Friday

November 28, 2014

November 28, 2014

Total # Posts: 9,616

**physics**

Vo = 136km/h = 136,000m/3600s=37.78 m/s V^2 = Vo^2 + 2a*d d = (V^2-Vo^2)/2a d = (0-(37.78^2))/-500 = 2.85 m.
*June 8, 2014*

**Math Personal Finance**

See previous post.
*June 8, 2014*

**Math Personal Finance**

P = Po(1+r)^n = 2000 r = (6%/12)/100% = 0.005 n = 12comp/yr. * 5yrs. = 60 compounding periods. Po(1.005)^60 = 2000 Po = 2000/(1.005^60) = $1482.74=Initial deposit or present value.
*June 8, 2014*

**Math Personal Finance**

P=Po(1+r)^n. Calculate int. for 1 year. r = (5%/12)/100% = 0.00417 = Monthly % rate expressed as a decimal. n = 12comp./yr. * 1yr. = 12 Compounding periods. P = 7000(1.00417)^12 = $7358.13 I = P-Po = 7358.13 - 7000 = $358.13 APY = (I/Po)*100% = (358.13/7000)*100% = 5.116%
*June 8, 2014*

**Engg Mechanics**

Block A: 200*sin30 = 100 N. = Force parallel to the plane. Fn = 200*cos30 = 173.2 N. = Normal force Block B: 100*sin30 = 50 N. = Force parallel to the plane. Fn = 100*cos30 = 86.6 N. = Normal force. P = Wa+Wb = 200 + 100 = 300 N.
*June 8, 2014*

**electronics**

Given: i = 6 mA. t = ? 1. Locate 6 mA on the y-axis. 2. Draw a hor. line from 6 mA to the curve. 3. Draw a ver. line from the intersection of the hor. line and curve to the x-axis. 4.Read the time, t , at the intersection of the ver. line and x-axis. Note: When t is given, ...
*June 8, 2014*

**physics**

a. The force of static friction decreases. b. The force of friction increases. The max. possible value of the force of friction increases. The max. angle at which block will slide decreases.
*June 8, 2014*

**Physics**

Upward: a = g = -9.8 m/s^2(slowing down) Highest point: a = g = -9.8 m/s^2. Downward: a = g = 9.8 m/s^2. The data shows that the acceleration due to gravity is present during the entire flight of the object. If g goes to zero, the object would not come down.
*June 8, 2014*

**Physics**

a. V^2 = Vo^2 + 2g*d V^2 = 4.9^2 + 19.6*21 = 435.61 V = 20.87 m/s. b. V = Vo + g*t t = (V-Vo)/g = (20.87-4.9)/9.8 = 1.63 s.
*June 7, 2014*

**physics**

a. False.
*June 7, 2014*

**Science**

2. Slower object: KE = 0.5m*V^2. Faster object: KE = 0.5m*(2V)^2 = 0.5m*4*V^2. 0.5m*4V^2/0.5m*V^2 = 4 Times. 3. KE = 0.5m*V^2 9. MA = 20/100 10. W = F*d. d = 0.
*June 7, 2014*

**Physics**

a. Vo=36mi/h * 1600m/mi * 1h/3600s=16 m/s. V^2 = Vo^2 + 2a*d d = (V^2-Vo^2)/2a d = (0-256)/-10 = 25.6 m. b. d = 0-1024)/-10 = 102.4 m.
*June 7, 2014*

**MULTIPLY. WRITE EACH PRODUCT IN SCIENTIFIC NOTATIO**

post.
*June 6, 2014*

**MULTIPLY. WRITE EACH PRODUCT IN SCIENTIFIC NOTATIO**

24. (3*10^6)(2*10^12) = 6*10^(6+12)=6*10^18 26. (6*10^9)(5*10^4) = 30*10^(9+4) = 30*10^13 = 3*10^14
*June 6, 2014*

**MULTIPLY. WRITE EACH PRODUCT IN SCIENTIFIC NOTATIO**

34. 5^10/5^7 = 5^(10-7) = 5^3 36. 76^11/76^5 = 76^(11-5) = 76^6
*June 6, 2014*

**electronics**

Rt = R2*R3/R1 = 2.11k R2*8.2k/2.2k = 2.11k 3.727R2 = 2.11k R2 = 0.566k = 566 ohms. NOTE: I'm assuming that Rt = 2.11k at 0oC.
*June 6, 2014*

