# Posts by Henry

Total # Posts: 13,538

**Math**

9. d = 25mi/in * 3.4in =

**Math**

1. (8+12+14)/8 2. 45/60. 3. d = 226mi/2h * 8h = 4. 82.15/31gal. 5. Cost = 0.65/Lb * P. 1.95 = 0.65P. P = ?. 6. 6/27 = 2/9. 7. Y/22 = 3/11. Multiply both sides by 22: 8. 1/6 = 8/x Cross-multiply:

**Math**

Let S1 = S2 = S3 = S4 = 10 units. A1 = L*W = 10 * 10 = 100 sq. units. S1 = S3 = 1.5*10 = 15. S2 = S4 = 0.5*10 = 5. A2 = L*W = 15 * 5 = 75 sq. units. A2/A1 = 75/100 = 0.75 = 75%. P2/P1 = (2L+2W)/(4*10) = (2*15+2*5) = 40/40 = 1.00 = 100%.

**Physics-Work, Energy, and Power**

At the top of hill: PE = M*g*h = 56.9 * 9.8 * 8.21 = 4578 J. At bottom of hill: KE = PE - Fk*d. 0.5M*V^2 = 4578 - 11.7*31.7 0.5*56.9*V^2 = 4207, V^2 = 148, V = 12.2 m/s.

**Physics**

R = (1.72*10^-80/2.82*10^-8) * 0.527 Ohms = 0.321 Ohms.

**Physics**

R = Ro + a(T2-T1)*Ro. 55.6 = 31.8 + a(53-28)31.8. 55.6 = 31.8 + 795a, a = 0.03/deg. C = Temp. coefficient.

**physics-sound**

Fd = (Vs+Vd)/(Vs-Vp) * Fp. Fd = (343+17)/(343-41) * 800. = 953.6 Hz. = Freq. heard by driver of the truck. Change = 953.6 - 800 = 153.6 Hz.

**physics**

A. Fg = (Vs+Vg)/(Vs-Vp) * Fp. 578 = (343+0)/(343-Vp) * 475. 578 = 343/(343-Vp) * 475, Divide both sides by 475: 1.21 = 343/(343-Vp), 416-1.21Vp = 343, Vp = 60.3 m/s. B. Fg = (Vs-Vg)/(Vs+Vp) * Fp. Fg = (343-0)/(343+60.3) * 475. Since the distance between the people on the gnd. ...

**Physics**

M1*V1 + M2*V2 = M1*V + M2*V 0.08*50i+0.06*50j = 0.08V+0.06V. 4i + 3j = 0.14V. 5[36.9o] = 0.14V, V = 35.7m/s[36.9o]. KE before the collision: KE1 = 0.5M1*V1^2 + 0.5M2V2^2. i^2 = 1, j^2 = 1. KE after the collision: KE2 = 0.5M1*V^2 + 0.5M2*V^2. KE(Lost) = KE1-KE2.

**Math**

A. Y/X = 6/3 = 2/1, Y = 2x. 2/3 = 2x, X = 1/2 * 2/3 = 2/6 = 1/3. B. F = M*a, 25 = M*1.25, M = 28 kg.. F = M*a. 35 = 28*a, a = ?.

**Science**

a. V^2 = Vo^2 + 2g*h1. Vo = 0. b. V^2 = Vo^2 + 2g*0.5h1. Vo = 0.

**Science**

1. Vo = 15m/s[60o]. Yo = 15*sin60 = 13 m/s. Y^2 = Yo^2 + 2g*h. 0 = 13^2 - 19.6h. h = ?. 2. h = 4.9*t^2. 1.8 = 4.9t^2, t = ?. d = V*t. d = 8*t.

**Science**

1. h = 4.9t^2. 12.5 = 4.9t^2. t = ?. d = V*t. d = 8*t. 2. Y = Yo + g*t. Yo = 0, g = 9.8 m/s^2, t = Value cal. in prob. 1.

**Maths**

Wage = 40*18.60 + 6.5*1.5*18.60 + 4*2*18.60 =

**maths**

P = 6 + 6 + 6 = 18 cm. P/2 = 18/2 = 9 cm. S1/2 = S2/2 = S3/2 = 6/2 = 3 cm. At = sqrt(9*3*3*3) = 15.6 cm^2 = 5pi = Area of triangle. Ac > 5pi = 12pi.

**Physics**

M*g = 10 * 9.8 = 98 N. = Wt. of box. Fp = 98*sin 0 = 0. = Force parallel with the surface. Fn = 98*Cos 0 - T*sin45 = 98 - 0.707T = Normal force. Fk = u*Fn = 0.5(98-0.707T) = 49-0.354T. T*Cos45-Fp-Fk = M*a 0.707T-0-(49-0.354T) = 10*1, 0.707T+0.354T = 59, T = 55.6 N.

