Monday

May 25, 2015

May 25, 2015

Total # Posts: 10,621

**physics**

Tr = 1.5 + 1 = 2.5 s. = Rise time or time to reach max ht. V = Vo + g*Tr = 0 Vo -9.8*2.5 = 0 Vo = 24.5 m/s. = Initial velocity. h max = Vo*t + 0.5g*t^2 = Vo*2.5 - 4.9*2.5^2 = 2.5Vo -
*February 2, 2015*

**physics**

d = 3.70 m. h = 1.67 m. V^2 = Vo^2 + 2g*h = 0 + 19.6*1.67 = 32.73 V = 5.72 m/s. = Velocity at bottom of incline. d = 0.5g*t^2 = 1.67 m. 4.9t^2 = 1.67 t^2 = 0.341 t = 0.584 s. = Time to reach bottom. V = Vo + a*t a = (V-Vo)/t V = 5.72 m/s Vo = 0 Solve for a.
*February 1, 2015*

**Physics**

Not enough INFO.
*February 1, 2015*

**Physics**

See Related Questions: Fri, 1-31-14, 10:23 PM.
*February 1, 2015*

**physics**

Vo = 1520m/s[35.5o] Yo = 1520*sin35.5 = 883 m/s. KE = 0.5M*Yo^2 = 0.5*25*883^2 KE = 0.5M*Yo^2 = 0.5*25*883^2=9,746,113 J. PE = KE = Mg*h = 9,746,113 25+9.8*h = 9,746,113 245h = 9,746,113 h = 39,780 m.
*February 1, 2015*

**PYHSICS**

With Arrow A = 9i B = -4 a. A+B = 9i + (-4) Q2. A+B = sqrt(9^2+(-4^2) = 9.85 Units. Tan Theta = 9/-4 = -2.25 Theta = -66.0 + 180 = 114o, CCW(Q2). b. A-B = 9i - (-4) = 9i + 4 A-B = sqrt(9^2+4^2) = 9.85 Units. Tan Theta = 9/4 = 2.25 Theta = 66o c. B-A = -4 - 9i X = -4 Y = -9
*February 1, 2015*

**physics**

Fg = 440 Hz = Freq. of the generator. Vg = 40 m/s = Velocity of the generator. Vs = 343 m/s = Velocity of sound. Vr = 0 = Velocity of the receiver(person listening to the sound). Fr = ((Vs+Vr)/(Vs-Vg))*Fg Fr = ((343+0)/(343-40))*440 = (343/303)*440 = 498.1 Hz. = Freq. heard by...
*February 1, 2015*

**Physics**

Impulse = M*(V-Vo) = 1295*(-1.5-25) = 34,318 kg*m/s
*February 1, 2015*

**physics**

Correction: Impulse = M*(V-Vo) = 0.42*(-27-12) = -16.4 kg*m/s
*February 1, 2015*

**physics**

V = Vo + a*t = -27 m/s. 12 + a*0.016 = -27 0.016a = -27 - 12 = -39 a = -2438 m/s^2 Impulse = M*V = 0.42 * (-27) = -11.34 F = M*a = 0.42 * (-2438) = 1024 N.
*February 1, 2015*

**physics**

See previous post: Wed, 1-28-15, 6:19 PM.
*February 1, 2015*

**Chemistry**

The primary route for making copper iodide is by reacting potassium iodide with copper sulfate: 2CuSO4 + 4KI + 2Na2S2O3 ==> 2CuI + 2K2SO4 + 2NaI + Na2S4O6 Identify potential green chemistry and green engineering challenges of the reaction.
*February 1, 2015*

**Physics**

Dc = Dp 0.5a*t^2 = 8.93 * 2.88 0.5a*2.88^2 = 25.72 Solve for a.
*January 30, 2015*

