Thursday

February 11, 2016
Total # Posts: 11,500

**Physics**

a. V=Vo + a*t=10.3 + 0.37*7.01=12.9 m/s.
*September 15, 2015*

**Trigonometry**

X1 = Distance from closest point to bottom of mountain. X1+13 = Distance from farthest point to bottom of mountain. Tan9.5 = h/X1, h = X1*Tan9.5. Tan4.5 = h/(X1+13), (X1+13)*Tan4.5. h = X1*Tan9.5 = (X1+13)*Tan4.5 X1 = 0.507(X1+13) = 0.507X1+6.59. X1-0.501X1 = 6.59. 0.5X1 = 6....
*September 15, 2015*

**Math**

1. 28x - 6(3x-5) = 60. 28x - 18x + 30 = 60. 10x = 30. X = 3. 2. 3(x-2)-8x < 44. 3x-6-8x < 44. -5x < 50. X > -10. Note: When I divided by (-5), I reversed the inequality sign.
*September 15, 2015*

**physics**

d = 3*10^8m/s * 8.3min*60s/min = 1.494*10^11 m.
*September 15, 2015*

**Physics**

A. d = V*t = 1700 m. 6.6*t = 1700. t = 258 s. B. T = 258 + 1700m/2s. C. T/60 = ? D. d = 2 * 1.7km = 3.4 km. E. D = 0.(The end = The beginning). F.
*September 15, 2015*

**Physics**

d = 3cm/yr * T = 5.0*10^7 cm. T = 1.67*10^7 years.
*September 14, 2015*

**Physics**

M*g = 10 * 9.8 = 98 N. = Wt. of object = Normal(Fn). Fs = u*Fn = 0.7 * 98 = 6.86 N. = Force of static friction. F-Fs = M*a. F-6.86 = 10*0 = 0. F = 6.86 N. Angle = 0o(parallel to the surface).
*September 14, 2015*

**College Physics**

Range = Vo^2*sin(2A)/g = 300 m. Vo = 700 m/s. g = 9.8 m/s^2. A = ?
*September 14, 2015*

**PHYISCS**

V^2 = Vo^2 + 2g*h. V = 0. g = -9.8 m/s^2. h = 4.6 m., max. Vo = ?.
*September 14, 2015*

**precalc**

Cot 2pi/5 = 1/Tan2pi/5 = 1/Tan72o = 0.325.
*September 14, 2015*

**Physics**

V1 = a*t = 4 * 6 = 24 m/s. d1 = 0.5a*t^2 = 0.5*4*6^2 = 72 m. d2 = V*t = 24 * 8 = 192 m. V = Vo + a*t = 0. a = -Vo/t = -24/10 = -2.4 m/s^2. V^2 = Vo^2 + 2a*d = 0. d3 = -(Vo^2)/2a = -(24^2)/-4.8 = 120 m. d = d1+d2+d3.
*September 14, 2015*

**Inverse Functions**

X = 1450 Ft. Y = 850 Ft. Tan A = Y/X.
*September 14, 2015*

**physique**

X = Vp = 320 km/h. Y = Vw = 65km/h. Vr = sqrt(X^2 + Y^2) = Resultant velocity.
*September 14, 2015*

**Physics**

V^2 = Vo^2 + 2g*h. V = 0. Vo = 8.2 m/s. g = -9.8 m/s^2. h = ? V = Vo + g*Tr. V = 8.2 m/s. Vo = 0. g = -9.8 m/s^2. Tr = ? = Rise time. Tf = Tr = Fall time. Tr+Tf = Time to return to starting point. V^2 = Vo^2 + 2g*h. Vo = 8.2 m/s. g = +9.8 m/s^2. h = 50 m. V = ?.
*September 14, 2015*

**Physics**

X = Vc = 3.5 m/s. Y = Vb = 9.5 m/s. Tan A = Y/X = 9.5/3.5 = 2.71429. A = 69.8o Tan 69.8 = 106m/d. d = 106/Tan69.8 = 39.1 m. Downstream.
*September 14, 2015*