**Physics**

M*g = M*9.8 = 9.8M = Wt. of skier. Fp = 9.8M*sin30 = 4.9M. = Force parallel with slope. Fn = 9.8M*Cos30 = 8.49M = Normal force. Fk = u*Fn = u*8.49M. Fp-Fk = M*a. 4.9M-u*8.49M = M*2 = 2M, u*8.49M = 2.9M, u = 0.342.

**Math**

A. Y = mx + b. Y = (-2/3)X + 4. B. ax + by = c. (0,-2), (3,0). m = (0-(-2))/(3-0) = 2/3. Y = mx + b. Y = (2/3)x - 2, -2x/3 + y = -2, Multiply both sides by -3: 2x -3y = 6. STD. form. C. m = -3, P(1,2). Y = mx + b. 2 = -3*1 + b, b = 5. Eq: Y = -3x + 5.

**math**

Tan45 = h/d, h = d*Tan45. Tan75 = h/(d-50), h = (d-50)*Tan75. Therefore, h = d*Tan45 = (d-50)*Tan75. d*Tan45 = (d-50)*Tan75. divide both sides by Tan75: 0.268d = d-50, d-0.268d = 50, d = 68.3 m. h = d*Tan45 = d = The hor. distance between point A and the foot of the tower.

**Math**

A. Ao = 7.5*15 + 585 = $697.50 = Initial amount. Y = 697.5 - 7.5x Y = -7.5x + 697.5. P1(X1,Y1) = (15,585). Y-Y1 = m(X-X1). Point-slope form. Y-585 = -7.5(X-15). B. Y = mx + b. Slope-intercept form. Y = -7.5x + 697.5. C. Y = -7.5*52 + 697.5.

**Algebra**

1st hiker walks X mi/h. 2nd hiker walks (x+2.2) mi/h. d1 + d2 = 22 miles. sx + 5(x+2.2) = 22. X = ?.

**Math**

L = 3.5 Ft. W = 18 1n. = 1.5 Ft. h = 2 Ft. As = 2(W*h) + 2(L*h) + 2(L*W). As = Surface area.

**Trigonometry**

X^2 + Y^2 = C^2. x^2 + 7^2 = 25^2, x^2 = 25^2 - 7^2 = 576, x = 24. If Cos C is negative, X is negative. Cos C = (-24)/25 = -0.96, C = 163.7o CCW from +x-axis. = 73.7o W. of N. P(-24,7).

**Algebra**

X is unknown until the Eqs has been solved. N or any letter can be used for the unknown. 1. 3x + 7 = -15. Isolate 3x by subtracting 7 from both sides: 3x+7-7 = -15-7, 3x = -22, divide both sides by 3: X = -22/3 = -7 1/3. 2. -2x - 6 = 12. -2x-6 + 6 = 12 + 6, -2x = 18, X = 18/-2...

**Math**

If a boat was 36' long and i built a model of it 18" long what scale would the model be

**physics**

Density = 0.72*Density of water. Density = 0.72 * 1000kg/m^3 = 720kg/m^3.

**Maths**

Cost of a cup = $X. Cost of a plate = $Y. Y = 4x.

**Science**

F = M*a. 5 = 10*a, a = 0.5 m/s^2. a = (V-Vo)/t. 0.5 = 2/t, t = 4 s.

**Math**

Exponential growth(n^10) is greater.

**Álgebra**

360o/8sections * 3sections =

**physics**

V^2 = Vo^2 + 2a*d. 0 = (29.17)^2 - 500*d, d = ?.

**Algebra**

1 st integer = X. 2nd integer = x+1. 3x + 4(x+1) = 95. x = ?.

**maths**

P1 = 2L + 2W = 2*15 + 2*10 = 50 M. = P2. W/L = 4/5. W = 0.8L. P2 = 2W + 2L. P2 = 2*0.8L + 2L, 50 = 1.6L + 2L, L = 13.9 M. W = 0.8L = 0.8*13.9 = 11.1 m.

**Math**

X = The # of min. used. C = Total cost. Plan A: C = 0.1x. Plan B: C = 0.075x + 10.

**Physics**

30o W. of N. = 120o CCW from + x-axis. The system is in equilibrium: A*Cos120 = - B*Cos (0). A = -B*Cos(0)/Cos120 = 2B. A*sin120 = -(-200), A = 200/sin120 = 231 Lbs. A = 2B, 231 = 2B, B = 115.5 Lbs.

