Sunday

August 28, 2016
Total # Posts: 11,954

**Physics**

Note: The Y component of the velocity decreases with time and is zero at the maximum height. The velocity increases as the stone falls. c. Dx = Xo*t = 13.91 * 1.76 = 24.48 m. d. Y = Yo + g*Tr = 0 12.35 - 9.8Tr = 0 9.8Tr = 12.35 Tr = 1.26 s. = Rise time or time to reach max ht...
*April 19, 2015*

**Physics**

Vo = 18.6m/s[41.6o] Xo = 18.6*Cos41.6 = 13.91 m/s Yo = 18.6*sin41.6 = 12.35 m/s a. X = Xo=13.91 m/s and does not change. Dx = Xo*t = 13.91 * 1.06 = 14.74 m. b. Y = Yo + g*t = 12.35 - 9.8*1.06 = 1.962 m/s h = Yo*t + 0.5g*t^2 Yo=12.35 m/s, t = 1.06 s, g=-9.8 m/s^2. Solve for h. ...
*April 19, 2015*

**physics**

M1*V1 + M2*V2 = M1*V + M2*V 11.2i - 18 = 4*V + 5*V = 9V V = -2 + 1.24i = 2.36m/s[31.9o] N. of W. = 2.36m/s[148.1o] CCW Momentum = M1*V + M2*V = 4*2.36[31.9] + 6*2.36[31.9] = 9.44[31.9] + 14.2[31.9o]=23.6m/s[31.9o] N. of W.
*April 19, 2015*

**Math**

I = Po*r*t = 875*(0.1425/12)*8 = $83.13
*April 19, 2015*

**science**

V = 60 mi/h * 1613m/mi * 1h/3600s = 26.9 m/s. V = Vo + a*t V = 26.9 m/s. Vo = 0 a = 2m/s^2 Solve for t.
*April 19, 2015*

**physics**

d = 110 - Vo*t = 110 - 20*0.7 = 96 m.
*April 19, 2015*

**algebra**

Multiply Eq2 by -4 and add Eq1 and Eq2: 2x + 4y + 2z = 144 -16x - 4y - 2z = -480 Sum: -14x = -336 X = 24 Multiply Eq1 by -2; add Eq1 and Eq3: -4x - 8y - 4z = -288 +x + 3y + 4z = 144 Sum: -3x - 5y = -144. X = 24. Solve for y: -3*24 - 5y = -144 -72 - 5y = -144 -5y = -144 + 72...
*April 18, 2015*

**science**

See previous post: Thu, 1-22-15, 12:20 AM.
*April 18, 2015*

**Algebra**

x = -1/3 x + 1/3 = 0 x = 3/7 x - 3/7 = 0 (x+1/3)(x-3/7) = 0 x^2 - 3x/7 + x/3 - 1/7 = 0 Multiply by 21: 21x^2 -9x + 7x - 3 = 0 21x^2 - 2x - 3 = 0
*April 17, 2015*

**Algebra**

Use same procedure as previous problem.
*April 17, 2015*

**Algebra**

x^2 - 2x + 2 = 0 X = (-B +- sqrt(B^2-4AC))/2A X = (2 +- sqrt(4-8))/2 X = (2 +- sqrt(-4))/2 X = (2 +- sqrt(4(-1))/2 X = (2 +- 2sqrt(-1)/2 X = (2 +- 2i)/2 = 1 +- i = 1+i, and 1-i. Check: (1+i)^2 - 2(1+i) + 2 = 1 + 2i - 1 - 2 -2i + 2 = 0
*April 17, 2015*

**Algebra**

(x+2)^2 + (x-7)(x-2) = 16 x^2 + 4x + 4 + x^2 - 2x -7x + 14 = 16 2x^2 - 5x + 18 = 16 2x^2 - 5x + 2 = 0 A*C = 2*2 = 4 = (-1)*(-4). Choose the pair of factors whose sum = B 2x^2 - (x+4x) +2 = 0 (2x^2-x) - (4x-2) = 0 x(2x-1) - 2(2x-1) = 0 (2x-1)(x-2) = 0 2x-1 = 0 2x = 1 X = 1/2 x-...
*April 17, 2015*

**Help with Psychology**

To help Karen overcome her fear of thunderstorms, a behavior therapist asks her to relax and imagine watching light rain with distant thunder. The therapist is using... a) client-centered therapy b) cognitive therapy c) systematic desensitization d) aversive conditioning ** ...
*April 17, 2015*

**Physics**

db = 10*Log W/Wo = 100 10*Log W/10^-12 = 100 Log W/10^-12 = 10 Log W - Log 10^-12 = 10 Log W = 10 + Log 10^-12 = 10-12 = -2 W = 10^-2 = 0.01 Watts.
*April 17, 2015*

**physics**

D = 5 + 10[30o] X = 5 + 10*Cos30 = 13.66 m. Y = 10*sin30 = 5 m. D = sqrt(X^2+Y^2) = sqrt(13.66^2 + 5^2) = 14.5 m.
*April 17, 2015*