**Math**

Vr=50[315oCCW] + 150[30oCCW] = Resultant velocity. X = 50*Cos315 + 150*Cos30 = 165.3 km. Y = 50*sin3i5 + 150*sin30 = 39.64 km. Tan A = Y/X = 39.64/165.3 = 0.23981. A = 13.5o. Vr=X/Cos A=165.3/Cos13.5 = 170km[13.5o] CCW. = 170km[N76.5oE].
*September 14, 2015*

**phsyics**

Incomplete.
*September 13, 2015*

**math**

X = ? Y = 10+2 = 12 m. L = 15 m. Sin A = Y/L = 12/15 = 0.80, A = 53.1o X = L*Cos A = 15*Cos53.1 = 9 m.
*September 13, 2015*

**Physics 111**

We need to know the width of the river.
*September 13, 2015*

**Physics**

V1 = a*t = 15 * 2.8 = 42 m/s. d1 = 0.5a*t^2 = 0.5*15*2.8^2 = 58.8 m. V^2 = V1^2 + 2a*d = 0. d = -V1^2/2a = -(42^2)/-7 = 252 m. Final position = 7800 + 252 = 8052 m. mark.
*September 13, 2015*

**Physics**

a. d1 = V*t=2m/s. * 1s.=2 m. Head start. d2 = 2 + d1. 0.5a*t^2 = 2 + V1*t. 0.5*0.2*t^2 = 2 + 2*t. 0.1t^2 -2t - 2 = 0. Use Quadratic Formula. t = 21 s. b. d = 2 + 2*21 = 44 m.
*September 13, 2015*

**Physics**

V^2 = Vo^2 + 2g*h. V = 0 @ max h. Vo = 19.8 m/s. g = -9.8 m/s^2. h = ?
*September 13, 2015*

**Physics**

h = 0.5g*t^2 = g = 9.8 m/s^2.
*September 13, 2015*

**physics**

See previous post: Sun, 9-13-15, 4:17 PM.
*September 13, 2015*

**Physics**

d = 0.5a*t^2 = 0.5*17*11^2 =
*September 13, 2015*

**physics**

d = 0.5g*t^2 = 0.55 m. 4.9t^2 = 0.55. t^2 = =.112. t = 0.335 s. = Fall time. Dx = Xo * t = 12 m. Xo * 0.335 = 12. Xo = 35.8 m/s.
*September 13, 2015*

**Physics**

V = Vo = 8m/s.
*September 13, 2015*

**Sicience**

Density = 20g/315.4mL = g/mL.
*September 13, 2015*

**physics**

V = Vo + g*Tr. V = 0 Vo = 8.7 m/s. g = -9.8 m/s^2. Tr = ? = Rise time. Tf = Tr = Fall time. Tr+Tf = Time in air.
*September 13, 2015*

**Math (Trigonometry)**

X1 m.= Naeem's distance from base of tower. X2 = distance between Zola and Naeem. X1+X2 = Zola's distance from base of tower. Tan 50 = h/X1 = 180/X1. X1 = 180/Tan50 = 151 m. Tan32 = 180/(X1+X2) = 180/(151+X2). (151+X2) = 180/Tan32 = 288. X2 = 137 m.
*September 13, 2015*

**Physics**

6. V = 2.2 + 5.225 = 7.425 m/s. 7. a. 4 sig. fig. b. A = 47 * 31.25 = 1469 cm^2. 8. a. P = 2L + 2W. b. A = L*W.
*September 13, 2015*

**Physics**

Vo = 30000m/3600s. = 8.33 m/s. V^2 = Vo^2 + 2a*d = 8.33^2 + 2*3*100 = 69.44 + 600 = 669.44. V = Sqrt(669.44) = 25.87 m/s.
*September 13, 2015*

**Physics**

kilo means 1000 units. 1450 m./1000 = 1.450 km.
*September 13, 2015*

**math**

See previous post: Sun, 9-13-15, 3:47 AM.
*September 13, 2015*

**Math**

X = Vw = -85 km/h. Y = Vp = 230 km/h. a. Vr = sqrt(X^2 + Y^2) = b. Tan A = Y/X = 230/-85 = -2.70588. A =-69.7o=69.7o N. of W. = 20.3o W. of N. = Bearing.
*September 13, 2015*

