Sunday

January 22, 2017
Total # Posts: 13,051

**Physics**

V = Vo + a*t. Vo = 60,000m/3600s = 16.7 m/s. V = 90,000/3600s. = 25 m/s. t = ?.

*November 14, 2015*

**physics**

d = 5.5rev * 6.28rad/rev = 34.6 radians. V = 34.6rad./3s. = 11.5 rad/s.

*November 14, 2015*

**Physics**

d1 = r*t = 160 km. 80*t = 160. t1 = 2 hrs. d2 = r*t = 160 km. 100*t = 160. t2 = 1.6 hrs. r(avg.) = (d1+d2)/(t1+t2) = 320/3.6 = 88.9 km/hr.

*November 14, 2015*

**maths**

4/4 - (1/4+1/4) = 4/4 - 2/4 = 2/4 = 1/2 Left. 1/2 - 1/3 = 3/6 - 2/6 = 1/6 Left for wife. (1/6)/(2/6) * 100,000 = 1/2 * 100,000 = 50,000 = Wife's amount.

*November 14, 2015*

**Math**

A = X Degrees. B = (4x+20) Degrees. x + (4x+20) = 90o. X = ?

*November 14, 2015*

**Math**

Correction: Math-99Henry. should be Math Henry.

*November 13, 2015*

**High school /physic**

Vf = Vo + g*t. Vf^2 = Vo^2 + 2g*h.

*November 13, 2015*

**Math**

See previous post: Thu, 11-12-15, 11:06 PM.

*November 13, 2015*

**Math**

P = Income-Expense=20x-((20x/2)+18)=200. 20x - (10x+18) = 200. 20x - 10x - 18 = 200. 10x = 218. X = 21.8 or 22 Copies.

*November 13, 2015*

**University of North Dakota**

Correction: Change 49.5 to 495.

*November 13, 2015*

**University of North Dakota**

Acc. #1: Po1 = $X @ 9.5%. Acc. #2: Po2 = $(33,000-X) @ 1.5%. Po1*r1*t = Po2*r2*t. X*0.095*1 = (33,000-X)*0.015*1. 0.095x = 49.5 - 0.015x. 0.11x = 495. X = $4500. 33000-X = 33000-4500 == $28,500.

*November 13, 2015*

**Math**

d = 16t^2 = 1551 Ft. t^2 = 96.9375. t = 9.8 s. V = 64h?? 62

*November 12, 2015*

**math**

V = 0.3125 * 18 * 15 = in^3. Wt. = V*D = V * 0.698 = The surface area(As) is the area of the 6 sides: As = 2*W*h + 2*L*h + 2*L*W. L = 18 1n. W = 15 in. h = 0.3125 in.

*November 12, 2015*

**Math homework**

6/6-1/6 = 5/6 = Remainder. 5/6-2/5 = 25/30-12/30 = 13/30 Left.

*November 12, 2015*

**algebra**

Y = x^2+5x-36 = 0 -36 = -4*9, -4+9 = 5 = B. (x-4)(x+9) = 0.

*November 12, 2015*

**PHYSICS**

D = 350rev/60s * 0.64s * 6.28rad/rev =

*November 12, 2015*

**PHYSICS**

Do you think we might need to know the angles?

*November 12, 2015*

**Physics**

V^2 = Vo^2 + 2a*d = 0. a = -(Vo^2)/2d. It will be negative.

*November 11, 2015*

**Physics**

C = pi*2r = 3.14 * 2*0.98 = 6.15 m. 0.44rev/s. * 6.15m/rev = 2.71 m/s. = Tangential speed.

*November 11, 2015*

**physics**

Wo = M*g = 2 * 9.8 = 19.6 N. Fn = 19.6-20*sin20 = Normal force. Fk = u*Fn = Force of kinetic friction. Net force = 20*Cos20 - Fk.

*November 11, 2015*

**Pre-Alegebra**

3. Irrational. 4. D^2 = 15^2 + 8^2, D = 17. 5. You can test each choice by taking the cube root or raising it to the 1/3 power: 100^(1/3) = 4.64 125^(1/3) = 5 = A perfect cube. 6. Whole number/Integer. 7. 9/10. 8. Sqrt(35) = 5.92 In. Cut the 6-in. piece. 10. Riley: Sqrt(36) = ...

