Wednesday

August 31, 2016
Total # Posts: 11,959

**Physics**

When t = 0, X = 0-0 +6 = 6 m. When t = 3 s., X = 3^2 - 4*3 + 6 = 3 m. d = 6-3 = 3 m.
*April 2, 2015*

**physics**

See previous post: Wed, 4-1-15, 6:39 AM.
*April 2, 2015*

**Physics**

If they were wired in series, All lights will go out when one bulb burns out. Therefore, they are connected in parallel.
*April 2, 2015*

**Physics 101**

Correction: a = (Fp-Fk)/M = (122.6-63.65)/25 = 2.36 m/s^2
*April 2, 2015*

**Physics 101**

M*g = 25 * 9.8 = 245 N. = Wt. of the box Fp = 245*sin30 = 122.5 N. = Force parallel to the incline. Fn = 245*Cos30 = 212.2 N. = Force perpendicular to the incline. Fk = u*Fn = 0.30 * 212.2 N. = 63.65 N. = Force of kinetic friction. a = (Fp-Fk)/M*sin30 = (122.5-63.65)/12.5 = 4....
*April 2, 2015*

**Finances**

a. P = Po(1+r)^n Po = $600 r = 1.5%/100 = 0.015 = Monthly % rate expressed as a decimal. n = 12comp./yr. * 2yrs. = 24 Compounding periods. Solve for P. Monthly Payment = P/24 a. Simple Interest: P = Po + Po*r*t = 600 + 600*0.015*24 Solve for P. Monthly payment = P/24 b. Same ...
*April 1, 2015*

**Physic**

V = Vo + a*t V = 6.5 m/s Vo = 0 t = 22 s. Solve for a. d = 0.5a*t^2 t = 22 s. Solve for d
*April 1, 2015*

**Physic**

V = Vo + a*t v = 22 m/s Vo = 5 m/s t = 18 s. Solve for a.
*April 1, 2015*

**Physic**

M*g = 637 N. g = 9.8 m/s^2 M = Mass in kg Solve for M.
*April 1, 2015*

**Geometry**

X = 3-2.5 = 0.5 m. Y = 1.5 m. Tan A = Y/X = 1.5/0.5 = 3.0 A = 71.57o Tan 71.57 = h/3 Solve for h.
*April 1, 2015*

**math**

See previous post: Tue, 3-31-15, 5:18 PM.
*April 1, 2015*

**physics**

M*g = 150 N = Wt. of the boy = Normal force(Fn). Fk = u*Fn = 0.05 * 150 = 7.5 N. = Force of kinetic friction. Fap-Fk = M*a Fap-7.5 = M*0 = 0 Fap = 7.5 N. = Force applied.
*April 1, 2015*

**algebra**

V = Vo + g*t = 128 - 32*2 = 64 Ft/s after 2 s. V = 128 - 32*4 = 0 After 4 s. Vavg = (0+4)/2 = 2 Ft/s
*April 1, 2015*

**precalculus**

U(3,1), V(6,-1) A. Tan A = (-1-1)/(6-3) = -2/3=-0.66666 A = -33.7o = 33.7 S. of E.
*March 31, 2015*

**business math**

P = Po(1+r)^n Po = $8,500 r = (6/4)/100 = 0.015 = Quarterly % rate expressed as a decimal. n = 4comp./yr. * 12yrs. = 48 Compounding periods. Solve for P. I = P-Po
*March 31, 2015*

**Physics**

Incomplete.
*March 31, 2015*

**Physics**

R = 120 Ohms. C = 40 uF = 40*10^-6 Farads. RC = 120 * 40*10^-6 = 4.8*10^-3 seconds t/RC = t/4.8*10^-3 = 208.3t 1/e^208.3t = 0.25 e^208.3t = 4 208.3t*Ln e = Ln 4 208.3t = 1.386 t = 0.00666 s. = 6.66 Milliseconds
*March 30, 2015*