**algeb**

3 years ago: Paul was X yrs. old. Vin was 3x yrs. old. At the present: Paul is x+3 yrs. old. Vin is 3x+3 yrs. old. x+3 + 3x+3 = 18. 4x = 12. X = 3. Paul = x+3 = 3+3 = 6 yrs. old. Vin = 3x+3 = 3*3 + 3 = 12 yrs. old. 3yrs. from now: Paul = 6+3 = 9 yrs. old. Vin = 12+3 = 15 yrs. ...
*September 13, 2015*

**Algebra**

d = (Vb-Vc)*T = 6km. (Vb-4)T = 6. Eq1: Vb*T - 4T = 6. (Vb+Vc)*T = 10 km. (Vb+4)*T = 10. Eq2: Vb*T + 4T = 10. Subtract the Eqs: Vb*T - 4T = 6. Vb*T + 4T = 10. -8T = -4. T = 0.5h. In Eq1, replace T with 0.5h: Vb*0.5 - 4*0.5 = 6. 0.5Vb = 8. Vb = 16 km/h.
*September 12, 2015*

**Physics**

Sin A = 0.05, A = 2.87o. Fp = Mg*sin2.87 = 1963 N. = Force parallel to the incline. V^2 = Vo^2 + 2a*d = 144. a = 144/2d = 144/400 = 0.36 m/s^2. F-1963 = 4000*0.36 = 1440. F = 3403 N.
*September 12, 2015*

**Algebra**

Y = 4x^2 + 3x - 1. Factor it: A*C = 4*(-1) = -4. 4x^2 + (4x-x) - 1. 4x^2+4x - (x+1). 4x(x+1) - (x+1). Y = (x+1)(4x-1). All real values of x will give a real output. Therefore, the domain is all real values of x. Interval notation: -infinity < X < +infinity. So x must be ...
*September 12, 2015*

**Algebra**

X = Ferns' score. 2x-28 = D. + E. score. x + 2x-28 = 272. 3x = 300. X = 100. 2x-28 = 2*100 - 28 = 172 = D. + E. Y = D. score. Y+8 = E. score. Y + Y+8 = 172. 2y = 164. Y = 82. Y+8 = 82 + 8 = 90.
*September 12, 2015*

**algebra**

P1 = Po(1+r)^n. Po = $25,000. r = (5.35%/4)/100% = 0.01338 = Quarterly % rate expressed as a decimal. n = 4comp./yr. * 5yrs. = 20 Compounding periods. P2 = Po(1+r)^n. r = 2.6%/100% = 0.026 = Annual % rate. n = 1comp./yr. * 5yrs. = 5 Compounding periods. P1+P2 = Total Amt.
*September 12, 2015*

**math**

2/5-3/7 - 1/6*3/2 + 1/14*2/5 = 14/35-15/35 - 3/12 + 2/70 = -1/35 - 1/4 + 1/35 = -1/4.
*September 12, 2015*

**Physics**

T = 150/60 = 2.5 h. d = V*T = 24 * 2.5 = 60 mi. Mile Marker = 132 + 60.
*September 12, 2015*

**physics**

Assume the car is coasting up the incline. At the bottom of the incline, the KE is at its' max value; and PE is zero. KE + PE = Constant. At bottom of incline, assume KE = 20 J., PE = 0. KE + PE = 20 + 0 = 20 J. Half-way up the incline, KE = 10 J. and PE = 20-10 = 10 J. KE...
*September 11, 2015*

**Physics-Trajectory**

A. Speed = 0. B. X = 12*5^3 - 2*5^2 = Y = 12*5^2 - 2*5 = Speed = sqrt(X^2 + Y^2) = C. D. Tan A = Y/X. A = ?
*September 11, 2015*