*November 11, 2015*

**Physics**

V^2 = Vo^2 + 2g*h. Vo = 0. g = 9.8 m/s^2. h = 2.1 m. V = ?

*November 11, 2015*

**physics**

Vf^2 = Vo^2 + 2g*h = 0. h = -Vo^2/2g + 1.5 = Meters above gnd. Vo = 20 m/s. g = -9.8 m/s^2.

*November 11, 2015*

**nwfsc**

V = I * R = 20 * 7 = 140 Volts. R = V/I = 140/14 = 10 Ohms.

*November 10, 2015*

**Math**

Libby: $60. Jess: 3/4 * 60 = $45. Stephanie: 1/3 * 45 = $15. Total:

*November 10, 2015*

**Math**

P(1, 2). Slope = 3. (1, 2), (x, y). m = (y-2)/(x-1) = 3. (y-2) = 3(x-1).

*November 9, 2015*

**math**

I've solved several of your problems. Where is the end?

*November 9, 2015*

**math**

Same procedure as your previous post.

*November 9, 2015*

**math**

I = Po*r*t = $1200. r = (18%/12)/100% = 0.015 = Monthly % rate expressed as a decimal. t = 1 Mo. Po*0.015*1 = 1200. Po = $80,000.

*November 9, 2015*

**math**

P = Po + Po*r*t = 2Po. Po + Po*0.1*t = 2Po Divide by Po: 1 + 0.1t = 2. 0.1t = 1. t = 10 Years.

*November 9, 2015*

**math**

I = Po*r*t. I = $3600. Po = $20,000. r = 0.12. t = ?.

*November 9, 2015*

**math**

P = Po + Po*r*t.

*November 9, 2015*

**Physics**

a. Vo = 60000m/3600s. = 16.7 m/s. Vf = 40000m./3600s. = 11.1 m/s. b. a = (Vf-Vo)/t = (11.1-16.7)/8 = -0.699 m/s^2.

*November 9, 2015*

**AP Physics**

Wb = M*g = 55 * 9.8 = 539 N. Fp = 539*sin40 = 347 N. = Force parallel to the ramp. Fn = 539*Cos40 = 413 N. = Normal force. Fp-Fk = M*a. 347-Fk = M*0 = 0. Fk = 347 N. = Force of kinetic friction. u = Fk/Fn.

*November 8, 2015*

**physics**

180N[W60oN] + [140N[90o] + 180N[E60oN]+ Ws = 0. Ws=-(-180*Cos60+180*sin60+140i+180*Cos60 +180*sin60) = -(155.9i+140i+155.9i) = -452i = -452 = 452 N. Downward. = Wt. of the sign. The horizontal components cancelled: 180*Cos60 - 180*Cos60 = 0.

*November 8, 2015*

**physics**

a = M1/M2 * a1 = 2.9/6.2 * 2.7 = 1.26 m/s^2.

*November 8, 2015*

**Science**

Mass = 200Lbs * 0.454kg/Lb = 90.8 kg. Wb = M*g = 90.8 * 9.8 = 889.84 N. = Normal force(Fn). a = 10Ft/s^2 * 1m/3.3Ft = 3.03 m/s^2. Fk = u*Fn = 0.25 * 889.84 = 222.5 N. = Force of kinetic friction. Fap-Fk = M*a. Fap = Applied or required force. Fap = ?.

*November 8, 2015*

**physics**

See previous pose: Sat, 11-7 15, 6:23 PM.

*November 8, 2015*

**physics**

F1 + F2 + F3 = 0. F3 = -(F1+F2) F3 = -(18[E60oN]+6.5[E45oS]) = -(18*Cos60+18*sin60 +6.5*Cos45+6.5*sin45 = -(9+15.6i + 4.6-4.6i) = -(13.6+11i) = -13.6 - 11i = 21.6N.[39o] Below the negative X-axis. f4

*November 8, 2015*

**ap physics**

Incomplete.

*November 8, 2015*

**physics**

Vf^2 = Vo^2 + 2a*d.

