Sunday

February 7, 2016
Total # Posts: 11,480

**physics**

Power=Fx * d/t = 70.9*Cos48 * 19.9/4.3 = Note: J/s = Watts.
*January 17, 2015*

**physical science**

When they are acting in the same direction. Fmax = 5 + 7 = 12 N.
*January 17, 2015*

**Physics Class**

0.450 ms = 0.000450 s. V = Vo + a*t = -36.7 36.7 + a*0.000450 = -36.7 0.00045a = -36.7-36.7 = -73.4 a = -163,111 m/s^2 F = M*a = 0.140 * (-163111) = 22,836 N.
*January 16, 2015*

**Physics**

X = 1.84 m/s = Velocity of the current. Y = 1.1 m/s = Velocity of the boat. A. Vb^2 = X^2 + Y^2 = 1.84^2 + 1.1^2 = 4.60 Vb = 2.14 m/s. B. Tan A = Y/X = 1.1/1.84 = 0.59783 A = 30.9o, CCW = 59.1o E. of N. Tan59.1 = d/115 d = 115*Tan59.1 = 192 m. Downstream.
*January 16, 2015*

**Algebra**

a. Month 1: Profit = (4.31-1.50)/cup * 1283cups = $3605.23 Profit = (0.83-0.76)/donut * 216Donuts = $15.12 Month 2: Profit = (4.31-1.50)/cup * 4210cups = $11,830.10 Profit = (0.83-0.76)/donut * 317Donuts = $22.19
*January 15, 2015*

**physics**

V^2 = Vo^2 + 2g*h V = 0 g = -9.8 m/s^2 h = 6.57 m. Solve for Vo.
*January 15, 2015*

**Math**

(32 + 56) 32 = (1,32), (2,16), (4,8). 56 = (1,56), (2,28), (4,14), (7,8). CF = 1,2,4,and 8. GCF = 8. 8(4 + 7)
*January 15, 2015*

**physical science**

F(net) = F1 + F2 = 3.5 + (-4.5) = -1 N. = 1 N. Towards the Left.
*January 15, 2015*

**PHYSICS**

V^2 = Vo^2 + 2a*d = 0 d = -Vo^2/2a = -(25^2)/-5 = 125 m. Circumference = pi * 2r = 3.14 * 0.54 = 1.70 m. 125m * 1rev/1.70m = 73.5 Revolutions.
*January 15, 2015*

**physics**

See Related Questions: Tue, 9-10-13, 1:08 PM.
*January 15, 2015*

**math(5th grade)**

2.8pages/3.5min * 31.5min = 25.2 Pages.
*January 15, 2015*

**General/Academics**

Glad I could help.
*January 14, 2015*

**General/Academics**

0.75*85 + 0.25*F = 89 63.75 + 0.25F = 89 0.25F = 89-63.75 = 25.25 F = 101 = Finals score. Cannot be done! 0.75*85 + 0.25*100 = 88.75% = Max. possible score.
*January 14, 2015*

**physics**

Eq1:0.5M*V^2 = 198 J. Eq2: M*V = 13 M = 13/V In Eq1, replace M with 13/V: 0.5*(13/V)*V^2 = 198 6.5*V = 198 V = 30.5 m/s.
*January 13, 2015*

**Physics**

Vo = 1860m/s[67.4o] Xo = 1860*Cos67.4 = 714.8 m/s. Yo = 1860*sin67.4 = 1717.2 m/s. Y = Yo-g*Tr = 0 1717.2 - 9.8Tr = 0 9.8Tr = 1717.2 Tr = 175 s. = Rise time. Tf = Tr = 175 s. = Fall time Tr+Tf = 175 + 175 = 350 s. = 5.83 Min. in flight.
*January 13, 2015*

**Math**

Social studies: X Hours. 2 3/4 + 1 1/2 + x = 5 3/4 Hours. Solve for X.
*January 13, 2015*

**algebra**

-2(2+3)-2k = -2-4(2+2) -2(5)-2k = -2-4(4) -10-2k = -2-16 -10-2k = -18 -2k = -18+10 -2k = -8 Divide both sides by (-2): K = 4. 9(
*January 13, 2015*