**Physics**

I need to know the direction(angle) of the forces.
*September 11, 2015*

**Physics**

Tan A = Y/X = -4.0/3.90 = -1.02564. A = -45.7o = 45.7o CW from +x-axis. = 314.3o CCW from +x-axis.
*September 11, 2015*

**Physics**

A. X = 8*2 = 16 cm/s. Y = 30 cm/s. Tan A = Y/X = 30/16 = 1.875. A = 61.9o B. X = 8*3 = 24 cm/s. Y = 30 cm/s. Dx = 24cm/s * 3s = 72 cm. Dy = 30cm/s * 3s = 90 cm. D = sqrt(Dx^2 + Dy^2) =
*September 11, 2015*

**physics**

When they meet: Da + Db = 350 km. 60*T+0.5*a*T^2 + 45*T = 350. a = 5 m/s^2. T = ?
*September 11, 2015*

**physics**

a. 0.03L = 1 oz. L = 33.4 oz. M = 0.454kg/16oz * 33.4oz = 0.948 kg.
*September 11, 2015*

**physics**

V = Vo + g*Tr = 0 @ max ht. Tr = -Vo/g = -17.5/-9.8 = 1.79 s. = Rise time. Tf = Tr = 1.79 s. = Fall time. Tr + Tf = 1.79 + 1.79 = 3.58 s = Time in flight. d = Vc*(Tr+Tf) = 17.5 * 3.58 = 62.7 m.
*September 11, 2015*

**physics**

X = Vc = 1.19 m/s. Y = Vs = 1.18 m/s Vr = sqrt(Vc^2 + Vs^2) = 1.68 m/s = Resultant velocity. Tan A = 1.18/1.19 = 0.99160 A = 44.8o a. d1=2.16/sin44.8 = 3.07 km. = 3070 m. Vr*T = 3070. 1.68*T = 3070. T = 1827 s. = 30.5 Min. b. d2 = 2.16/Tan44.8 = 2.175km = 2175 m. = Distance ...
*September 11, 2015*

**Math**

A = 2pi/3 = 360/3 = 120o. 120-360 = -240o.
*September 11, 2015*

**Physics**

Since the h is the same, the Fall time is the same: t2 = t1 = 2.8 s. d = 2 * 100 = 200 m.
*September 11, 2015*

**math Help**

1. (2/3)/1/4 = 2/3 * 4/1 = 8/3 = 2 2/3. 2. 2/3*5/2 = 10/6 = 1 2/3. 3. 9/4 * 5/2 = 45/8 = 5 5/8 Cups. 4. 1/4 * 1/4 = 1/16. 5. 1/16 * 1/4 = 1/64
*September 11, 2015*

**math**

The length of the shadow is proportional to the height of pole (Direct variation). h = 24/20 * 15m. =
*September 11, 2015*

**physics**

Incomplete.
*September 11, 2015*

**Trigonometry**

d = 79.5mi[N37o 10'W] = 79.5mi[127.17o] CCW. X = 79.5*Cos127.17 = Miles west. Y = 79.5*sin127.17 = Miles north.
*September 11, 2015*

**geometry**

The slope does not change. Multiply all coordinates by 3.5: A(7, 7), B(10.5, 28). m = (28-7)/(10.5-7) = L = Sqrt((10.5-7)^2 + (28-7)^2) =
*September 11, 2015*

**Physics**

D=-27i + 70 + 64.5i + 28.5=98.5 + 37.5i D = sqrt(98.5^2 + 37.5^2) =
*September 11, 2015*

**Physics**

d = V*T = 16km. 250*T = 16. T = 0.064 h. to drive the full length. d1 = 230*t1 = 8 km. t1 = 0.035 h to cover 1st half of the length. t2=0.064 - 0.035 = 0.0292 h to cover the last 8 km. Vavg = d2/t2 = 8/0.0292 = 274 km/h.
*September 10, 2015*

**Physics**

V^2 = Vo^2 + 2a*d. V = 6.5 m/s. Vo = 8.3 m/s. d = 5.6 m. a = ?. It should be negative.
*September 10, 2015*