*November 7, 2015*

**physics**

V = Vo + a*t = 41.1m/s[26.5o]. 3.87[40.3o] + a*8.35 = 41.1[26.5]. 2.95+2.50i + a*8.35 = 36.8 + 18.3i. a*8.35 = 36.8 + 18.3i - 2.95 - 2.50i. a*8.35 = 33.9 + 15.8i = 37.4m/s[25o]. a = 4.48m/s^2[25o], CCW from +X-axis.

*November 7, 2015*

**physics**

Vf = Vo + a*t = 59.1m/s[43o]. Vo + 2.1[31.7o]*7.75 = 59.1[43]. Vo + 16.3[31.7o = 59.1*Cos43+59.1*sin43. Vo+16.3*Cos31.7+16.3*sin31.7=43.2+40.3i. Vo + 13.9 + 8.6i = 43.2 + 40.3i. Vo=43.2 + 40.3i - 13.9 - 8.6i=29.3+31.7i Vo = 43.2m/s[47.3o], CCW from +X-axis.

*November 7, 2015*

**physics**

P = E*I = 1.0 W. 1.58I = 1.0 I = 0.633A. A*h = 1. 0.633*h = 1. h = 1.58 Hours.

*November 7, 2015*

**Calculus**

Find the Absolute Maximum and Absolute Minimum of f on (0,3]. f(x)=(x^3-4x^2+7x)/x Multiple choice question I know the minimum is (2,3) but the maximum is either nothing or (0,7) but I can't tell which one

*November 7, 2015*

**science**

Velocity in kg/h???.

*November 7, 2015*

**physics**

Wc = M*g = 330 * 9.8 = 3234 N. = Normal force, Fn. Fap = 410 + 240 = 650 N. = Force applied. Fap-Fk = M*a. Fap-Fk = M*0 = 0. Fk = Fap = 650 N. = Force of kinetic friction. u = Fk/Fn.

*November 6, 2015*

**Physics**

V = Vo + g*T. Vo = 0. g = 10 m/s^2. h = ho - 0.5g*Tf^2 = 0. 5Tf^2 = 80. Tf^2 = 16. Tf = 4 s. = Fall time. Use T in the range of 0 to 4 seconds. (T,V,h). (0,0,80). (1,10,75). (2,20,60). (3,30,35). (4,40,0).

*November 6, 2015*

**math**

1. 6x4 in., P = 2*6 + 2*4 = 20 in. 2. 6*2 in., P = 2*6 + 2*2 = 16 in. 3. 4x2 in., P = 2*4 + 2*2 = 12 in. =

*November 5, 2015*

**Physics**

Units of time are seconds, minutes, and hours. Vf = Vo + a*t = 0. t = -Vo/a = -22/-6.1 = 3.61 s.

*November 5, 2015*

**Precalc**

a. P = Po(1+r)^n Po = $60,000. r = (4%/4)/100% = 0.01 = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 10yrs. = 40 Compounding periods. P = ? c. Same procedure as a.

*November 5, 2015*

**physics**

Vf = Vo + a*t = 3 + 1*5 = 8 m/s. d = Vo*t + 0.5a*t^2.

*November 5, 2015*

**universitu**

See Related Questions: Sun, 2-17-13, 5:09 AM.

*November 5, 2015*

**Calculus**

use tangent line approximation (linear approximation) to estimate The cube root of 1234 to 3 decimal places. Hint: the equation should be y=f'(x0)(x-x0)+f(x0) 11^3=1331 can be easily computed using binomial theorem. I used linear approximation and got 10.733, but it is not...

*November 4, 2015*

**Physics**

a. 0.5g*t^2 = 4780. 4.9t^2 = 4780. t^2 = 975.5 t = 31.2 s. Dx = 175m/s * 31.2s = 5465.8 m.

*November 3, 2015*

**math**

2. X = -30 km/h = velocity of the wind. Y = -295 km/h.=velocity of the aircraft. Q3. tan A = y/x = -295/-30 = 9.83333. A = 84.2o S. of W. = direction. Vr = y/sin A=-295/-sin84.2 = 296.5 km/h.

*November 3, 2015*

**physics**

Vf^2 = Vo^2 + 2g*h. g = 9.8 m/s^2.