**Physics**

V = 43.3 m/s. a = -5.61 m/s^2 d = 24 + 55.9 = 79.9 m. V^2 = Vo^2 + 2a*d = 43.3^2 Vo^2 - 11.22*79.9 = 1874.9 Vo^2 = 1874.9 + 21036 = 22,911 Vo = 151.4 m/s. = Initial velocity. V = Vo + a*t V = 43.3 m/s. Vo = 151.4 m/s. a = -5.61 m/s^2 Solve for t.
*January 30, 2015*

**Physics**

V^2 = Vo^2 + 2a*d V = 333 m/s. Vo = 370 m/s. d = 12 cm = 0.12 m. Solve for a. It will be negative.
*January 30, 2015*

**physics**

h = Vo*t + 0.5g*t^2 = 1.5 m. Vo*0.18 + 4.9*0.18^2 = 1.5 0.18Vo + 0.159 = 1.5 0.18Vo = 1.5 - 0.159 = 1.34 Vo = 7.45 m/s = Initial velocity at top of window. V^2 = Vo^2 + 2g*h = 7.45^2 0 + 19.6*h = 55.5 h = 2.83 m. Above the window.
*January 30, 2015*

**Green Chemistry**

The primary route for making copper iodide is by reacting potassium iodide with copper sulfate: 2CuSO4 + 4KI + 2Na2S2O3 ==> 2CuI + 2K2SO4 + 2NaI + Na2S4O6 Identify potential green chemistry and green engineering challenges of the reaction.
*January 30, 2015*

**physics**

V^2 = Vo^2 + 2a*d V = 0 Vo = 125 m/s a = -9.8 m/s^2 Solve for d, the length.
*January 30, 2015*

**physics**

0.5g*t^2 = 2 4.9t^2 = 2 t^2 = 0.408 Tf = 0.639 s. = Fall time. Tr = Tf = 0.639 s. = Rise time. Tr+Tf = 0.639 + 0.639 = 1.28 s. = Time in air.
*January 29, 2015*

**physics**

M*g = 74.6kg * 9.8N/kg = 731.1 N. = Wt. of the skier. Fp = 731.1*sin39.0 = 460.1 N. = Force parallel to the incline. Work = Fp*d = 460.1 * 63.3 = 29,124 J.
*January 29, 2015*

**Physic**

V^2 = Vo^2 + 2g*h V = 0 Vo = 78.5 m/s. g = -9.8 m/s^2. Solve for h.
*January 29, 2015*

**Physics**

Circumference = pi*Dia. = 3.14 * 50 = 157.1 cm. 25rad * 1rev/6.28rad = 3.98 revs. d = 157.1cm/rev * 3.98revs = 625.4 cm.
*January 29, 2015*

**Need Help Physics**

Vbc = Vb + Vc = 6.44i Vb + 1.09 = 6.44i Vb = -1.09 + 6.44i Tan Ar = Y/X = 6.44/-1.09 = -5.90826 Ar = -80.4o = Reference angle. A = -80.4 + 180 = 99.6o, CCW = 9.6o W. of N.
*January 29, 2015*

**Maths**

PQ = X1 = 15 m. X2 = hor. distance from Q to kite. Tan45 = h/X2 h = X2*Tan45 Tan35 = h/(X1+X2) h = (X1+X2)*Tan35 = (15+X2)*Tan35 X2*Tan45 = (15+X2)*Tan35 X2 = 10.5 + 0.7X2 0.3X2 = 10.5 X2 = 35 m. h = X2*Tan45 = 35 * Tan45 = 35 m. h = (X1+X2)*Tan35 = 50*Tan35 = 35 m.
*January 29, 2015*

**physics**

e. None of the above.
*January 29, 2015*

**physics**

d = Vo*t + (-Vo^2/2a) = 25*0.564 + -(25^2)/-8 = 92.2 m.
*January 29, 2015*

**physicss**

V = Vo + a*t = -52 m/s 44 + a*0.003 = -52 0.003a = -52 - 44 = -96 a = -0.032 m/s^2 F = M*a
*January 29, 2015*