**physics**

Yes, when an object is DROPPED, the initial velocity is 0. When an object is THROWN up or down, the initial velocity is not 0. V^2 = Vo^2 + 2g*h V^2 = 0 + 19.6*7 = 137.2 V = 11.71 m/s.
*June 5, 2014*

**Math**

1. ) = 30 + 0.1(x-200) F(x) = 30 + 0.1x - 20 The above Eq is valid for all real values of X that are equal to or greater than 200: X => 200. When x = 200 minutes, the cost would be $30. None of the given choices meets this requirement. 2. Cost = 30 + 0.1x - 20 = $100 0.1x...
*June 5, 2014*

**math**

See previous post: Fri,5-23-14,12:35 PM.
*June 5, 2014*

**electrical circuit instrumentation**

R1 = 6 ohms. R2 = 8 ohms. Re=R1*R2/(R1+R2) = 6*8/(6+8)=3.429 ohms = Equivalent resistor. I1 = E/R1 = 36/6 = 6 Amps. I2 = E/R2 = 36/8 = 4.5 Amps. I = I1+I2 = 6 + 4.5 = 10.5 Amps. P1 = E^2/R1 = 36^2/6 = 216 Watts. P2 = E^2/8 = 36^2/8 = 162 Watts. P = P1+P2 = 216 + 162 = 378 ...
*June 5, 2014*

**mechanical physics**

Incomplete.
*June 5, 2014*

**Physics**

1. h = 0.5g*t^2 = 6 m. 4.9t^2 = 6 t^2 = 1.224 Tf = 1.107 s. = Fall time. d = Xo * Tf = 24 m. Xo * 1.107 = 24 Xo = 21.7 m/s = Initial velocity. 2. Fg = m*g = Force of gravity = Wt. of car. Fk = u*Fg = Force of kinetic friction. Fn = Fap-Fk = Net force. Fap = Force applied.
*June 5, 2014*

**science**

1. V = Vo + g*t = 30 - 10*2 = 10 m/s. 2. hmax = (V^2-Vo^2)/2g hmax = (0-(30)^2)/-20 = 3. V = Vo + g*Tr = 0 @ hmax. Tr = -Vo/g = -30/-10 = 3 s. = Rise time.
*June 3, 2014*

**math**

I=2.37*10^-5 watts/cm^2=Sound Intensity level. Io =10^-16 W/cm^2 = Reference sound Intensity level. db = 10*Log(I/Io) db = 10*Log(2.37*10^-5/10^-16) = 114
*June 3, 2014*

**physics**

Fx = 100*Cos37 = 79.86 N. Wb = m*g = 10kg * 9.8N/kg = 98 N. = Weight of buggy. a=(Fx-Fk)/m = (79.86-60)/10=1.99 m/s^2
*June 3, 2014*

**physics**

d1 = d2 25t = 0.5a*t^2 25 = 0.5a*t 0.002t = 25 t = 12,500 s. = 3.5 h.
*June 3, 2014*

**physics**

d = 300km[135o] V1 = 300km[135o]/0.75h = 400 km/h[135o] without wind. Vp + Vw = 400o[135] Vp + 80[-90o] = 400[135] Vp = 400[135] - 80[-90] Vp=400*cos135+i400*sin135 - (80*Cos(-90)+i80*sin(-90)) Vp = -282.84+i282.84 - (0-i80) Vp = -282.84 + i362.84 = 460km[128o] = Velocity and ...
*June 3, 2014*

**Algebra 1 (Reiny)**

1. T = T1*T2/(T1+T2) = 2*3/(2+3)=1.2 h. 2. H = 0.4 * 300 = 120 mL in current solution. W = 0.6 * 300 = 180 mL of water in current solution. W/(W+H) = 0.7 W/(W+120) = 0.7 W = 0.7W + 84 0.3W = 84 W = 280 mL of water in new solution. 280-180 = 100 mL of water added to new solution.
*June 3, 2014*

**geometry**

A1 = 150o = Minor Arc. A2 = 360-150 = 210o = Major Arc. X = (A2-A1)/2 = (210-150)/2 = 30o=Angle formed by 2 tangents. a
*June 3, 2014*