**algebra**

Y = 16/20 * 38 = 30.4 in. = Length of short side. Cost = 6*Area = 6 * (38*30.4) = 20/16 = 38/30.4 = 1.25
*January 13, 2015*

**math**

Your Eq is correct. Length = X Ft. (2/5)x + 600 = 2000. Multiply both sides by 5/2: x + 1500 = 5000 X = 3500 Ft.
*January 12, 2015*

**physics**

F = 1500beats/30s = 50 Beats/s = 50cycles/s = 50 Hz. P = 1/F = 1/50 = 0.02 s.
*January 12, 2015*

**Physics**

Incomplete.
*January 12, 2015*

**physics**

Conservation of Linear Momentum: M1V1 + M2V2 = M1*V3 + M2*V3 M1 = 160,000 kg V1 = 0.30 m/s M2 = 115,000 kg V2 = -0.120 m/s Solve for V3, the final velocity.
*January 12, 2015*

**physics**

h = Vy*t + 0.5g*t^2 = 840 m. 0*t + 4.9t^2 = 840 Solve for t.
*January 12, 2015*

**physics**

Conservation of Linear Momentum: M1V1 + M2*V2 = M1V3 + M2*V4 77.5V1 + 21*0 = 77.5*2.18 + 21*2.18 77.5V1 + 0 = 214.73 V1 = 2.77 m/s.
*January 12, 2015*

**physics**

M1V1+M2V2 = 85*2 + 55*2.7[22o] = 170 + 148.5[22o] = 170 + 148.5*Cos22 + 148.5*sin22 = 170+137.7 + 55.63i = 307.7 + 55.63i = 313[10.25o] kg-m/s.
*January 12, 2015*

**Physics**

F = M*g = 0.55kg * 9.8N/kg = 5.39 N. K = 5.39N./0.36m = 14.97 N/m
*January 12, 2015*

**algebra--1 question**

Slope = (Y2-Y1)/(X2-X1)=(1+3)/(1+2)=4/3.
*January 12, 2015*

**Physic- help on direction (Not that good at Englis**

Use the slope that was calculated from your lab data to calculate g using the given equation.
*January 12, 2015*

**physics**

M*g = 89.7kg * 9.8N/kg = 879.1 N. = Wt. of box = Normal force(Fn). Fs = u*Fn = 0.5 * 879.1 = 439.5 N. = Force of static friction. Fap-Fs = M*a Fap-439.5 = M*0 = 0 Fap = 439.5 N. = Force applied.
*January 12, 2015*

**physics**

1. Draw a hor. line. 2. Draw a ver. line to the right-end of the hor. line and label it h for height. 3. From the left-end of the hor. line, draw a hypotenuse to the top of the ver. line. The angle it makes with the hor. is 12o. 4. Draw a 2nd hyp. to the right of the first and...
*January 12, 2015*

**physics**

E = 43 Volts. R1 = 13 Ohms R2 = 24 Ohms R3 = 41 Ohms I = I1 + I2 + I3 = E/R1 + E/R2 + E/R3 = 43/13 + 43/24 + 43/41 = 3.31 + 1.79 + 1.05 = 6.15 Amps. = Total current drawn from the 43-volt source. P = E*I = 43 * 6.15 = 264.5 Watts = Total power taken from the 43-volt source. P1...
*January 12, 2015*

**Physics-Check**

d1 = V1*T = 5m/s * 10min*60s/min=3000 m. d2 = V2*T = 4.5 * 10min*60s/min = 2700 m d1-d2 = 3000 - 2700=300 m. Between them. Or d=(5-4.5)m/s * 10min*60s./min=300 m. Between them.
*January 12, 2015*

**Mechanics**

Open circuit.
*January 11, 2015*

**math**

1 - 1/x = 2 - 2/x -1/x + 2/x = 1 1/x = 1 X = 1 3 - 3/x = 3 - 3/1 = 3 - 3 = 0
*January 11, 2015*

**physics**

V = Vo + a*t Vo = 2.5 m/s. a = 1.5 m/s^2 t = 4 s. Solve for V(m/s).
*January 11, 2015*