**physics- work**

Work = F*d = Mg*d = 10*9.8*100 = 9800 J.
*September 10, 2015*

**physics**

My answer is e also.
*September 10, 2015*

**algebra**

Incomplete.
*September 10, 2015*

**physics**

a. V = Vo + g*t. Vo=1.5 m/s, g=9.8m/s^2 b. d = Vo*t + 0.5g*t^2. c. Vo = -1.5 m/s, d. Vo = -1.5 m/s.
*September 10, 2015*

**Physics**

d = V*t = 5050m/s * 0.0914s = 461.57 m. 461.57/91.4 = 5.05 Football fields.
*September 10, 2015*

**Math**

Cos30 = sin60.
*September 10, 2015*

**Math**

A. Tan t = h/Y, h = Y*Tan t B. Tan38 = h/3. h = 3*Tan38 = 2.34 m.
*September 10, 2015*

**Math**

csc30 = 1/sin30 = 1/0.5 = 2.0. sec30 = 1/Cos30 = 1/(sqrt3/2) = 2/sqrt3 = 1.155. Question: Did you show all of the problem?
*September 10, 2015*

**Math**

d = V*T = 4 mi. 3.5*T = 4. T = 1.143 h. To walk the 4 miles. 1.143 - 40/60 = 0.476 h. To walk the last 2 miles. V = 2mi/0.476h = 4.2 mi/h. = Speed for last two miles.
*September 10, 2015*

**Math**

a = 7, b = 10, c = 15. a^2 + b^2 = c^2. 49 + 100 = 149. But c^2 = 225. So it is NOT a rt. triangle.
*September 10, 2015*

**MATH (factoring)**

Recall: a^2-b^2 = (a+b)(a-b). In your case: a^2 = (x+y)^2. b^2 = 64. Factored: ((x+y)+8)((x+y)-8).
*September 10, 2015*

**Precal**

%CV = 1.25Vo. Where Vo is initial velocity Change = 1.25Vo-Vo = 0.25Vo. %Change = (0.25Vo/Vo)*100% = 25%.
*September 10, 2015*

**Algebra 1**

X mockingbirds. 3X finches. x + 3x = 48. X = ?
*September 10, 2015*

**PHYSICS**

X = Vw = 8 m/min. Y = Vs = 20m/min. Tan A = Y/X = 20/8 = 2.500. A = 68.2o N. of E. = 21.8o E. of N. Direction = 21.8o W. of N. to offset affect of the wind. C is the closest answer.
*September 10, 2015*

**Physics**

t = 7.1min * 60s/min = 426 s. d = V*t = 2.4m/s * 426s = 1022.4 m.
*September 10, 2015*

**science**

13. Tr = 6/2 = 3 s. = Rise time. Tf = Tr = 3 s. = Fall time. a. V = Vo + g*Tr = 0 @ max h. Vo = -g*Tr = -(-9.8)*3 = 29.4 m/s. = Initial velocity. b. h = Vo*Tr + 0.5g*Tr^2. h = 29.4*3 - 4.9*3^2 = 44.1 m. c. Since Tr is 3 seconds, the ball is on its' way down after j seconds...
*September 9, 2015*

**Physics**

d = 16.1 + 24 + 36.9 = 77 mi. T = 16.1/56 + 24/46 + 36.9/37.3 = Hours. Vavg = d/T.
*September 9, 2015*

**Physics**

Vo=a*t = 217 * 0.0854 = 18.53 m/s[58.3o] a. Xo = 18.53*Cos58.3. b. Yo = 18.53*sin58.3
*September 9, 2015*

**Physics**

t = 54mi * 1h/39mi = 1.385 h. T = 108/52 = 2.077 h for entire distance 2.077 - 1.385 = 0.692 h remaining. Vavg = 54/0.692 = 78 mi/h.
*September 9, 2015*

**Quick physics check**

1. D = 42-18 = 24 cm. 2. D = 9-18 = -9 cm. 3. 42-18 = 24 cm. 4. 6-4-2 = 0. 5. d = 12 + 22 = 34 Miles. 6. D = sqrt(12^2 + 22^2) = 25 miles.
*September 9, 2015*