**physics**

a. Vo = 25m/s[35o] Xo = 25*Cos35 = Yo = 25*sin35 = b. See previous prob. c. See previous prob.
*January 29, 2015*

**physics**

Vo = 20m/s[30o] Xo = 20*Cos30 = 17.3 m/s. Yo = 20*sin30 = 10 m/s. Y = Yo + g*Tr Y = 0 Yo = 10 m/s. g = -9.8 m/s^2 Solve for Tr(Rise time). Tf = Tr = Fall time. Dx = Xo * (Tr+Tf)
*January 29, 2015*

**physics**

h = 0.5g*t^2 h = 1.40 m g = +9.8 m/s^2 Solve for t(Fall time). Xo = 0.600 m/s Dx = Xo * t
*January 29, 2015*

**physics**

Mass = 5.0L * 1oz/0.03L * 0.454kg/16oz = 4.73 kg.
*January 28, 2015*

**Physics**

V = 115km/h = 115000m/3600s = 31.94 m/s. Momentum = m*V Direction: East.
*January 28, 2015*

**physics**

d = V*t = 8 miles (Vs-Vc)t = 8 Vs = 1.7 m/h Vc = 0.7 m/h Solve for t.
*January 28, 2015*

**physics**

V = Vo + a*t V = 0 Vo = 8.2 m/s t = 5.8 s. Solve for a (It will be negative). F = m*a
*January 28, 2015*

**Physics**

V^2 = Vo^2 + 2g*h V = 0 Vo = 78.5 m/s. g = -9.8m/s^2 Solve for h.
*January 27, 2015*

**Physics**

h = 15 m Xo = 478 m/s? = Initial hor. speed. h = 0.5g*t^2 = 15 m 4.9t^2 = 15 t^2 = 3.06 Tf = 1.75 s.Fall time. Dx = Xo*Tf = 478m/s * 1.75s = 836 m. From bottom of cliff.
*January 25, 2015*

**physics**

V = 11.5m/s[236o] - 6.7m/s[270o] V = 11.5*Cos236 + 11.5*sin236 - (-6.7i) = -6.43 - 9.53i + 6.7i = -6.43 - 2.83i Tan Ar = Y/X = -2.83/-6.43 = 0.44074 Ar = 23.8o = Reference angle. A = 23.8 + 180 = 203.8o, CCW = 66.2o W. of S. = Direction. V = X/Cos2o3.8 = -6.43/-.915 = 7.03 m/s...
*January 25, 2015*

**physics**

h = 0.5g*t^2 = 2 m. 4.9t^2 = 2 t^2 = 0.408 Tf = 0.639 s. = Fall time. Xo*Tf = 4.4 m Xo * 0.639 = 4.4 Xo = 6.89 m/s.
*January 25, 2015*

**phys**

V = Vo + g*t Vo = 0 g = +9.8 m/s^2 t = 3.5 s. Solve for V.
*January 25, 2015*

**physics**

F = 225N/m * 0.15m = 33.75 N.
*January 25, 2015*

**physics**

V^2 = Vo^2 + 2a*d = 15^2 37^2 + 2a*0.31 = 15^2 2a*0.31 = 15^2 - 37^2 a = (15^2-37^2)/0.62 = -1845.2 m/s^2 A. V = Vo + a*t t = (V-Vo)/a = (15-37)/-1845.2 = 0.012 s B. V^2 = Vo^2 + 2a*d V = 0 Vo = 37 m/s a = -1845.2 Solve for d(thickness).
*January 25, 2015*

**physics**

M1*V1 + M2*V2 = M1*V + M2*0.76 24*60.8 + 12*15.2 = 24*V + 12*0.76. Solve for V(cm/s).
*January 25, 2015*