**Math**

P = Po*(1+r)^n Po = $4000 = Initial principal. r = (13.8%/12)/100% = 0.0115 = Monthly % rate expressed as a decimal. n = 12Comp/yr. * 4yrs. = 48 Compounding periods. Solve for Principal(P).
*January 11, 2015*

**Physics**

a. d = Circumference = pi*2r = 3.14 * 18.54 = 58.25 m. b. T = 1min/861rev = 60s/861rev = 0.0697 s./rev c. V=861rev/min * 58.25m/rev * 1min/60s =
*January 11, 2015*

**physic**

V = 160km/2h = 80 km/h
*January 11, 2015*

**physics**

M*g = 550 N. = Normal force(Fn). Fs = u*Fn = 0.16 * 550 = 88 N. = Force of static friction. Fap-Fs = M*a Fap - 88 = M*0 = 0 Fap = 88 N. = Force applied.
*January 11, 2015*

**trigonometry**

X = 36 m. Tan60 = Y/X Solve for Y. h = Y + 2.3
*January 11, 2015*

**trigonometry**

Y = 5 m. = ht. A = 20.67o L = Length of rope. sin A = Y/L Solve for L.
*January 11, 2015*

**trigonometry**

The wire and the post form a rt. triangle with the gnd. L = Hyp. = Length of wire. X = 7 m. A = 56.67o L = X/Cos A
*January 11, 2015*

**physics**

M = 1970 kg V = 102km/h = 102,000m/3600s. = 28.33 m/s. KE = 0.5*M*V^2
*January 11, 2015*

**physics**

d = 1.7km * 1000m/km = 1700 m. T = 10.7min * 60s./min = 642 s. V = d/T = 1700m/642s. = 2.65 m/s. KE = 0.5*M*V^2
*January 11, 2015*

**Math**

1. F(x) = -3(x^2+4x+4)=-3x^2 - 12x - 12 2. (x-2)(x+1)/(3x+1)(x+1) = (x-2)/(3x+1) F(x)=(x+1)(x-2)/(x+1)(3x+1)=(x-2)/(3x+1)
*January 10, 2015*

**Math**

1. F(x) = (x+1)(x-2)/(x+1) = x-2 = g(x) 2. F(x) = (x+3)(x-4)/x+3 = x-4. g(x) = (x-2)(x-4)/(x-2) = x-4 = F(x).
*January 10, 2015*

**Math**

(3x+1)/2(x^2-1) + 2(x+1)/2(x^2-4x+3) = (3x+1)/2(x+1)(x-1) + 2(x+1)/2(x-1)(x-3) = (3x+1)/2(x+1)(x-1) + (x+1)/(x-1)(x-3) LCD = 2(x+1)(x-1)(x-3) (3x+1)(x-3)/2(x+1)(x-1)(x-3) + (x+1)(x+1)/2(x+1)(x-1)(x-3) (3x^2-9x+x-3)/LCD + (x^2+2x+1)/LCD Add the Numerators: (4x^2-6x-2)/LCD = 2(...
*January 9, 2015*

**Physics**

See Related Questions: Thu, 10-2-14, 2:52 AM.
*January 9, 2015*

**physics**

1. 38Lbs/95Lbs * 882N. = 352.8 N. on Mars. 2. 95Lbs/38Lbs * 333N = 832.5 N. on Earth.
*January 9, 2015*

**Physics**

100km/h = 100,000m/3600s = 27.8 m/s. 85km/h = 85,000m/3600s = 23.6 m/s. 25km/h = 25,000m/3600s = 6.94 m/s. M1*V1 + M2*V2 = M1*V3 + M2*V4. 1500*(-27.8)+1200*23.6=1500*(-6.9)+1200V Solve for V.
*January 8, 2015*

**Physics**

A. d = V*T = 57.5 m V*15.3 = 57.5 V = 3.76 m/s B. V = Va + Vp = 3.76 1.75 + Vp = 3.76 Vp = 2.0 m/s
*January 8, 2015*

**physics**

Range = Vo^2*sin(2A)/g = 12 m. Vo^2*sin(28.4)/9.8 = 12 Vo^2*sin(28.4) = 117.6 Vo^2 = 117.6/sin(28.4) = 247.3 Vo = 15.72 m/s.
*January 8, 2015*