*November 3, 2015*

**Science**

10cm3 * 10.5g/cm^3 = wt. in grams. mass = g/1000 = kilograms.

*November 2, 2015*

**physics**

140rev/60s * 20s =

*November 2, 2015*

**science**

T1*Cos26.5 - T2*Cos49.5 = 0 T1*Cos26.5 = T2*co49.5. T1 = 0.73T2. T1*sin26.5 + T2*sin49.5 = 155. Replace T1 with 0.73T2: 0.73T2*sin26.5 + T2*sin49.5 = 155. 0.324T2 + 0.760T2 = 155. 1.084T2 = 155. T2 = 143 N. T1 = 0.73*143 = 104 N.

*November 1, 2015*

**Physics**

Work = F*d = 23*Cos39 * 8 =

*November 1, 2015*

**physics**

Fs = M*g = 14 * 9.8 = 137 N. Fp = 137*sin 0 = 0 = Force parallel to the gnd. Fn = 137*Cos 0 - 25*sin30 = 124.5 N. = Normal Force. Fk = u*Fn = Force of kinetic friction. a = (25*Cos30-Fp-Fk)/M. Vf^2 = Vo^2 + 2a*d Vo = 1 m/s. d = 18 m. Vf = ?

*November 1, 2015*

**Physics**

See Relate Questions: Tue, 2-7-12, 1:11 AM.

*November 1, 2015*

**science**

Vf = Vo + g*Tr = 0. 24.5 + (-9.8)Tr = 0. -9.8Tr = 24.5 Tr = 2.5 s. = Rise time. Tf1 = Tr = 2.5 s = Time to fall back to roof. h = Vo*t + 4.9*t^2 = 24 m. 24.5*t + 4.9t*2 = 24. 4.9t^2 + 24.5t -24 = 0. Use Quad. Formula. t = 0.839 s. = Time to fall from top of roof to gnd. T = Tr...

*November 1, 2015*

**physics**

Fb = M*g = 13 * 9.8 = 127.4 N. Fp = 127.4*sin 0 = 0. = Force parallel to the floor. Fn = 127.4 - 86.4*sin64. = Normal force. Fk = u*Fn = Force of kinetic friction. a = (F*Cos64-Fp-Fk)/M.

*November 1, 2015*

**physics**

F1 = 2123 N. F2 = 1930 N. M = 1245 kg. a = (F1-F2)/M.

*November 1, 2015*

**Physics**

Vo = 90,000m/3600s = 25 m/s. d1 = Vo*t = 25m/s * 0.75s. = 18.75 m. d2 = 40 - 18.75 = 21.25 m. = Required stopping distance. Vf^2 = Vo^2 + 2a*d. Vf = 0. Vo = 25 m/s. a = -10 m/s^2. If d is => 21.25 m, the car will hit the barrier.

*October 31, 2015*

**physics**

4. Incomplete. Vo = 35m/s[33o]. Xo = 35*Cos33 = 29.4 m/s. Yo = 35*sin33 = 19.1 m/s. 5A. Yf^2 = Yo^2 + 2g*h. Yf = 0. g = -9.8 m/s^2. h = ? B. Dx = Vo^2*sin(2A)/g. A = 33o. g = +9.8 m/s^2. Dx = ?

*October 31, 2015*

**SPR,PHYSICS**

Vs = Db/Dw * Vb = 2/3 * Vb = 2Vb/3. Db/Dw * Vb = 2Vb/3. Db/Dw = 2/3 = 0.667. Db=0.667Dw = 0.667 * 0.998=0.665 g/cm^3. = Density of the block. Notes: Vs = Volume submerged. Db = Density of the block. Dw = Density of water. Vb = Volume of the block.

*October 30, 2015*

**force**

Wb = M*g = 10*9.8 = 98 N. Fp = 98*sin 0 = 0 = Force parallel to the surface. Fn = 98*Cos 0 - 40*sin30 = 78 N. = Normal force. Fk = u*Fn = 0.3 * 78 = 23.4 N. = Force of kinetic friction. a=(Fap*Cos30-Fp-Fk)/M=(34.6-0-23.4)/10 = 1.12 m/s^2.