**College Algebra**

Length = L units. Width = (2L-2) units. A = L(2L-2) = 82cm^2 2L^2 - 2L - 82 = 0 L^2 - L - 41 = 0 Use Quadratic formula: L = 6.92 cm. Width = 2L - 2 = 2*6.92 - 2 = 11.84 cm. The diagonal divides the rectangle into 2 congruent rt. triangles. D^2 = L^2 + W^2 = 6.92^2 + 11.84^2=...
*June 2, 2014*

**College Algebra ..Please help**

1. Width = W ft. Length = (W+20) ft. A = W*(W+20) = 8,000 Ft^2 W^2 + 20W - 8000 = 0 C=-8,000 = -80*100. Sum = -80+100=20=B. (w+100)(w-80) = 0 w+100 = 0, W = -100 w-80 = 0, W = 80 Ft = Width. w+20 = 80+20 = 100 Ft = Length. 2. A = L*W = 16, L = 16/W. Eq2: L^2 + W^2 = 68 In Eq2...
*June 2, 2014*

**LOOK AT THE GIVEN TRIANGLES??**

a. P1 = (4x+2) + (7x+7) + (5x-4)=16x + 5 P2 = (x+3) + (x+7) + (2x-5) = 4x+5 b. P1-P2 = (16x+5) - (4x+5) = 16x+5 - 4x-5 = 12x.
*June 2, 2014*

**physics**

d = 3.6 + 0.5 = 4.1 miles. T1 = 3.6/9 + 12/60 = 0.6 hr. To travel 4.1 miles from starting point. V1 = d/T1 = 4.1/0.6 = 6.83 mi/h. V2=9.0 mi/h. when returning to starting point. V = (V1+V2)/2 = (6.83+9.0)/2 = 7.92 mi/h = Average velocity for entire trip.
*June 2, 2014*

**physics**

T^2 = 4pi^2(L/g) = 3.4^2 39.48(2.87/g) = 11.56 113.3/g = 11.56 11.56g = 113.3 g = 9.80 m/s^2
*June 2, 2014*

**Kirr**

P = Po(1+r)^n = $25,000 r = (1.825%/365)/100% = 5*10^-5 = Daily % rate expressed as a decimal. n = 365Comp./yr. * 4yrs. = 1460 Compounding periods. Po(1.00005)^1460 = 25000 Po = 25,000/1.00005^1460 = $23,240.
*June 2, 2014*

**algebra math!**

9pts/6wks = 1.5 Points/week.
*June 1, 2014*

**physics**

Wb = m*g = 40kg * 9.8N/kg = 392 N. = Weight of box. Fk=u*(Wb-F*sin31)=0.45(392-173*sin31) = 136.3 N.= Force of kinetic friction. Fn=F*CosA-Fk = 173*cos31 - 136.3 = 12 N. = Net force. a = Fn/m = 12/40 = 0.30 m/s^2.
*June 1, 2014*

**Physics**

hmax = L*sin A = 5*sin40 = 3.21 m. At top of incline: KE + PE = mg*hmax-Fk*L 0 + PE = mg*3.21-10*5 PE = mg*3.21-50. At bottom of incline: KE + PE = mg*3.21-50 KE + 0 = mg*3.21-50 KE = mg*3.21-50 = 35 J. m*9.8*3.21-50 = 35 31.5m-50 = 35 31.5m = 85 m = 2.7 kg. = Mass of box.
*May 30, 2014*

**physics**

V = Vo + g*Tr = 0 at max. ht. Tr = -Vo/g = -40/-9.8 = 4.08 s. = Rise time. T = 4.08-1 = 3.08 s. h1 = Vo*t + 0.5g*t^2 h1 = 40*3.08 - 4.9*3.08^2 = 76.7 m. h2 = 40*4.08 - 4.9*4.08^2 = 81.6 m. h2-h1 = 81.6-76.7 = 4.93 m. = Distance traveled in the last second.
*May 30, 2014*

**physics**

h = Vo*t + 0.5g*t^2 h = 0 + 5*3^2 = 45 m. = Height of object above gnd. (45-1)/4m = 11th floor.
*May 30, 2014*

**physics**

V^2 = Vo^2 + 2a*d V^2 = 0 + 10*10 = 100 V = 10 m/s.
*May 30, 2014*

**algebra**

I = 2.42*10^-5 Watts/cm^2. = Sound intensity. Io = 10^-16 Watts/cm^2 = Reference sound intensity. dB = 10^Log(2.42*10^-5/10^-16) = 114.
*May 29, 2014*