**Algebra 1**

x + y = 9 x - y = 1 Sum: 2x + 0 = 10 X = 5 In Eq1, replace x with 5: 5 + y = 9 Y = 4 Solution set: (x,y) = (5,4).
*January 8, 2015*

**physics**

Wt. = M*g = 40kg * 9.8N/kg = 392 N. L = 6 + (1m/50N * 392N) = 13.84 m.
*January 8, 2015*

**6TH GRADE MATH**

Platypus: 10 min. Whale: 7(10+x) = 112
*January 8, 2015*

**Math**

a. X = 2. F(x) = -3(2+-2)^2 - (2+1) = -51 g(x) = -3*2^2 - 13*2 - 13 = -51 The student should finish the assignment.
*January 8, 2015*

**Math**

5/(x-1) + 2/(x+1) = -6 LCD = (x-1)(x+1) 5(x+1)/(x-1)(x+1)+2(x-1)/(x+1)(x-1) =-6 (5x+5)/(x-1(x+1) + (2x-2)/(x+1)(x-1)=-6 Add the numerators: (7x+3)/(x+1)(x-1) = -6 7x+3 = -6(x+1)(x-1) 7x+3 = -6(x^2-1) 7x+3 = -6x^2 + 6 6x^2 + 7x = 6 - 3 = 3 6x^2 + 7x - 3 = 0 Use AC method of ...
*January 8, 2015*

**Math 1 please help**

1. Shirts: A = (12-3)x = 9x x = The number of shirts bought. Jeans: A = (19-3)x = 16x
*January 8, 2015*

**science**

See Related Questions: Tue, 7-17-12, 8:08 AM.
*January 7, 2015*

**Physical Science**

Correct!
*January 7, 2015*

**Physical Science**

(M1+M2)V = M2*V2 80.16V = 0.16*20 Solve for V.
*January 7, 2015*

**Physics**

M*g = 100 N. Fn = 100*Cos0 - Fap*sin45 = Normal force Fn = 100 - 0.707Fap Fk=u*Fn = 0.75(100-0.707Fap)=75-0.53Fap Fap-Fk = M*a Fap-75 + 0.53Fap = M*0 = 0 1.53Fap = 75 Fap = 49 N. = Force applied. Fn = 100-0.707Fap = 100 - 0.707*49 = 65.4 N. = Normal force.
*January 7, 2015*

**math**

Eq1: Do mean Y = (3/2)x - 1? If so, then Y = 3x/2 - 1 Multiply both sides by 2: 2y = 3x - 2 Eq1:-3x + 2y = -2 Eq2: +x + y = 4 Multiply Eq2 by 3 and add the Eqs: -3x + 2y = -2 +3x + 3y = 12 Sum: 5y = 10 Y = 2. In Eq2, replace y with 2 and solve for x: x + 2 = 4 X = 2. Solution ...
*January 7, 2015*

**physics**

Wt. = M*g M = 23.5 kg g = 9.8m/s^2 Solve for Wt.
*January 7, 2015*

**physics**

Mass = 4.6g/1000 = 0.0046 kg a. KE1 = 0.5*M*V1^2 = 3.13 J. b. KE@ = 0.0023*74.8^2 = 12.87 J.
*January 7, 2015*

**Physics**

M*g = 500 N. = Normal force(Fn). Fk = u*Fn = 0.3 * 500 = 150 N. = Force of kinetic friction. Fap-Fk = M*a Fap-150 = M*0 = 0 Fap = 150 N. = Force applied.
*January 7, 2015*

**Math**

In a rt. triangle, the sum of the square of the two shorter sides is equal to the square of the 3rd side. X = 959 Miles Y = 1,011 Miles Z = 1,033 Miles X^2 + Y^2 = Z^2 959^2 + 1011^2 = 1033^2 1,941,802 > 1,067,089 It is not a rt. triangle.
*January 7, 2015*

**physics**

Circumference = pi * 2r = 3.14 * 2*399 = 2507 m. V = 40.5m/s * 6.28rad/2507m = 0.1015 rad/s.
*January 7, 2015*