**physics**

M1*V1 + M2*V2 = M1*V + M2*V 0.205*7.8 + 0.029*15 = 0.205*V + 0.029*V Solve for V.
*March 30, 2015*

**Physical Science**

Mass = 1000sacks * 20kg/sack = 20,000kg d = 1000sacks * 1m/sack = 1000 m. Power = F*d/t = Mg * d/t = 20000*9.8 * 1000/1800s = 108,889 J/s = 108,889 Watts
*March 30, 2015*

**math**

1. Po = 0.8 * 1687.00 = $1349.60 = Amt. financed. P = Po(1+r)^30 r = 16/12/100 = 0.0133 = Monthly % rate expressed as a decimal. P = 2008.05 I = P-Po = 2008.05 - 1349.60 = $658.45 2. Round and then estimate. ESTIMATE = 458 + 43 = 501.
*March 29, 2015*

**math**

1. Correct. 2. I = Po*r*t = 1100*(0.08/12 )*1 = $7.33 3. State Tax = 0.015 * 515 = $7.73 Medicare Tax = 0.0145 * 515 = $7.47 4. P = Po + Po*r*t = 619 + 619*(0.13/12)*1 = 625.71 5. 830.78-144 = $686.78 The given bal. is not compatible with the loan: 144/mo * 24mo = $3456 3456-...
*March 29, 2015*

**Physics**

44.2 + Dg = 212 m Dg = 212 - 44.2 = 167.8 m = Distance of green car. Vor = 23km/h = 23,000m/3600s = 6.39 m/s. Vr = 46km/h = 46,000m/3600s = 12.78 m/s. Vor*T1 = 44.2 6.39T1 = 44.2 T1 = 6.92 s. = Time to pass. Vr*T2 = 76.3 12.78*T2 = 76.3 T2 = 5.97 s. = Time to pass. a. Vog*T1...
*March 29, 2015*

**physics**

1. P = Mg * Vavg = 85*9.8 * 8/2 = 3332 J/s = 3332 Watts 2a. Momentum = M1*V1 + M2*V2 = 1000*30 + 3400*20i = 30,000 - 68,000i,Q4 = sqrt(30,000^2+68000^2) = 74,324 kg-m/s 2b. Total final momentum = Total initial momentum = 74,324 kg-m/s. 2c. M1*V1 + M2*V2 = M1*V + M2*V 30,000 - ...
*March 29, 2015*

**math**

X Qtrs (152-x) Dimes 25x + 10(152-x) = 2705 Cents 25x + 1520-10x = 2705 15x = 2705 - 1520 = 1185 X = 79
*March 28, 2015*

**physics**

The max range occurs at 45o. Vo = 25.8m/s[45o] Xo = 25.8*Cos45 = 18.2 m/s Yo = 25.8*sin45 = 18.2 m/s a. Y = Yo + g*Tr = 0 Tr = -Yo/g = -18.2/-9.8 = 1.86 s = Rise time. Tf = Tr = 1.86 s = Fall time. Tr+Tf = 1.86+1.86 = 3.72 s. = Time in air. b. Dx = Xo*(Tr+Tf)=18.2 * 3.72 = 67....
*March 28, 2015*

**Maths**

I = Po*r*t Po = $500 r = 0.06 = APR expressed as a decimal. t = 10 years. Solve for I.
*March 28, 2015*

**Physics**

a. M*g = 56 * 9.8 = 548.8 N. = Wt. of refrigerator = Normalforce(Fn). Fs = u*Fn = 0.58 * 549 = 318.3 N. = Force of static friction. b. Fap-Fs = M*a Fap - 318.3 = M*0 = 0 Fap = 318.3 = Force applied.
*March 27, 2015*