**physics**

M1*V1 + M2*V2 = M1*V + M2*V 2100*12 + 2780*0 = 2100V + 2780V 25,200 = 4880V V = 5.16 m/s. See Related Questions: 10-30-14, 9:40 PM.
*January 25, 2015*

**Physics**

When they are acting in the same direction: 5 + 7 = 12 N.
*January 25, 2015*

**Physics**

V = 22.6km/18min * 1min/60s = 0.0209 km/s = 20.9 m/s.
*January 25, 2015*

**College Physics**

d = V*t = 0.12 m t = 0.12/V = 0.12/7*10^7=1.71*10^-5 s.
*January 24, 2015*

**Physics**

Dc = 19t = Distance of the car after t1. Dm = Vo*t + 0.5a*t^2 = Distance of motorcycle after t1. Dm = Dc + 52 Vo*t + 0.5a*t^2 = 19t + 52 19*t + 2.5*t^2 = 19t + 52 19t - 19t + 2.5*t^2 = 52 t^2 = 20.8 t = 4.56 s. So the motorcycle caught-up 4.56 s after it started to accelerate.
*January 24, 2015*

**Physics**

V1/V2 = M2/M1 = 56kg/96kg = 0.583
*January 24, 2015*

**science**

a. Fx = 3000*Cos30 = 2598 N. b. Fy = 3000*sin30 = 1500 N. = Lifting force.
*January 24, 2015*

**Physics**

See previous post: Fri, 1-23-15, 5:42 PM.
*January 24, 2015*

**Physics**

V^2 = Vo^2 + 2a*d = 0 Vo^2 = -2a*d = -2*(-7.2)*72 = 1036.8 Vo = 32.2 m/s. = Speed before braking.
*January 23, 2015*

**math**

Let g = 10 m/s^2. d = 0.5g*t^2 = 5*9^2 = 405 m.
*January 23, 2015*

**physics**

Vo = 50m/s @ A Degrees Xo = 50*Cos A Yo = 50*sin A Y^2 = Yo^2 + 2g*h = 0 @ max ht. Yo^2 = -2g*h = -2*(-9.8)*100 = 1960 Yo = 44.27 m/s = Ver. component of initial velocity. Yo = 50*sin A = 44.27 sin A = 44.27/50 = 0.88544 A = 62.3o. = Angle of projection. Y = Yo + g*Tr = 0 Tr...
*January 23, 2015*

**science**

d1 + d2 = 12 m. d1 = Vo*t + 0.5g*t^2 d2 = 0.5g*t^2 16*t - 4.9*t^2 + 4.9*t^2 = 12 16*t = 12 t = 0.75 s.
*January 22, 2015*

**science**

a = (V-Vo)/t = 30-0)/10 = 3 m/s^2
*January 22, 2015*

**Physics**

P = F * d/t = M*g * d/t = 5*9.8 * 12/15 = 39.2 J/s = 39.2 Watts.
*January 22, 2015*

**physics**

Vo[51o] Xo = 7 m/s Vo = ? Xo = Vo*Cos51 = 7 m/s Vo = 7/Cos51 = 11.12 m/s = Initial velocity. Yo = Vo*sin51 = 11.12*sin51 = 8.64 m/s.
*January 22, 2015*

**physics**

Increase = (26-(-16))C*1.7*10^-5/C*30cm = 42*1.7*10^-5 * 30cm =0.02142 cm.
*January 22, 2015*

**math**

1. (7a^5-a^3) - (a^5-5a^3-1) Remove the parenthesis: 7a^5-a^3 - a^5+5a^3+1 Combine like-terms: 6a^5 + 4a^3 + 1
*January 22, 2015*

**Math**

(3x+4)*Log2 4 = (3x+4)*2 = 6x + 8.
*January 22, 2015*

**math**

h(t) = -4t^2 + 30t + 1.4 Tr = -B/2A = -30/-8 = 3.75 s. = Rise time. Tf = Tr = 3.75 s. = Fall time. Tr+Tf = 3.75 + 3.75 = 7.5 s. To get under the ball.
*January 22, 2015*