**physics**

h = L*sin A = 5.5*sin30 = 2.75 m. Fn = mg*Cos A = 66*Cos 30 = 57.16 N. = Normal force = Force perpendicular to the ramp. Fk = u*Fn = 0.22 * 57.16 = 12.57 N. = Force of kinetic friction. KE + PE = mg*hmax-Fk*L KE + PE = 66*2.75-12.57*5.5 = 112.4 J. 0 + PE = 112.4 PE = 112.4 J. ...
*May 29, 2014*

**physics**

1. W = Fk*d 25 * 12 Joules. 2. W = 175 * 12 Joules. 3. Wn = (175*12)-(25*12)
*May 29, 2014*

**Physics**

a. X = 45 N. Y = -22 N. F^2 = 45^2 + (-22)^2 = 2,509 F = 50.09 N. b. d = 22m/s * 3s = 66 m. Work = F*d = 50.09 * 66 = 3306 Joules.
*May 29, 2014*

**Physics! Help**

F1 + F2 = 96.2N.[55.4o] + 11.6N.[41.2o] X = 96.2*cos55.4 + 11.6*cos41.2=63.35 N
*May 29, 2014*

**ALGERBRA**

What are the "2 ways?" Perhaps some punctuation would help.
*May 29, 2014*

**Physics**

F = m*a = 1350 N. Impulse=m*V = m*(a*T) = 1350*6.2*10^-3= 8.37 kg-m/s.
*May 29, 2014*

**Physics**

D=2m[270o] + 16m[180o] + 24m[90o]+36m[180o]. Add the two 180o vectors directly: D = 2[270o] + 52[180o] + 24[90o] X = 2*cos270+52*cos180+24*cos90 = -52 m. Y = 2*sin270+52*sin180+24*sin90 = 22 m. D^2 = (-52)^2 + 22^2 = 3188 D = 56.5 m.
*May 29, 2014*

**Physics**

a. d = V*T = 420 m. (12.5-2.70) * T1 = 420 T1 = 420/(12.5-2.70) = 42.9 s. b. T2 = 420/(12.5+2.7) = 27.6 s. T = T1+T2 = 42.9 + 27.6 = 70.5 s. c. T1 = 420/12.5 = 33.6 s. T2 = T1 = 33.6 s. T = 33.6 + 33.6 = 67.2 s.
*May 29, 2014*

**physics**

V=100mi/h * 1600m/mi * 1h/3600s = 44.44m/s V = Vo + a*t = 0 a = -Vo/t = -44.444/10^-3=-44,444 m/s^2 F = m*a = 0.145 * (-44,444) = - 6444 N.
*May 28, 2014*

**physics**

Da = 69.5*T Db = 84.3*T B will have to travel 394 m(0.394km) farther than A to catch up: Db = Da + 0.394km 84.3 * T = 69.5 * T + 0.394 84.3T-69.5T = 0.394 14.8T = 0.394 T = 0.02662 h. = 1.60 Min. to catch up.
*May 28, 2014*

**math**

Slope = Tan A = Y/X = 1/6 A = 9.46o
*May 27, 2014*

**math**

Slope = Y/x = 16 Y/3 = 16 Y = 48 m.
*May 27, 2014*

**physics**

W = F*d*cos A = 457 * 6.2 * Cos 8 = 2806 J.
*May 26, 2014*

**college physics**

mgh2 - mgh1 = mg(h2-h1) = mg(3.1) = 64.7*9.8(3.1) = 1966 Joules.
*May 26, 2014*

**college physics**

a. Always negative.
*May 26, 2014*

**Physics**

The force of the golf club acts in parallel with the motion of the ball which is 28o. So my answer is 1650 N. @ 28o. Since they did not ask for the direction of the force, I did not include the angle. Your answer is not the full force acting on the ball; it is the hor. ...
*May 26, 2014*

**Physics**

a=(V-Vo)/t a = (55-0)/1.5*10^-3=36670 m/s^2 F = m*a = 0.045 * 36,670 = 1650 N.
*May 26, 2014*

**Business mat**

7x + 3 = 7x + b b = 7x-7x+3 = 3
*May 24, 2014*

**Math**

Eq1: Y = x^2+3x-4 Eq2: Y = 2x+2 In Eq1, replace Y with 2x+2 and solve for X: 2x+2 = x^2+3x-4 x^2+3x-2x = 2+4 x^2+x = 6 x^2+x-6 = 0 C = -6 = -2*3. (x-2)(x+3) = 0 x-2 = 0, X = 2. x+3 = 0, X = -3. Solution: X = -3, and 2.
*May 23, 2014*