**PHYSICS**

a. Vsc = Vs + Vc = 1.24i + 0.553, Q1. Tan A = Y/X = 1.24/0.553 = 2.240 A = 66o CCW. V = Y/sin A = 1.24/sin66 = 1.36m/s d1 = 2.24km/sin66 = 2.45 km = 2450 m. to cross d1 = V*t t = d1/V = 2450/1.36 = 1802 s. = 30 Min. to cross. b. d2 = 2450*Cos66 = 997 m. Downstream.
*March 27, 2015*

**math**

4x - 3y = 8
*March 26, 2015*

**PHYSICS**

Correction: h = 0.5g*t^2 = 2.7 4.9t^2 = 2.7 t^2 = 0.551 Tf1 = 0.74 s = Fall time to board. Tr = Tf1 = 0.74 s = Rise time. h = 0.5g*t^2 = 3 + 2.7 = 5.7 4.9t^2 = 5.7 t^2 = 1.16 Tf2 = 1.08 s = Fall time from max ht. to the water. Dx = Xo*(Tr+Tf2) = 2.9 m. Xo = 2.9/(0.74+1.08) = 1...
*March 26, 2015*

**PHYSICS**

h = 0.5a*t^2 = 3+2.7 = 5.7 m. 4.9t^2 = 5.7 t^2 = 1.16 Tf = 1.08 s. = Fall time. Dx = Xo*Tf = 2.9 m Xo * 1.08 = 2.9 Xo = 2.69 m/s = Hor. component of initial velocity. Y^2 = Yo^2 + 2g*h = 0 Yo^2 = -2g*h = -2*(-9.8)*2.7 = 52.92 Yo = 7.27 m/s = Ver. component of initial velocity...
*March 26, 2015*

**Physics**

r1 = 8.8 m/s = Your speed. r2 = 0.8 * 8.8 = 7.04 m/s = Your bro. speed. 8.8t = 110 t = 12.5 s 7.04t = 110-d 7.04*12.5 = 110-d d = 110-88 = 22 m. Head start.
*March 26, 2015*

**algebra**

OOPS!! Please disregard the above calculations and answer.
*March 25, 2015*

**algebra**

(x-5)/(x-9) = (x-5)(x+5)/(x-9)(x+5) = (x-5)(x+5))/(x^2+5x-9x-45) = (x-5)(x+5))/(x^2-4x-45) = (x-5)(x+5)/(x+5)(x-9) = (x-5)/(x-9)
*March 25, 2015*

**math**

Do you mean 240/80 = 3? If so, your answer is correct.
*March 25, 2015*

**physic**

Since their is no acceleration, the force of friction(Fk) is = the hor. force applied: Fap - Fk = M*a Fap - Fk = M*0 = 0 Fk = Fap = 200 N.
*March 25, 2015*

**physics**

Dw/1000 = 0.78 Dw = 780kg/m^3 = Density of wood.
*March 25, 2015*

**physics**

Db/Dw = 400/1000 = 2/5
*March 25, 2015*

**physics**

Vo = 29.2m/s[31.7o] Xo = 29.2*Cos31.7 = 24.84 m/s Yo = 29.2*sin31.7 = 15.34 m/s Y = Yo + g*Tr = o Tr = -Yo/g = -15.34/-9.8 = 1.57 s. = Rise time. Dx = Xo*Tr = 24.84m/s * 1.57s = 38.9 m.
*March 25, 2015*

**physics**

M*g = 0.53 * 9.8 = 5.2 N. = Wt. of the mass. K = 1/0.039 * 5.2N = 133.2N/m
*March 25, 2015*

**physics**

Dx = Vx * Tf = 20.6 m. 27.7 * Tf = 20.6 Tf = 0.744 = Fall time. h = 0.5g*Tf^2 g = +9.8 m/s^2 Solve for h.
*March 25, 2015*

**Math**

Correct.
*March 24, 2015*

**Math**

1. P = Po(1+r)^n Po = $5,000 r = (6%/4)/100 = 0.015 = Quarterly % rate expressed as a decimal. n = 4comp./yr. * 2yrs. = 8 Compounding periods. Solve for P.
*March 24, 2015*