**physics**

A. X = -23*125 = -2875 m. Y = -12*255 = -3060i. Disp. = sqrt(x^2+y^2). B. Tan A = Y/X. A = ?. = Direction.

**physics**

Vo = 17m/s[39o]. Xo = 17*Cos39 = 13.2 m/s. Yo = 17*sin39 = 10.70 m/s. A. d = Xo*t. 3.8 = 13.2*t, t = 0.288 s. B. h = Yo*t +0.5g*t^2. h = 10.7*0.288 - 4.9*0.288^2 = 2.67 m.

**Science**

A. Work = F * d. They did the same amount of work. B. Power is the rate of doing work: P = F * d/t. The one that took the least time used the most power.

**Vectors**

50o W. of N. = 140o CCW. Vr = Vb + Vc. 20[140] = Vb + (-8i), Vb = 20[140] + 8i, Vb = 20*Cos140+20*sin140 + 8i. Vb = -15.3 + 12.9i + 8i = -15.3 + 20.9i Tan A = Y/X = 20.9/(-15.3) = -1.36601. A = -53.8o = 53.8o N. of W. = 126.2o CCW = 36.2O W of N.

**Vectors**

B. sin50 = 1.7/d. d = 2.22 km.

**Vectors**

Your last question doesn't make sense.

**Vectors**

A. Vr = Vb + Vc. 20[50o] = Vb - 8i, Vb = 20[50]+8i, Vb = 20*Cos50+20*sin50 + 8i, Vb = 12.9 + 15.3i + 8i = 12.9 + 23.3i. Tan A = Y/X =23.3/12.9 = 1.80620, A = 61o.

**Math**

3n = 5n-14. n = ?.

**math**

The opposite angles of a parallelogram are equal: A = C = 35 + 25 = 60o. 360-120 = 240o. B = D = 240/2 = 120o.

**Maths**

$5400 * 1m/$13.50 = 400 m. Cost = 400m * 28.50/m =

**Algebra**

Po1*r1*t + Po2*r2*t = 2730. Po1*0.06*1 + Po2*0.09*1 = 2730, 0.06Po1 + 0.09Po2 = 2730. Eq1: 6Po1 + 9Po2 = 273,000, Eq2: Po1 + Po2 = 39000. Multiply Eq2 by -6 and Add: +6Po1 + 9Po2 = 273000. -6Po1 - 6Po2 = -234,000. Sum: 3Po2 = 39,000, Po2 = $13,000 Invested in acc. #2. Po1 = 39...

**Physics**

post it.

**Physics**

V = 80m/s[-20O]. Vx = 80*Cos(-20) = 75.2 m/s. Vy = 80*sin(-20) = -27.4 m/s. A. Range = V^2*sin(2A)/g. Range = 80^2*sin(2*20)/9.8 = 420 m. B. Tan20 = h/420, h = 153 m. h = Vy*t + 0.5g*t^2. 153 = -27.4*t + 4.9t^2, 4.9t^2 - 27.4t - 153 = 0, Using the Quadratic Formula: t = 9.04 s...

**Math**

L = 23.5 ft. W = 11 ft. h = 2/3 ft. 1 yd^3. = 27 ft^3. V = L*W*h. Yards = V/27.

**physics**

Incomplete.

**electronic**

Incomplete.

**physical science**

R = pL/A. p = 110*10^-8 Ohm.m = 110*10^-6 ohm.cm. L = 220 cm. r = 0.056 cm. A = pi*r^2 = 3.14*0.056^2 = 9.85*10^-3 cm^2. R = (110*10^-6*220)/(9.85*10^-3) = 2.46 Ohms.

**pyhsical science**

R = pL/A A = pi*r^2 = pi*(2r)^2 = pi*4r^2. R2 = 2pL/(pi*4r^2) = pL/Pi*2r^2.

**Maths**

Wt. of box = 20 N. Fp1 = 4*Cos20 = 3.76 N. = Force acting Parallel to ramp(upward). Fp2 = 20*sin A = Force acting parallel to ramp(downward). Fp2-Fs-Fp1 = 0. 20*sin A-5-3.76 = 0. Equilibrium. A = ?. 9 20*sin A-5-3.76 = 0. 20*sin A = 8.76, A = 10.8o. = Angle of elevation.

**algebra**

1. a^2 + b^2 = c^2. 15^2 + b^2 = 20^2, 225 + b^2 = 400. b^2 = 400-225 = 175, b = sqrt(25*7) = 5*sqrt7. 2. 9^2 + 12^2 = c^2. c^2 = 81 + 144 = 225, c = sqrt(225) = 15.