**math**

1. Y = 4x + 12 = $44. 4x = 44-12 = 32 X = 8 Months = Oct. of same yr. 2.C 3. 8 < x + 3 X > 8-3 X > 5. 7 > 5 Answer = b = 7.
*May 23, 2014*

**Math**

Bought X boxes. Sold (x-5) boxes. Cost = $Y per box Selling price = $(y+0.1) per box. P = $6 * 1.25 = 7.50 = Sale price. (x-5)(y+0.1) = 7.50 xy + 0.1x - 5y - 0.5 = 7.5 Eq1: xy + 0.1x - 5y = 8 Eq2: xy = $6.00 Y = 6/x In Eq1, replace Y with 6/x: 6 + 0.1x - 30/x = 8 0.1x - 30/x...
*May 23, 2014*

**math**

Martin is X yrs. old. Joan is (x-3) yrs. old. 4x/5 = x-3 Multiply both sides by 5: 4x = 5x-15 X = 15 (4/5) * 15 = 12 yrs. = Joan's age.
*May 23, 2014*

**math**

Answer = b.
*May 23, 2014*

**math**

I = 2.42*10^-5 watts/cm^2 = Intensity. Io = 1*10^-16 watts/cm2 = Reference. db = 10*Log(I/Io) db = 10*Log(2.42*10^-5/1*10^-16) db = 10*Log(2.42*10^11) = 114.
*May 23, 2014*

**math**

Thu sales = $X Sat sales = $2X Other 4 days = $X/2 X/(x+2X+X/2) = X/(3 1/2)x = X/(7x/2) = x * 2/7x = 2/7.
*May 23, 2014*

**physics**

KE = 0.5m*V^2 = 900,000 J. 0.5*2000*V^2 = 900,000. Solve for V in m/s.
*May 21, 2014*

**Algerbra 2**

V = Vo - Vo*r*T V = 16,400 - 16,400*0.24*2 = $8528
*May 21, 2014*

**Math**

Do you mean: A(7,3.-4), B(-3,1,0). If so, X = (7+(-3))/2 = 2. Y = (3+1)/2 = 2. Z = (-4+0)/2 = -2. (x,y,z) = (2,2,-2).
*May 21, 2014*

**Math**

See previous post.
*May 21, 2014*

**Math**

C(h,k), P(x,y). x - h = x+5 x-x-h = 5 -h = 5 h = -5 y-k = y-2 y-y-k = -2 -k = -2 k = 2 C(-5,2). r = sqrt(9) = 3 Units.
*May 21, 2014*

**math**

-15-(-4) + (-12) + -8 -9 = -15 + 4 -12 -17 = -44 + 4 = -40.
*May 21, 2014*

**Science**

B. 1.
*May 21, 2014*

**Science**

1d. 2a.
*May 20, 2014*

**Math**

Incomplete.
*May 20, 2014*

**Science**

W = F*d = 200 * 45.5
*May 19, 2014*

**physics**

T^2 = 4pi^2(L/g) = (2.45)^2 39.48(1.25/g) = 6.00 6g = 49.35 g = 8.23 m/s^2
*May 19, 2014*

**Math**

csc R = 13/12 = 1/sin R 1/Sin R = 13/12 sin R = 12/13 = Y/r. X^2 + Y^2 = r^2 X^2 + 12^2 = 13^2 X^2 = 13^2 - 12^2 = 25 X = 5. Tan T = Y/X = 12/5
*May 19, 2014*

**Algebra**

Tan A = 5.67 A = 80 Degrees.
*May 19, 2014*

**Geometry**

A(0,0), B(a,0). AB is a hor. line, because Y is constant. AB = a - 0 = a. D(0,b), C(a,b). DC is a hor. line. DC = a - 0 = a. AB = DC, Because quantities that are = to the same or = quantities are = to each other. A(0,0), D(0,b). AD is a Vertical line, because X is constant. AD...
*May 19, 2014*

**algebra**

h max = ho + (V^2-Vo^2)/2g h max = 80 + (0-(64^2))/-64 = 144 Ft above gnd.
*May 19, 2014*

**maths**

Multiply numerator and denominator by 2^-(1-n): 1/2^(1-n) = 2^-(1-n)/1 = 2^(-1+n) = 2^(n-1)
*May 19, 2014*