**Physics**

V = a*t
*March 24, 2015*

**Physics**

Vpw = Vp + Vw = -245 m/s Vp + 38i = -245 Vp = -245-38i(Q3). Tan A = Y/X = -38/-245 = 0.15510 A = 8.82o S. of W. = 188.82o CCW.
*March 24, 2015*

**Math**

1. I = P - Po = 516.50*6 - 3000 = $99. I = 99/0.5yr = $198/yr. APR = (198/3000) * 100 = 6.6%
*March 24, 2015*

**Math**

1. % Increase = (6.58-4.80)*100/4.80 = 37.083 2. 1125 + 0.262*1125 = $1419.75 1419.75 - 1125 = $294.75 Saved when paid annually.
*March 24, 2015*

**physics**

E = emf = 24V. R1 = 5 Ohms R2 = 9 Ohms R3 = 12 Ohms a. 1/Req = 1/R1 + 1/R2 + 1/R3 1/Req = 1/5 + 1/9 + 1/12 = 0.3944444 Req = 2.535 Ohms b. It = I1+I2+I3 = E/Req = 24/2.535 = 9.47A I2 = 12/9 * I3 = 1.33I3 I1 = 12/5 * I3 = 2.4I3 I3 + 1.33I3 + 2.4I3 = 9.47 4.733I3 = 9.47 I3 = 2....
*March 23, 2015*

**Math**

A. 2x + 3y = 7 -x + 2y = -7 Multiply Eq2 by 2 and add the Eqs. 2x + 3y = 7 -2x + 4y = -14 sum = 7y = -7 Y = -1 In Eq1, replace y with -1 and solve for x: 2x + 3*(-1) = 7 2x = 10 X = 5 P(5,-1) B. P(5,-1), Q(-9,2). D^2=(-9-5)^2 + (2-(-1))^2=196 + 9 = 205 D = 14.3 Y = mx + b m...
*March 23, 2015*

**physics**

a1 = 10,000/500 = 20 m/s^2 a2 = 14,000/780 = 17.95 m/s^2 500 kg boat moves faster.
*March 23, 2015*

**physics**

M*g = 20 * 9.8 = 196 N. = Wt. of box. Fp = 196*sin3 = 10.26 N. = Force parallel to the incline. Fn = 196*Cos3 = 195.7 N. = Force perpendicular to the incline. Fk = u*Fn = 0.1 * 195.7 = 19.57 N. = Force of kinetic friction. a = (Fap-Fp-Fk)/M = (50-10.26-19.57)/20 =1.01 m/s^2 V^...
*March 23, 2015*

**Physics**

Fx - Fk = M*a Fex*Cos24 - 25 = M*0 = 0 Fex*Cos24 = 25 Fex = 27.4 N. = Force exerted Work = Fx*d = 27.4*Cos24 * 18 = 450.6 Joules
*March 22, 2015*

**physics**

a = Fap-Fk)/M = (150-95)/46 = 1.20 m/s^2
*March 22, 2015*

**pre algebra**

2.Add all like-terms: 2x^2-6x^2 = -4x^2 -4x+3x = -x 8 + 7 = 15 Results: -4x^2 - x + 15 3. Add all like-terms:
*March 22, 2015*

**Vector**

a. Tan A = Y/X = 10/26 = 0.38462 A = 21.04o E of N. Tan A = d/150 d = 150*Tan 21.04 = 57.7 m = Distance downstream. b. To offset the affect of the current, the swimmer must head 21.04o W. of N. Vsc = Vs + Vc = 26i Vs + 10 = 26i Vs = -10 + 26i Tan A = 26/-10 = -2.600 X = -68.96...
*March 22, 2015*