**Algebra**

Eq1: A = x * y = 48ft^2. Eq2: P = 2x + 2y = 28 Ft.

**Scienc**

2. Acceleration. 3. A. 6. D. Friction opposes motion. 7. A. Velocity includes magnitude and direction. 8. A. 10. C. 11. F = M*a = 3 * 7 = 21 N.

**Math**

A. V(t) = Vo - 899t. Vo = Initial value. B. V(t) = Vo + 899t.

**Science**

M*g = 50 * 9.8 = 490 N. = Wt. of box. Fp = 490*sin30 = 245 N. = Force parallel with incline. Fn = 490*Cos30 = 424 N. = Normal force. Fs = u*Fn = = 424u. = Force of static friction. Fp-424u = M*a. 245-424u = M*0 = 0, U = ?.

**Physics**

1 gal. = 8 Lbs. = 3.63 kg. M = 10,000gal * 3.63kg/gal. = 36,320 kg. PE = M*g*h = 36,320*9.8*61.5 Joules.

**physics**

V*a*(T2-T1) = 136. 1.53*a*(15.1-92.2) = 136, a = ?.

**physics**

a = 13*10^-6/oC = Temp. coeff. h*a*(T2-T1) = 0.197. h*13*10^-6(40-(-9)) = 0.197, h*0.000637 = 0.197, h = 309.3 m.

**physics**

a = 13*10^-6/deg.C = Temp. coeff. 358*a*(T2-T1) = 358*13*10^-6*(22-1.99) = 0.093 m. = 9.3 cm.

**Math 115**

C = 0.79P, P>100.

**physics**

d = 0.5a*t^2. 8 = 0.5a*4^2, a = 1 m/s^2. V = Vo + a*t.= 0 + 1*4 = 4 m/s.

**physics**

(14*10^-6/19*10^-6) * 3600cm = 2653 cm.

**physics**

M*g = 532 * 9.6 = 5214 N. = Wt. of piano. Work = F*d = 372 * 14 = 5208 J. PE = Mg*h = 5208. 5214h = 5208, h = 1.0 m.

**Physics**

M = 28.44 kg. F1 = 3964N[63.74o]. F2 = 130N[180o]. Fx = 3964*Cos63.74 - 130*Cos180 = 1624 N. Fx = M*a. a = Fx/M = 1624/28.44 = 57 m/s^2.

**Physics**

Fr = 5.9[28o] + 5[90o]. X = 5.9*Cos28 = 5.2Units. Y = 5*sin28 + 5*sin90 = 7.35 Units. Fr = sqrt(x^2+y^2) = Tan A = Y/X = 7.35/5.2 = 1.41346, A = 54.7o = The direction.

**Math**

1. x + y = 9. 2xy = 40, xy = 20, X = 20/y. x + y = 9. 20/y + y = 9, Multiply by Y: 20 + y^2 = 9y, y^2 - 9y + 20 = 0, (x-4)(x-5) = 0, Y = 4, and 5. x + y = 9. Solution set: (x,y)=(4,5),(5,4). So the numbers are 4, and 5. 2. 1st number = X. 2nd number = x+4. x^2 + (x+4)^2 = 208...

**Physics**

Vo = 5m/s[60o]. Xo = 5*cos60 = 2.5 m/s. Yo = 5*sin60 = 4.33 m/s. 1. Y = Yo + g*t. 0 = 4.33 - 9.8t, t = ?. 2. Y^2 = Yo^2 + 2g*h. 0 = 4.33^2 - 19.6h, h = ?. 3. Range = Vo^2*sin(2A)/g. Range = 5^2*sin(120)/9.8 =

**science**

The Elasticity of the balloon. The balloon can be stretched beyond its' normal size and can return to normal if the elastic limit is not exceeded.

**physics**

a. (185+182+183+182+182+184+183)/7 =

**Physics**

I drew out image of the situation but still need help. Can someone help with this problem about swinging on wrecking ball when at 1st: wrecking ball is in line with birdbath when at a peak in its trajectory, and in line with front door at the lowest point in the trajectory. ...

**math**

M*g = 35 * 9.8 = 343 N. = wt. of carton. Fp = 343*sin40 = 220.5 N. = Force parallel to the incline. Fn = 343*cos40 = 262.8 N. = Normal force. Fp-Fs = M*a. 220.5 - Fs = 35*0, Fs = 220.5 N. = Force of static friction.

**math**

Correction: 326 = 0.816T2, T2 = 400 N.

**math**

60o W. of N. = 150o CCW from +x-axis. 45o E. of N. = 45o CCW from +x-axis. The system is in equilibrium: 326*cos150 = -T2*cos45. 326 = 0.742T2, T2 = 400 N. Fb = -(T1*sin150+T2*sin45). Fb = -(326*sin150+400*sin45) = -446 N. = 446 N. Downward. M*g = 446. M = 446/g = 446/9.8 = 46...