**Physics**

See previous post: Wed, 9-9-15, 5:11 PM.
*September 9, 2015*

**Physics**

See previous post: Wed, 9-9-15, 5:19 PM.
*September 9, 2015*

**Algebra grade 7**

L = sqrt 64 = 8 Ft.
*September 9, 2015*

**math**

x LOwer level sold. (35,000-X) Upper level sold. 10x + 5(35,000-x) = $325,000. X = ?
*September 9, 2015*

**physics**

F = 61[98o] + 63[50o]. X = 61*Cos98 + 63*Cos50 = -8.49 + 40.50 = 32.0 N. Y = 61*sin98 + 63*sin50 = 60.41 + 48.26 = 108.7 N. F = sqrt(X^2+Y^2).
*September 9, 2015*

**Physics**

X = 52.6 m/s. Y = 36.8 m/s. Speed = sqrt(X^2+Y^2). Tan A = Y/X = 36.8/52.6 = 0.69962 A = 34.98o CCW = 55.02o From vertical.
*September 9, 2015*

**Physics**

Glad I could help!
*September 9, 2015*

**Physics**

1. 0.5g*t^2 = 2010 m. 4.9t^2 = 2010. t = 20.3 s. = Fall time. D = Xo*t. = 193m/s * 20.3s = 3909 m. 2. V=sqrt(Xo^2+Yo^2)=sqrt(193^2+58^2) = 202 m/s. 3. Vo*t + 0.5g*t^2 = 2010 m. 58*t + 4.9*t^2 = 2010. 4.9t^2 + 58t - 2010 = 0. Use Quadratic Formula. t = 15.2 s. = Fall time. D = ...
*September 9, 2015*

**Physics**

1. F = M*a. a = (V-Vo)/t = (18-0)/0.05 = 360 m/s^2. F = 0.6kg * 360 = 216 N. 2. Xo = 18*Cos58 = 9.54 m/s. Yo = 18*sin58 = 15.3 m/s. Y = Yo + g*Tr = 0. 15.3 - 9.8*Tr = 0. Tr = 1.56 s.= Rise time. h=Yo*Tr + 0.5g*t^2=15.3*1.56-4.9*1.56^2 = 11.9 m. 0.5g*Tf^2 = 11.9 -10 = 1.9 4.9Tf...
*September 9, 2015*

**Physics**

V^2 = Vo^2 + 2a*d. V = 0. Vo = 12 m/s. d = 10 m. a = ?
*September 9, 2015*

**physics**

0.5g*t^2 = 20 m. g = 9.8 m/s^2. t = ?
*September 9, 2015*

**physics**

F = 80*Cos60.
*September 8, 2015*

**Physics**

Vp + 64[225] = 470[90o] = 470i. Vp + 64*Cos225 + i64*sin225 = 470i. Vp - 45.3 - 45.3i = 470i. Vp = 45.3 + 515.3i = 517km/h[85o]. = velocity of the plane. Direction = 85o N. of E. = 5o E. of N.
*September 8, 2015*

**physics**

0.5a*t^2 = 35. t = ?
*September 8, 2015*

**algebra1**

Speed = 0.10m/s * 3.3Ft/m * 3600s/h =
*September 8, 2015*

**Physics**

V1*t + 0.5g*t^2 = 32. V1*1.8 + 4.9*1.8^2 = 32. 1.8V1 = 32 - 15.88 = 16.12. V1 = 9.0 m/s. = Velocity at the beginning of the last 32 m. h = (V1^2-Vo^2)/2g + 32 = (9^2-0)/19.6 + 32 = 36.13 m.
*September 8, 2015*

**Physics**

V = Vo + g*t. Vo = - 6.3 m/s. g = +9.8 m/s. t = 3 s. V = ?
*September 8, 2015*

**Physics**

V = Vo + g*t = 0 14 - 9.8t = = 0. 9.8t = 14. t = 1.43 s. To reach max h. h = Vo*t + 0.5g*t^2=14*1.43 + 4.9*1.43^2 = 10 m.
*September 8, 2015*