.
*November 1, 2014*

**Physics**

Momentum = m1*V1-m2*V2 = 0.062*9-0.006*4 = 0.558 - 0.024 = 0.534 m1*V = 0.534 0.062*V = 0.534 V = 8.61 m/s.
*October 31, 2014*

**Physics**

Fr = F1 + F2 + F3 + F4 = 0 100N[37o] + 200N[270] + 250N[200] + F4=0 X = 100*Cos37+200*Cos270+250*Cos200 = -155 N. Y = 100*sin37+200*sin270+250*sin200 = -225 N. Tan Ar = Y/X = -225/-155 = 1.45161 Ar = 55.44o = Reference angle. A = 55.44 + 180 = = 235.4o F1+F2+F3 = Y/sin A = -...
*October 31, 2014*

**Physics**

a.m2*V2-m1*V1 = 0.250*10[71o]-0.250*28.4[58.8o] = 2.50[71o] - 7.1[58.8o] X = 2.50*Cos71 - 7.1*Cos58.8 = -2.86 Y = 2.50*sin71 - 7.1*sin58.8 = -3.71 M^2=X^2 + Y^2 = (-2.86)^2 + (-3.71)^2 = 21.9 M = 4.68 = Magnitude of change in momentum. b. Tan Ar = Y/X = -3.71/-2.86 = 1.29720 ...
*October 31, 2014*

**Physics**

(m1+m2)*V = m1*V1 (1424+1424)*V = 1424*11 2848V = 15,664 V = 5.5 m/s.
*October 31, 2014*

**Science Physics**

a. Vw = L*W*h = 85 *30 * (15-3) = 30,600 Ft^3. b. 30,600Ft^3 * 62.4Lbs/Ft^3 = 1.909,440 Lbs. = 955 Tons = Wt. of water displaced. Load = 955 - 160 = 795 Tons.
*October 31, 2014*

**Algebra 1B**

Width = W in. Length = (2W + 4) in. V = L * W * h = 2720 n^3. (2W+4) * W * 17 = 2720 2W^2+4W)*17 = 2720 34W^2 + 68W - 2720 = 0 W^2 + 2W - 80 = 0 -80 =-8*10. (W-8)(W+10) = 0 W-8 = 0 W= 8 W+10 = 0 W = -10 Use positive value of W: W = 8.
*October 31, 2014*

**physics**

T^2 = 4*pi^2*(L/g) T^2 = 39.48(0.65/9.8). Solve for T.
*October 31, 2014*

**algebra 2**

h = 16 - (16-12)/0.5h h = 16 - 8in/h Eq: h(t) = 16 - 8t
*October 31, 2014*

**physics**

a = 0 @ constant speed.
*October 30, 2014*

**physics**

Vo = 17 m/s V = 40 m/s t = 10.7 s. a = (V-Vo)/t. Solve for a.
*October 30, 2014*

**physics**

Vo = 0 a = 4.5 m/s^2 t = 7.9 s. d = ? V = Vo + a*t = 0 + 4.5*7.9 = 35.6 m/s. V^2 = Vo^2 + 2a*d = 35.6^2 Vo^2 + 2a*d = 35.6^2. Solve for d.
*October 30, 2014*

**physics**

V = ? Vo = 0 a = 0.557 m/s^2 d = 5.85 m. V^2 = Vo^2 + 2a*d. Solve for V.
*October 30, 2014*

**physics**

Vo = 0 V = ? t = 5.4 s. a = (V-Vo)/t = 2.57 m/s^2 (V-0)/5.4 = 2.57. Solve for V.
*October 30, 2014*

**physics**

D = Vo*t + 0.5g*t^2 D = 208*2.6 - 4.9*2.6^2 =
*October 30, 2014*

**Fluid Mechanics**

A flat-bottom river barge is 30.0 ft wide, 85.0 ft long, and 15.0 ft deep. (a) How many ft3 of water will it displace while the top stays 3.00 ft above the water? (b) What load in tons will the barge contain under these conditions if the empty barge weighs 160 tons in dry dock?
*October 30, 2014*

**Physics**

m*g = 0.5kg * 3.8N/kg = 1.9 N.
*October 30, 2014*

**math**

T = Time in seconds. h = Ht. in feet. (T, h) (0,100) Platform. (0.5,136) (1.0,164) (1.5,184) (2.0,196) V(2.5,200) Max. point. (3.0,196) (3.5,184) (4.0,164) (4.5,136) (5.0,100). Platform.
*October 29, 2014*

**math**

T = (d/2)/r1 + (d/2)/r2 = 35 min=2100 s. = (d/2)/4 + (d/2)/3 = 2100 = d/8 + d/6 = 2100 Multiply both sides by 24: 3d + 4d = 50,400 7d = 50,400 d = 7,200 m.
*October 29, 2014*

**Physics**

V^2 = Vo^2 + 2g*h = 0 @ max ht. h max = -(Vo^2)/2g = -(2.3^2)/-19.6 = 0.270 m. Since the max ht. is only 0.27 m, the ball cannot be spiked at o.8 m. Please check the numbers for accuracy.
*October 29, 2014*

**physics**

a. Balll #1: V = Vo + g*Tr = 0 @ max Ht. Tr = -Vo/g = -25/-9.8 = 2.55 s. = Rise time. Tf = Tr = 2.55 s. = Fall time. Tr+Tf = 2.55 + 2.55 = 5.10 s. = Time to hit gnd. Ball #2: h = 0.5g*t^2 = 20.0 m. 4.9*t^2 = 20 t^2 = 4.08 Tf = 2.02 s. = Fall time = time to hit gnd. b. Ball #1...
*October 29, 2014*

**phy**

Fs = m*g = 72.7kg * 9.8N/kg = 712.5 N. = Force of the student. P = F * d/t = 712.5 * 11.9/10.1 = 839.4 J./s = 839.4 Watts = 0.8394 Kilowatts.
*October 29, 2014*

**Physics**

Vo = 60km/h = 60,000m/3600s = 16.67 m/s. V = 15km/h = 15,000m/3600s = 4.17 m/s. a = (V^2-Vo^2)/2d = (4.17^2-(16.67^2))/900 = -0.289 m/s^2. d = (Vf^2-Vo^2)/2a d = (0-(16.67^2)/-0.579 = 480 m. To stop. 480-450 = 30 m. Further before coming to rest.
*October 28, 2014*

**Physics**

I believe there is an error in the problem. The problem should read: The speed of a train is reduced from 60km/h to 15km/h(for example) at the same time as it travels a distance of 450 m. If the retardation is uniform, find how much further it will travel before coming to rest...
*October 28, 2014*

**Physics**

V = I*R = 0.0015A * 20 = 0.03 Volts.
*October 28, 2014*

**Physics**

See previous post: Tue, 10-28-14, 12:16 AM.
*October 28, 2014*

**Math**

V = Vo + g*Tr = 0 @ max Ht. Tr = -Vo/g = -55/-32 = 1.72 s. = Rise time. Tf = Tr = 1.72 s. = Fall time. Tr+Tf = 1.72 + 1.72 = 3.44 s. To hit gnd
*October 28, 2014*