**physics**

M1*V1 + M2*V2 = M1*V3 + M2*V4 0.4*0 + 0.50*V2=0.40*1.4[37o] + 0.5*1.86 0 + 0.50V2 = 0.447 + 0.337i + 0.93 0.5V2 = 1.377 + 0.337i V2 = 2.754 + 0.674i V2 = sqrt(2.754^2+0.674^2) m/s
*March 21, 2015*

**sicence**

V^2 = Vo^2 + 2g*h Vo = 0 g = +9.8 m/s^2 h = 4 m. Solve for V.
*March 21, 2015*

**physics**

Va = 120revs/10s * 6.28rad/rev = 75.36 rad/s. = Angular velocity. a = (V-Vo)/t = (75.36-0)/10=7.54 rad/s^2
*March 21, 2015*

**physics**

V = Vo + a*t V = 40 rad/s Vo = 72 rad/s a = -4 rad/s^2 Solve for t
*March 21, 2015*

**physics**

a = (Fe-Fa)/M = (31.5*10^6-0.18*10^6)/2.764*10^6 = 31.32*10^6/2.764*10^6 = 11.33 m/s^2
*March 21, 2015*

**Physics**

Incomplete.
*March 21, 2015*

**maths**

V = As*h = 1760 * h = 12,320 cm^3 Solve for h.
*March 21, 2015*

**physics**

2*10^30kg * 1proton/2*10^-27kg = 1*10^57 Protons
*March 21, 2015*

**Medical physics**

Silver: V = L*W*h = 3*3*3 = 3^3 = 27 cm^3 Wt. = V*D = 27cm^3 * 10g/cm^3=270 grams Aluminum: V = 4*4*4 = 64 cm^3 Wt. = 64cm^3 * 2.96g/cm^3 = 189.4 grams
*March 21, 2015*

**math _ HELP !!**

(-3,5), (-1,4) Y = mx + b m = (y2-y1)/(x2-x1) = (4-5)/(-1-(-3)) = -1/2 = The slope. Y = (-1/2)(-3) + b = 5 3/2 + b = 5 b = 5 - 3/2 = 10/2 - 3/2 = 7/2 Eq: Y = (-1/2)x + 7/2
*March 21, 2015*

**physics**

Ts = (768kV/1kV) * 200 = 153,600 Turns.
*March 20, 2015*

**physics**

See previous post: Thu, 3-19-15, 12:01 AM.
*March 19, 2015*

**physics**

Vo = 10m/s[30o] Xo = 10*Cos30 = 8.66 m/s Yo = 10*sin30 = 5 m/s 1. Y = Yo * g*Tr = 0 Tr = -Yo/g = -5/-9.8 = 0.51 s. = Rise time. h = ho + -(Yo^2)/2g = 19 + -(5^2)/-19.6 = 20.3 m. Above gnd. h = 0.5g*t^2 = 20.3 4.9t^2 = 20.3 t^2 = 4.14 Tf = 2.03 s. = Fall time. Dx = Xo*(Tr+Tf...
*March 19, 2015*

**physics**

Vo = 20m/s[40o] Above the hor.? Xo = 20*Cos40 = 15.32 m/s. Yo = 20*sin40 = 12.86 m/s. 1. Y^2 = Yo^2 + 2g*h = 0 h = 35 + -(Yo^2)/2g h = 35 + -(12.86^2)/-19.6 = 43.44 m Above the green. Y = Yo + g*Tr = 0 Tr = -Yo/g = -12.86/-9.8 = 1.31 s. = Rise time. h = 0.5g*t^2 = 43.44 4.9t^2...
*March 19, 2015*

**physics**

V*D = 400cm^3 * 0.998g/cm^3 = 399.2 Grams = 0.399 kg F = M*g = 0.399 * 9.8 = 3.91 N. = Upthrust
*March 19, 2015*