*October 30, 2015*

**Physics**

I had this same problem... θ > arctan(μs) is the answer, you're welcome fam.

*October 29, 2015*

**math**

d1 = 10km/h * (20/60)h = 3.33 km., N. d2 = 5km/h * (35/60)h=2.92 km, NE.(45o)= 2.06 + 2.06i. Disp.=2.06 + 2.06i + 3.33i=2.06 + 5.39i = sqrt(2.06^2 + 5.39^2)

*October 29, 2015*

**physics**

Disp. = 7m[90o] + 9m[E45oN] + 7m[W14oS] X = 9*Cos45 + (-7*Cos14) = -0.428 m. Y = 7 + 9*sin45 + (-7*sin14) = 11.7 m. Q2. Tan A = Y/X = 11.7/-0.428 = -27.26754. A = -87.9o = 87.9o N. of W. = Direction. Disp. = Y/sin A = 11.7/sin87.9 = 11.71 m

*October 29, 2015*

**physics**

Disp. = 7m[90o] + 9m[E45oN] + 7m[S76oW]. X = 9*Cos45 - 7*Cos76 = 1.84 m. Y = 7 + 9*sin45 - 7*sin76 = 6.57 m. Q1. Tan A = Y/X = 6.57/1.84 = 3.57168. A = 74.4o N. of E. = Direction. Disp. = Y/sin A = 6.57/sin74.4 = 6.82 m

*October 29, 2015*

**physics**

A quantity that shows magnitude AND direction.

*October 29, 2015*

**geometry**

x^2 + 3x = 180o. x^2 + 3x-180 = 0. Factor it. -180 = -12*15. -12 + 15 = 3 = B. (x-12)(x+15) = 0. X = 12. X = -15. Use positive value of X: X = 12. X^2 = 144o. 3x = 36o.

*October 29, 2015*

**physics**

See Related Questions: Fri, 3-7-14, 4:52 PM.

*October 29, 2015*

**Chemistry**

Methanol, a major industrial feedstock, is made by several catalyzed reactions, such as CO(g) + 2H2(g) ----> CH3OH (l) One concern about using CH3OH as an auto fuel is its oxidation in air to yield formaldehyde, CH2OH(g) which poses a health hazard. Calculate ÄG o at ...

*October 29, 2015*

**physics math science**

Vo = 200m/s[37o]. Xo = 200*Cos37 = 160 m/s. Yo = 200*sin37 = 120 m/s. A. Y = Yo + g*t = 120 + (-9.8)*2 = 100 m/s. = Ver. component of velocity @ 2 s. Tan A = Y/Xo = 100/160 = 0.625. A = 32o. V = Xo/Cos A = 160/Cos32 = 189 m/s[32o]. B. Yf = Yo + g*Tr Yf = 0, g = -9.8m/s^2, Tr...

*October 29, 2015*

**Calculus**

i did. Dv=128pi. But the answer is 8pi/25

*October 28, 2015*

**Calculus**

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.02 cm thick to a sphere with diameter 40 meters. (Recall that the volume of a sphere of radius r is V=(4/3)πr^3. Notice that you are given that dr=0.02.)

*October 28, 2015*

**math**

2 Years ago: Pam: X yrs. old. Sam: 2x/3 yrs. old. 3 yrs. later: Pam: x+3 yrs. old. Sam: 2x/3 + 3 yrs. old 2x/3 + 3 = 3/4(x+3). 2x/3 + 3 = 3x/4+9/4 8x + 36 = 9x + 27. X = 9. So Pam is 3 yrs. older than Sam. In 20 yrs.: x + (x+3) = 69. 2x = 66. X = 33. X+3 = 36. Today: Sam: 33...

*October 28, 2015*

**Physics**

F = 0.119m * 110N./m =

*October 28, 2015*

**Physics**

Wc = M*g = 1400 * 9.8 = 13,720 N. = Normal force, Fn. Fp = mg*sin 0 = 0. = Force parallel to the road. Fk = u*Fn = 0.0889 * 13,720 = 1220 N. a = (Fp-Fk)/M = (0-1220)/1400 = -0.871 m/s^2. Vf^2 = Vo^2 + 2a*d. Vf = 0. a = -0.871 m/s^2. Vo = 21.4 m/s. d = ?.