**Physics**

While swinging on the wrecking ball, a student notices that, at first, the wrecking ball is in line with her birdbath when it is at a peak in its trajectory, and in line with her front door at the lowest point in the trajectory. However, since drag is acting against the ...

**Physics**

I was able to figure out mass of ball = to 720kg and velocity of ball and her is 1.90m/s

**Physics**

While swinging on the wrecking ball, a student notices that, at first, the wrecking ball is in line with her birdbath when it is at a peak in its trajectory, and in line with her front door at the lowest point in the trajectory. However, since drag is acting against the ...

**Slopes**

m = (7-4)/(-5-0) = -0.6 Y = mx + b. 4 = -0.6*(6) + b, b = 7.6. Eq: Y = -0.6x + 7.6

**Math**

r = 226mi/2h = 113 mi/h. d = r*t =

**Math**

csc A = 4/3. sin A = 3/4, x^2 + 9 = 16, X = sqrt7 = 7^(1/2). Cot A = sqrt7/3. sin A + sin A * cot^2 A = 3/4 + 3/4*(7^(1/2))^2 = 3/4+3/4*7 = 24/4 = 6.

**Finance**

My answer is unsecured debt.

**math**

Income/wk. = $12/wash * XWASHES = $12X. Cost/wk. = ($12/h*40h)2 + $450/wk. = $1410. Profit/wk. = 12x - 1410 = $3.25/wash * xwashes. = $3.25x. Profit/wk. = 12x - 1410 = 3.25x, X = 161.14 or 162 Washes.

**algebra**

Vp*t = 1600. Vp*5 = 1600, Vp = 320 mi/h. in still air. (Vp-Vw)*5.5 = 1600. 5.5Vp - 5.5Vw = 1600. 5.5*320 - 5.5Vw = 1600, Vw = ?.

**Physics**

8 - 5 = 3 N. M*g = 3, M = 3/9.8 = 0.306 kg = Mass of water displaced. V*D = 306 grams. V*1 = 306, V = 306 cm^3. volume of water displaced = volume of candle holder. M*g = 8 N. M = 8/g = 8/9.8 = 0.816 kg = Mass of candle holder. D = 816g/306cm^3 = 2.67 g/cm^3.

**Science**

F = M*G = 500 * 10 = 5000 N. P = F*d/(t*0.4) = 5000*80/(10*0.4) = 100,000 Watts.

**Science**

F = M*a. a = F/M. = 15,000/500.

**Physics**

2. F4 =-(F1*sin90+F2*sin170+F3*sin34.8)F4 = -(70 + 8.68 + 34.2) = -113 N. = 113 N. Downward.

**Physics**

F1 = 70N[90o]. F2 = 50N[170o], CCW from +x-axis. F3 = 60N[Xo]. 1. The hor. component of F2 and F3 must be equal and opposite: F3*Cos X = -F2*Cos170. 60*Cos X = -50*Cos170, Cos X = 0.82067, X = 34.8o. 2.

**Physics**

F1 = 70N[90o]. F2 = 50N[170o], CCW from +x-axis. F3 = 60N[Xo]. F3*Cos X = -F2*Cos170. 60*Cos X = -50*Cos170, Cos X = 0.82067, X = 34.8o.

**Physics**

V = 1000mi/2wks * 1wk./7days * 1day/24h = 2.98 mi/h.

**Physics**

a. V = Vo + a*t. -45 = 36 + a*0.002, a = ? F = M*a. b. V = Vo + a*t. 45[30o] = 36 + a*0.002, 39 + 22.5i = 36 + a*0.002, 3 + 22.5i = a*0.002, a = 1500 + 11,250i = 11,350 m/s^2[82.4o]. F = M*a = 0.15 * 11,350[82.4o] = 1,702N[82.4o]. Magnitude = 1,702 N.

**math**

I don't see the scale.

**Physics**

Range = Vo^2*sin(2A)/g. Range = 65^2*sin(2*37)/9.8 = 414.4 m.

**Physics**

V = Vo + a*t. 55 = 70 + a*25, a = ?.

**Physics**

V = 68mi/72min * 60min/h =

**Physics**

V^2 = Vo^2 + 2g*h. 0 = 69^2 - 19.6h, h = ?. V = Vo + g*t. 0 = 69 - 9.8t, t = ?.

**physics**

V^2 = vo^2 + 2g*h. V^2 = 0 + 19.6*27, V = 23 m/s.