**physics**

Correct.
*March 19, 2015*

**physics help**

1. D1 = Vx * Tf h = 0.5g*Tf^2 Tf^2 = h/0.5g Tf = sqrt (h/0.5g) = Fall time of fired bullet. h = 0.5g*Tf^2 Tf^2 = h/0.5g Tf = sqrt (h/0.5g) = Fall time of the dropped bullet. Therefore, the fall times are equal. 2. Directly at it.
*March 19, 2015*

**physics**

Vo = 54m/s[42o] Xo = 54*Cos42 = 40.1 m/s. Yo = 54*sin42 = 36.1 m/s. a. Y = Yo + g*Tr = 0 Tr = -Yo/g = -36.1/-9.8 = 3.69 s. = Rise time. Y^2 = Yo^2 + 2g*h = 0 h = -(Yo^2)/2g = (36.1^2)/-19.6=66.5 m. 0.5g*t^2 = 66.5-14 = 52.5 4.9t^2 = 52.5 t^2 = 10.71 t = 3.27 s. = Fall time(Tf...
*March 19, 2015*

**maths**

364
*March 18, 2015*

**maths**

357
*March 18, 2015*

**algebra**

An amusement park charges $8 admission and average of 2000 visitors per day. A survey shows that , for each $1 increase in the admission cost, 100 fewer people would visit the park. Write an equation to express the revenue, R(x) dollars, in terms of a price increase of x ...
*March 18, 2015*

**algebra**

May i ask how you found it?
*March 18, 2015*

**algebra**

A rectangular lot is bordered on one side by a stream and on the other three sides by 600m of fencing. Find the dimensions of the lot if area is a maximum.
*March 18, 2015*

**algebra**

9.6 = 0.128^2/0.11x^3 Multiply both sides by x^3: 9.6x^3 = 0.128^2/0.11 = 0.14895 Divide both sides by 9.6: x^3 = 0.01552 Take cube root of both sides by raising to the 1/3 power: (x^3)^1/3 = 0.01552^(1/3) X = 0.24944
*March 18, 2015*

**physics**

d = V1*T1 + V2*T2 = 118*1.05 + 40*1.15 = 169.9 km. Vavg = d/(T1+T2) = 169.9/2.20 = 77.2 km/h.
*March 18, 2015*

**Physics**

F = M*a = 1100 * 4.6 = 5060 N. d = 0.5a*t^2 = 0.5*4.6*5^2 = 57.5 m. P = F*d/t = 5060 * 57.5/5 = 58,190 J./s = 58,190 W.
*March 18, 2015*

**physics**

d1 + d2 = 18 m. 0.5g*t^2 + Vo*t + 0.5g*t^2 = 18 4.9t^2 + Vo*t - 4.9t^2 = 18 Vo*t = 18 20t = 18 t = 0.9 s. d1 = 0.5g*t^2 = 4.9 * 0.9^2 = 3.97 m.
*March 17, 2015*

**math - slope-intercept**

The slope of the 3 perpendicular lines must be the negative RECIPRECAL of the first line. 1. Y = (2/3)x + 4 2. Y = (-3/2)x + 4 3. Y = (-3/2)x + 5 4. Y = (-3/2)x + 6
*March 17, 2015*

**math**

Ta = 8:00AM + 1 1/2h = 9:30 AM. = Time arrived. d = r*T = 20 km T = 20/r = 20/80 = 1/4 h. = 15 min. Travel time. Ta = 8:30AM + 15 min = 8:45 AM. = Time arrived. 9:30AM - 8:45AM = 8:90 - 8:45 = :45 Min = Difference in arrival time.
*March 17, 2015*

**Maths**

90T = 70(T+1) 90T = 70T + 70 20T = 70 T = 3.5 h.
*March 17, 2015*

**Calculus**

Compute dy/dx using the chain rule. y = (u/4) + (4/u) u = (x - x^7) dy/dx =
*March 17, 2015*