**maths**

a. 16 = 2^4 b. 2^m/8 = 2^m/2^3 = 2^(m-3)
*May 19, 2014*

**Algebra**

P = Po(1+r)^n P = $25,000 Po = ? = Initial principal. r = (8%/2)/100% = 0.04 = Semi-annual % rate expressed as a decimal. n = 2Comp./yr. * 3.5yrs. = 7 Compounding periods. P = Po(1.04)^7 = 25,000 Po = 25,000/1.04^7 = $18,997.95 Down.
*May 19, 2014*

**Physics**

a = (V-Vo)/t = (-46-30)/5*10^-3 = -15.2*10^3 m/s^2. F = m*a = 0.15 * (-15,200) = -2280 N.
*May 18, 2014*

**Math**

a. Bal = 739.65 - 0.05*739.65 + 179.39 = $882.06 b. P min = 0.05 * 882.06
*May 18, 2014*

**math**

Ab = Ab = (W*L)/2 = (6*9)/2 = 27 in^2 = Area of the base. V = (Ab*h)/3 = (27*4)/3 = 36 in^3.
*May 18, 2014*

**math**

As = 2(L*h) + 2(W*h) + 2(L*W) As = 2(2*12) + 2(9*12) + 2(2*9) =
*May 18, 2014*

**math**

I don't know how they came up with $5100; perhaps you should double check that number. In a discounted loan, the discounted amount(Int.) is paid up-front by deducting it from the amount borrowed: Ab = $2,000 = Amount borrowed = Face value. Ar = Amount received by borrower...
*May 18, 2014*

**physics**

Vo = 12m/s[50o] Xo = 12*Cos50 = 7.71 m/s. Yo = 12*sin50 = 9.19 m/s. Tr = -Yo/g = -9.19/-9.8 = 0.938 s. Tf = Tr = 0.938 s. = Fall time. Tr+Tf = 0.938 + 0.938 = 1.876 s. = Time to return to launching height. Range=Xo * (Tr+Tf) = 7.71 * 1.876=14.46m D max = 14.46 + 8m/s * 1.876s ...
*May 17, 2014*

**physics**

Incomplete.
*May 17, 2014*

**Physics**

P = (0.45m/3.6m) * 1s. = 0.125 s. = The period of the wave. Freq. = 1/P = 1/0.125 = 8 Cycles/s = 8 Hz.
*May 17, 2014*

**Physics**

d=V*T = 3*10^8 * (1.9*10^-3/2) = 2.85*10^5 m.
*May 17, 2014*

**Geometry**

As = 2*(W*h)) + 2*(L*h) + 2(L*W) a. As = Al + 2*(L*W) = 992 Ft^2 Al + 2*(20*12) = 992 Al = 992 - 480 = 512 Ft^2.=Lateral area. b. Al = 2(W*h) + 2(L*h) = 512 Ft^2 2(12*h) + 2(20*h) = 512 Solve for h.
*May 17, 2014*

**physics**

Circumference = pi*2r = 3.14 * 80 = 251.3 cm. = 2.513 m. V=300rev/min * 2.513m/rev * 1min/60s = 12.57 m/s. = Linear speed. Va = 12.57m/s * 6.28rad/2.513m=31.4rad/s = Angular speed.
*May 17, 2014*

**physics**

X = 100N Y = 100N Tan A = Y/X = 100p/100 = 1.0 A = 45o = Direction. Fr = X/Cos A = = 100/Cos45=141.4N[45o]= Resultant force. a=Fr/m=141.4[45o]/100kg=1.414m/s^2[45o].
*May 16, 2014*

**math**

Ans: D
*May 15, 2014*

**Pre cal**

1. AB+BC = 200km[110o] + 100km[37o] X = 200*cos110 + 100*cos37 = 11.46 km. Y = 200*sin110 + 100*sin37 = 248.1 km. D^2 = X^2 + Y^2 = (11.46)^2+(248.1)^2 = 61,685 D = 248.4 km 2. V = 530mph[N25oW] = 530mph[115o] CCW.
*May 15, 2014*

**Pre cal**

1. An airplane flies 200 km on course 110 degrees from point A to point B. At point B the pilot changes course to 37 degrees. The plane then flies 100 km to point C. Find the magnitude of the net displacement from A to C. 2. A plane is flying on a bearing of 25 degrees west of...
*May 15, 2014*

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