*October 28, 2015*

**physics**

Wt. = 450cm^3 * 0.998g/cm^3 = 449.1 g. = 0.449 kg of water displaced. M*g = 0.449kg * 9.8N/kg = 4.40 N.

*October 28, 2015*

**SCIENCE\PHYSICS**

Disp. = 0. d = 5 + 5 = 10 m.

*October 28, 2015*

**PHYSICS**

d = 5 + 5 = 10 m.

*October 28, 2015*

**PHYSICS**

Displacement = 0, because it ended at the starting point.

*October 28, 2015*

**PHYSICS**

a. Yf^2 = Yo^2 + 2g*h = 0. Yo^2 = -2g*h = -2*(-9.8)27.5 = 539. Yo = 23.2 m/s. = Vertical component of initial velocity. sin A = Yo/Vo = 23.2/47.5 = 0.48877. A = 29.3o. Xo = Vo*Cos A = 47.5*Cos29.3 = 41.4 m/s = Horizontal component of initial velocity = Total velocity at max ht...

*October 28, 2015*

**physics**

The velocity of the truck must be equal to the hor. component of the airplane's velocity: V = 180*Cos47.

*October 28, 2015*

**Calculus**

Let f(x)=x^4. If a=1 and dx=Δx=1/2, what are Δy and dy?

*October 28, 2015*

**Physics**

Wb = M*g = 200 N., M = 200/g = 200/9.8 = 20.4 kg. sin A = 3/10 = 0.30, A = 17.5o. Fp = 200*sin17.5 = 60 N. Fn = 200*Cos17.5 = 191 N. a. Work = F*d = 120 * 10 = 1200 J. b. Mg*h-0 = Mg*h = 200 * 3 = 600 J. c. Conservation Energy : Change in KE = Change in PE = 600 J. d. Fap-Fp-...

*October 27, 2015*

**maths**

P = Po(1+r)^n. r = (6%/2)/100% = 0.03 = Semi-annual % rate expressed as a decimal. n = 2comp./yr. * 1yr. = 2 Compounding periods. P = Po(1+0.03)^2 = Po*1.03^2 = 1.0609Po. I = P-Po = 1.0609Po-Po = 0.0609Po. I/Po * 100% = 0.0609Po/Po = 6.09% = Effective APR.

*October 27, 2015*

**Physic**

V = Vo^2 + 2g*h. Vo = 0, h = 20 m, V = ?

*October 27, 2015*

**physicsn**

Notes: 1. Fp = Force parallel with the hill. 2. Fn = Normal force. 3. Fk = Force of kinetic friction. 4. Ws = Wt. of skier.

*October 27, 2015*

**physicsn**

Ws = M*g = 9.8M. Fp = 9.8M*sin30 = 4.9M. Fn = 9.8M*Cos30 = 8.49M. Fk = u*Fn = u*8.49M. Fp-Fk = M*a. 4.9M-u8.49M = M*(-1.4) Divide by M: 4.9-8.49u = -1.4 -8.49u = -1.4-4.9 = -6.3 u = 0.74.

*October 27, 2015*

**Physics**

a. a=(Vf-Vo)/t = (5-0)/20 = 0.25 m/s^2. d1 = Vo*t + 0.5a*t^2 = 0 + 0.5*0.25*20^2 = 50 m.

*October 27, 2015*

**math**

a scooter is priced between $1000 and $2000. its price is a multiple of 10. All the digits in the price, except for the thousands digit, are even numbers. The value of the hundreds digit is 30 times the value of the tens digit. what is the price of the scooter?

*October 26, 2015*

**science**

Wb = M*g = 10 * 9.8 = 98 N. Fn = 98-40*sin30 = Normal force. Fk = u*Fn = Force of kinetic friction. a = (Fap*Cos30-Fk)/M.

*October 26, 2015*

**Physics**

a = F/m = 5/0.3 = 16.7 m/s^2. V = Vo + a*t. Vo = 20 m/s. t = 0.03 s. V = ?

*October 26, 2015*