**physics**

M*g = 300 N T = 100N[30o] Fn = M*g - T*sin A = 300 - 100*sin30
*January 7, 2015*

**algebra**

Oops! Posted pre-maturely. 2. Mark: 3y Linda: 3y - 6 In the end: Mark: 3y + 2y = 5y Linda: (3y-6) + 2y = 5y - 6
*January 7, 2015*

**algebra**

1. Sheila: 7y Craft sticks. Mark: 7y-4y = 3y Linda: 3y-y = 2y In the end: Sheila: 7y-2y-2y = 3y Mark: 3y + 2y = 5y Linda: 2y + 2y = 4y 2. Mark: 3y Linda: 3y - 6
*January 7, 2015*

**Math**

X = 8 Mi Y = -10 Mi a. Tan A = Y/X = -10/8 = -1.25 A = -51.3o = 51.3o S. of E. b. d^2 = X^2 + Y^2 = 8^2 + (-10)^2 = 164 d = 12.81 Miles. c. d = V*t Solve for t. d. A = -51.3 + 180 = 128.7o, CCW = 51.3o N. of W.
*January 6, 2015*

**physics please help**

See previous post: Tue, 1-6-15, 3:44 PM.
*January 6, 2015*

**Physics**

Ww = Vo*Dw = 4m^3 * 1000kg/m^3 = 4,000 kg. = Wt. of water displaced. = Wt. of the object in water.
*January 6, 2015*

**Physics**

Vb = L*W*h = 5 * 1 * 0.5 = 2.5 m^3 = Vol. of the box=Vol. of water displaced. Mass = Vb*Dw = 2.5m^3 * 1000kg/m^3 = 2500 kg.
*January 6, 2015*

**physics**

Vo = 157km/h = 157000m/3600s = 43.61 m/s V = 75.0km/h = 75000m/3600s = 20.83 m/s. a = -7.0 m/s^2 A. V = Vo + a*t Solve for t. B. V^2 = Vo^2 + 2a*d Solve for d.
*January 6, 2015*

**physics please help**

Tr = Tf = 4s./2 = 2 s. = Rise & fall time. h = Vo*Tf + 0.5g*Tf^2 = 0 + 4-9*2^2 = 19.6 m.
*January 6, 2015*

**Algebra**

2nd number = X 1st number = 3x/20 x + 3x/20 = 46 Multiply both sides by 20 and solve for x.
*January 5, 2015*

**Math**

I = Po*r*t = 10,000*(0.0225/52)*13 = 56.25 = Int. for 13 weeks. (56.25/13) * 52 = $225 = Int./yr. %I = (225/10,000)*100% = 2.25%
*January 5, 2015*

**Physics**

At 100 m. above gnd. PE max = Mg*h max = 80*100 = 8000 J. KE + PE = 8000 0 + 8000 = 8000 At 60 m. above gnd. PE = Mg*h = 80*60 = 4800 J. KE + PE = 8000 KE + 4800 = 8000 KE = 8000-4800 = 3200 J. The total energy, KE+PE, is constant at 8000 J.
*January 5, 2015*

**physics**

h = 4*sin37 = 2.41 m. V^2 = Vo^2 + 2g*h Vo = 0 g = 9.8 m/s^2 h = 2.41 m Solve for V.
*January 4, 2015*

**physics**

V^2 = Vo^2 + 2g*h V = 0 Vo = Initial velocity. g = -9.8 m/s^2 h = 1.28 m. Solve for Vo.
*January 4, 2015*

**Physics**

Tx = 20*Cos53 = 12 N. = Hor. component of tension. Tx-Fk = M*a 12-Fk = M*0 = 0 Fk = 12 N.
*January 4, 2015*

**physics**

h = 0.5g*t^2 h = 2 m. g = 9.8 m/s^2 t = Fall time. Solve for t. Dx = Xo * t Dx = Hor. distance from table. Xo = 10m/s = Initial hor. velocity. Solve for Dx.
*January 4, 2015*

**Math**

9(a^2+a+4) Answer: Neither. Difference of 2 squares = (a+b)(a-b) = a^2 - b^2
*January 4, 2015*