**physics**

KE = 0.5*M*Vo^2 = 1400 J. Vo^2 = 1400/0.5M = 1400/2 = 700 Vo = 26.5 m/s. = Initial velocity. KE = 0.5*M*Xo^2 = 120 J. Xo^2 = 120/0.5M = 120/2 = 60 Xo = 7.75 m/s=Hor. component of initial velocity. Xo = Vo*Cos A = 7.75 m/s. 26.5*Cos A = 7.75 Cos A = 0.29230 A = 73o Yo = Vo*sin ...
*March 17, 2015*

**physics**

a. V = 5.23m/1.52s. = 3.44 m/s. b. L = V*T = 3.44m/s * 0.84s = 2.89 m.
*March 17, 2015*

**Physics**

Vo = 40m/s[30o] Xo = 40*cos30 = 34.64 m/s. Yo = 40*sin30 = 20 m/s. Y = Yo + g*Tr = 0 Tr = -20/-9.8 = 2.04 s. = Rise time. Y^2 = Yo^2 + 2g*h = 0 h = -(Yo^2)/2g = -(20^2)/-19.6=20.4 m. h = 0.5g*t^2 = 20.4 4.9t^2 = 20.4 t^2 = 4.16 t = 2.04 s. = Fall time(Tf). Range = Xo * (Tr+Tf...
*March 17, 2015*

**physics**

Xo = 20 m/s Yo = 30 m/s Vo = sqrt(20^2+30^2) = 36.1 m/s Y = Yo + g*t = 30 + (-9.8)*2 = 10.4 m/s. = Ver. component of velocity 2 s later. X = Xo = 20 m/s and remains constant. V^2 = X^2 + Y^2 = 20^2 + 10.4^2 = 508.16 V = 22.5 m/s. = Total velocity.
*March 16, 2015*

**physics**

V = Vo + a*t V = 0 Vo = 29.3 m/s t = 4.9 s. Solve for a. It will be negative. F = M*a Solve for F. It will be negative.
*March 16, 2015*

**sciences**

Incomplete.
*March 16, 2015*

**Physics**

V = 100m/9.83s. = 10.17 m/s. KE = 0.5*M*V^2 = 0.5*65*10.17^2=3361 J.
*March 16, 2015*

**Science**

Fap = 24 N. = Force applied. a = 2.28 m/s^2? Fk = 14 N. (Fap-Fk) = M*a M = (Fap-Fk)/a = (24-14)/2.28 = 4.39 kg. = Mass of the box.
*March 16, 2015*

**Science**

The vertical component of velocity is zero.
*March 16, 2015*

**Physics**

V^2 = Vo^2 + 2a*d V = 0 Vo = 12 m/s a = -3 m/s^2 Solve for d.
*March 16, 2015*

**physics**

M*g = 17kg * 9.8N./kg = 166.6 N. = Wt. of box. Fn = 166.6 - 88.9*sin56 = 92.9 N. = Normal force. Fk = u*Fn = 0.24 * 92.9 = 22.3 N. = Force of kinetic friction. a = (F*Cos A-Fk)/M = F = 88.9 N. A = 56o M = 17 kg Solve for a.
*March 15, 2015*

**physics**

M1*V1 + M2*V2 = M1*V + M2*V 450*30i + 560*(-35) = 450V + 560V 13,500i - 19,600 = 1,010V b. V = -19.41 + 13.37i V = sqrt = (19.41^2+13.37^2) = 23.6 m/s. a. Momentum = M1*V + M2*V = 450*23.6 + 560*23.6 = 23,836 kg-m/s. c. Tan A = Y/X = 13.71/-19.41=-0.70634 A = -35.2o = 35.2o CW...
*March 15, 2015*

**science**

For a given power level, transmitting at a high voltage reduces the current transmitted which reduces the I^2R loss in the transmission lines. Step-down transformers are used to reduce the voltage to the required level for home use.
*March 15, 2015*

**science**

M = 30m^3 * 1.3kg/m^3 = 39 kg.
*March 15, 2015*