**Math**

X Girls. (x+45) Boys. 2x/3 girls passed the test. (4/5)(x+5) = (4x/5 + 4) boys passed test 49 Children failed the test. (x - 2x/3) = x/3 = Girls that failed the test. (X+45) - (4x/5+4) = x/5+41 Boys that failed the test. X/3 + (x/5+41) = 49 x/3 + x/5 = 49-41 = 8 Multiply both ...
*January 22, 2015*

**physics**

Incomplete.
*January 22, 2015*

**phyisics**

h = 1*sin30 = = 0.5 m. V1^2 = Vo^2 + 2g*h = 0 + 19.6*0.5 = 9.8 V1 = 3.13 m/s. M1*g = 25N. M1*9.8 = 25 M1 = 2.55 kg M2*9.8 = 40N. M2 = 4.08 kg M1*V1 + M2*V2 = M1*V + M2*V 2.55*3.13 + 4.08*0 = 2.55*V + 4.08*V 7.982 = 6.63V V = 1.20 m/s.
*January 22, 2015*

**Math**

1b.
*January 21, 2015*

**Physics**

X = 64.9 Y = ? Z = 87.6 = magnitude or hypotenuse. X^2 + Y^2 = Z^2 Solve for Y.
*January 21, 2015*

**Physics**

2.75*10^19electrons/30s = 9.167*10^17 electrons/s I = (9.167*10^17/62.42*10^17) * 1Amp = 0.147 Amps.
*January 21, 2015*

**physics**

Vo = 15.4m/s[34.7o] Xo = 15.4*Cos34.7 = 12.7 m/s. Yo = 15.4*sin34.7 = 8.77 m/s. Y = Yo + g*Tr = 0 @ max.ht. 8.77 - 9.8*Tr = 0 -9.8Tr = -8.77 Tr = 0.895 s. = Rise time. Tf = Tr = 0.895 s. = Fall time. Tr+Tf = 0.895 + 0.895 = 1.79 s. to hit gnd.
*January 21, 2015*

**physics**

h = Vo*t + 0.5g*t^2 Vo = 0 t = 2.88 s. g = 9.8 m/s^2 Solve for h.
*January 21, 2015*

**physics**

M2V2 = M1V1 0.00255*V2 = 10.7*4.26 Solve for V2.
*January 21, 2015*

**Math proportions**

(20min/150people) * 240people = 32 Min.
*January 21, 2015*

**Algebra 2**

g(x) = x^2 - 7x - 9 g(-1) = (-1)^2 - 7*(-1) - 9 = -1 F(g(-1)) = F(-1) F(x) = x - 2 F(-1) = -1 - 2 = -3.
*January 21, 2015*

**algebra**

m2 = m1 = -A/B = -1/1 = -1 P(0,0), m2 = -1 Y = mx + b = 0 -1*0 + b = 0 b = 0 Y = -1x + 0 Eq2: Y = -x
*January 21, 2015*

**7th grade math**

Both answers are correct!
*January 21, 2015*

**Algebra**

2(x+3) = 2x+3 2x+6 = 2x+3 2x-2x = 3-6 0 = -3? No
*January 21, 2015*

**algebra**

10in./30in. =
*January 21, 2015*

**Math (Greatest common factor)**

12 = (1*12), (2*6), (3*4). 18 = (1*18), (2*9), (3*6). 24 = (1*24), (2*12), (3*8), (4*6). CF = 1,2,3,6. GCF = 6. So we factor out 6 and the variables with the lowest exponents: 6m^2n^3(2m^3 + 3m - 4) I, normally, find the GCF by knowing that the GCF =< 12 (the smallest ...
*January 21, 2015*