**Pre-calc/trig**

Log3(x^2+18) = 5 x^2 + 18 = 3^5 x^2 + 18 = 243 x^2 = 243 - 18 = 225 X = 15
*January 4, 2015*

**Algebra 1**

Use Long-hand division. (6x^3 - 3x^2 + x - 40)/(x-2)=6x^2+9x+19, Remainder 2.
*January 4, 2015*

**mathe**

Use your calculator.
*January 4, 2015*

**physics**

V^2 = Vo^2 + 2g*h Vo = 0 g = 9.8 m/s^2 h = 2 m. Solve for V. Note: The 10 m/s velocity is in the hor. direction.
*January 4, 2015*

**Physics**

V^2 = Vo^2 + 2a*d V = 17 m/s Vo = 0 d = 45 m Solve for a. F = m*a
*January 3, 2015*

**Physics**

a = (Fap-Ff)/M = (56-12)/25 = 1.76 m/s^2
*January 3, 2015*

**physics**

A = 5m[45o], CCW. B = 7m[330o], CCW C = 4m[240o], CCW. X=5*Cos45 + 7*Cos330 + 4*Cos240 = 7.60 m Y=5*sin45 + 7*sin330 + 4*sin240=-3.43 m Tan A = Y/X = -3.43/7.60 = -0.45113 A = -24.28o = 24.28o S of E = 335.7o, CCW. Resultant = X/CosA = 7.60/Cos335.7 = 8.34m[335.7]
*January 3, 2015*

**physics**

See previous post: Sat, 1-3-15, 4:01 AM.
*January 3, 2015*

**physics**

a. h = 50 + 20 = 70 m Above gnd. V^2 = Vo^2 + g*h = 0 Vo^2 - 9.8*20 = 0 Vo^2 = 196 Vo = 14 m/s. = Initial velocity. V = Vo + g*Tr = 0 14 - 9.8Tr = 0 9.8Tr = 14 Tr = 1.43 s. = Rise time. h = 0.5g*t^2 = 70 m. 4.9t^2 = 70 t^2 = 14.3 Tf = 3.78 s. = Fall time. Tr+Tf = 1.43 + 3.78...
*January 3, 2015*

**physics**

a. P = 5m[150o] X = 5*Cos150 Y = 5*sin150 b. 120-90 = 30o Clockwise from positive x axis = 330o counter clockwise from positive x axis. Q = 3.6m[330o], CCW. X = 3.6*Cos330 Y = 3.6*sin330
*January 3, 2015*

**physics**

D = 5m[37o] + 10m[150o] X = 5*Cos37 + 10*Cos150 = -4.67 m. Y = 5*sin37 + 10*sin150 = 8.00 m. Tan Ar = Y/X = 8/-4.67 = -1.71501 Ar = -59.75o = Reference angle. A = -59.75 + 180 = 120.25o = Direction. D=Y/sin A = 8.00/sin120.25=9.26m[120.25] = Magnitude and direction.
*January 3, 2015*

**physics**

a. h = 0.5g*t^2 h = 43.4 m. g = 9.8 m/s^2 t = Fall time. Solve for t. b. Dx = Xo * t Dx = 57.3 m. = Hor. dist. Xo = Initial hor velocity. t = Fall time calculated in part a. Solve for Xo. c. X = Xo=Initial hor velocity(part b). Xo does not change. d. Y = Yo + g*t Yo = 0 g = 9....
*January 2, 2015*

**physics**

X^2 + Y^2 = r^2 X = 25 cm Y = Ver. Displacement r = 50 cm = Hypotenuse. Solve for Y.
*January 2, 2015*

**physics**

V = Vo + a*t V = -40 m/s Vo = 30 m/s t = 0.04 s. Solve for a(It is negative).
*January 2, 2015*

**physics**

11. Ws = 5000 N. = Normal force(Fn). Fk = u*Fn = 0.05 * 5000 = 250 N. = Eorce of kinetic friction. Fap-Fk = M*a Fap - 250 = M*0 = 0 Fap = 250 N. = Force applied. Work = Fap * d = 250 * 1000 = 250,000 J. = 2.5*10^5 J. 12. Same procedure as #11.13. 13. Ws = 6000 N. = Normal ...
*December 31, 2014*