**Math**

1. 6.4in * 2.54cm/in = 2. 4qts * 32oz./qt * 0.03L/oz. = 3. 10kg * 1Lb/0.454kg =
*January 21, 2015*

**physics**

A train whistle is heard at 280Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 270 Hz. What is the speed of the train before slowing down? What is the speed of the train after slowing down?
*January 21, 2015*

**physics**

M1*V1 + M2*V2 = M1*V3 + M2*V3 V3 = V4. 60*0 + 15*12 = 60*V3 + 15*V3 180 = 75V3 V3 = 2.4 m/s = Velocity after impact.
*January 21, 2015*

**physics**

a. Y = Yo + g*t = 0 + 9.8*1 = 9.8 m/s = Ver Component. X-component = 0 b. D = 0.5*g*t^2. Solve for D. Y = D, X = 0.
*January 21, 2015*

**physic**

a. V = a*t = 0.5m/s^2 * 4s. = 2 m/s. b. d = 0.5a*t^2 = 0.5*0.5*4^2 = 4 m.
*January 21, 2015*

**Physics**

T = 12s./10rev = 1.2 s/rev.=1.2 s/cycle
*January 20, 2015*

**Math**

sqrt(450) = sqrt(2*225) = 15*sqrt(2).
*January 20, 2015*

**ascience**

X = 20 mi/h Y = 10 mi/h Tan A = Y/X = 10/20 = 0.50 A = 26.56o V = X/Cos A = 20/Cos26.56 = 22.4mi/h[26.56o]
*January 20, 2015*

**Physics**

Do you mean (5,7)? If so, (x,y) = 5,7).
*January 20, 2015*

**physics**

V = Vo + g*t V = 0 Vo = Initial velocity. g = -9.8 m/s^2 t = 2.2 s. Solve for Vo.
*January 20, 2015*

**Math**

Yes.
*January 20, 2015*

**math**

Bearing of Q = 90-50 = 40o East of North
*January 20, 2015*

**Math**

8/25 + 1/5 + x = 4/5 X = 4/5 - 8/25 - 1/5=20/25 - 8/25 - 5/25 = 20/25 - 13/25 = 7/25 of the grid for Nate.
*January 19, 2015*

**physics**

Dx = 1710*Cos5.08 =
*January 19, 2015*

**physics**

196m[-18o] X = 196*Cos(-18) = Y = 196*sin(-18) =
*January 19, 2015*

**grade 7 math**

1. -450 km/h(South). 2. 675 km/h(North). 675T - (-450T) = 3600 km. Solve for T.
*January 19, 2015*

**grade 7 math**

A = -80 km/h(West). B = 110 km/h(East). 110T - (-80T) = 760 km 110T + 80T = 760 190T = 760 T = 4 h.
*January 19, 2015*

**grade 7 math**

d1 + d2 = 210 km 60*2 + r2*2 = 210 Solve for r2.
*January 19, 2015*

**grade 7 math**

9:15 - 8:00 = 1h and 15 min. = 1.25h head-start. d = r1*t = 45km/h * 1.25h = 56.25 km Head-start. r2*t = r1*t + 56.25 t = 9:15AM to 1PM = 3h and 45min = 3.75h r2 * 3.75 = 45*3.75 + 56.25 r2*3.75 = 225 r2 = 60km/h = Michael's speed. r1*t + 56.25 = 45*3.75 + 56.25 = 225 km...
*January 19, 2015*

**grade 7 math**

d1 = r1*t1 = 60km/h * 0.5h = 30 km after 0.5 h. r2*t = r1*t + 30 65 * t = 60*t + 30 65t - 60t = 30 5t = 30 t = 6 hours.
*January 19, 2015*

**Physics/Math**

a. Eo/Ei = 1.0 100/Ei = 1 Solve for Ei. b. Eo/Ei = 0.30 Eo/100 = 0.30 Solve for Eo. c. Same procedure as a and b.
*January 19, 2015*