**Math**

F = M*a = 0.04 * 2.5 = 0.1 N. a = F/M = 0.1/0.050 = 2.0 m/s^2
*April 16, 2015*

**exponential equations**

See previous post: Tue, 4-7-15, 7:04 AM.
*April 16, 2015*

**Physics**

h = 0.5g*t^2 g = 9.8 m/s^2. h = 0.75 m. Solve for t(Fall time). V^2 = Vo^2 + 2g*h Vo = 0 Solve for V. Dx = Xo*t Xo = 1.50 m/s t = Fall time. Solve for Dx
*April 16, 2015*

**math**

One Solution: Add to verify. -3x + 2y = 4 3x + 2y = 4 No Solution: Multiply Eq1 by (-1) and add. -3x + 2y = 4 -3x + 2y = 6 Infinite Solutions: Multiply Eq1 by 3 and compare. -3x + 2y = 4 -9x + 6y = 12
*April 16, 2015*

**College Physics**

See Related Questions: Thu, 8-26-14, 5:07 PM.
*April 16, 2015*

**math**

= 0.36603
*April 16, 2015*

**physics**

2nd Cable Direction = 30o E. of N.
*April 15, 2015*

**physics**

a = Fx/M = 15*Cos15/M
*April 15, 2015*

**physics**

V^2 = Vo^2 + 2a*d V = 0 Vo = 27 m/s d = 578 m. Solve for a. It will be negative. F = M*a
*April 15, 2015*

**physics**

M*g = 741 M = 741/g = 741/9.8 = 75.6 kg M*g = 5320 N. g = 5320/M = 5320/75.6 = 70.4 m/s^2
*April 15, 2015*

**Math**

Correction: You probably want the multiplication done without converting to base 10. I can't show it clearly here But it is long-hand multiplication. 1111 0101 0011110 0111100 1001011= Product = 75 base 10.
*April 15, 2015*

**Math**

1111 * 101 (1*2^3+1*2^2+1*2^1+1*2^0)*(1*2^2+0*2^1+1*2^0) = (8+4+2+1)*(4+0+1) = (15) * (5) = 75
*April 15, 2015*

**Physics**

a = F/M = 8/0.5 = 16 m/s^2 V^2 = Vo + 2a*d Vo = 0 a = 16 m/s^2 d = 4 m. Solve for V(m/s).
*April 15, 2015*

**math**

6*(1.2 * 1.2 * 1.2) = 6(1.2^3) = 10.37 in^3.
*April 15, 2015*

**angles**

ABD = ABC - DBC = 71 - 45 = 26
*April 15, 2015*

**Physics**

Vo = 7 m/s V = 3 m/s t = 8 s. V = Vo + a*t Solve for a. It should be negative. d = Vo*t + 0.5a*t^2 Solve for d.
*April 15, 2015*

**physics**

Wt. = M*g Newtons.
*April 15, 2015*

**Science**

Momentum = M*V
*April 14, 2015*

**Physics**

Y^2 = Yo^2 + 2g*h = 0 Yo^2 = -2g*h = -2*(-9.8)*10,000 = 196,000 Yo = 443 m/s.=Ver. component of initial velocity. Vo = Yo/sin60 = 443/sin60 = 512 m/s = Initial velocity = Speed at which the cannon is fired.
*April 14, 2015*

**math analysis**

Plane #1: d1 = 140mi/h[38o] * 2h = 280mi/h[38o] Plane #2: d2 = 180mi/h[350o] * 2h = 360mi[350] d = d2-d1 = 360[350o] - 280[38o] X = 360*Cos350 - 280*Cos38 = 134 mi Y = 360*sin350 - 280*sin38 = -235 mi D^2 = X^2 + Y^2 = 134^2 + (-235)^2 = 73,181 D = 270.5 Miles apart.
*April 14, 2015*

**Science**

1. Amplitude and wavelength. 4. Freq. 8. The freq. 9. To transmit signals.
*April 14, 2015*

**Physics**

See previous post: Sat, 4-11-15, 4:23 AM.
*April 14, 2015*

**physics**

Work = F*h = Mg*h = 400*9.8 * 60 = 235,200 J. = Output energy. Input energy = Output/Eff. = 235,200/0.8 = 294,000 J.
*April 14, 2015*

**Allegbra**

Use Quadratic Formula and get: 11.24
*April 13, 2015*

**Science**

For a given power output, the higher the voltage the lower the current. The lower current means less I^2*R loss. P = E*I = 100 * 10 = 1000 W. P = 1000 * 1 = 1000 W. Note: When E is increased to 1000 volts, the current is reduced to 1A.
*April 13, 2015*

**physics**

Fb = 80 kHz = Freq. emitted by the bat. Fh = 78 kHz = Freq. heard. Vs = 343 m/s = Velocity of sound in air. a. The bat is moving away. b. Fh = (Vs-Vr)/(Vs+Vb) * Fb = 78 kHz (343-0)/(343+Vb) * 80 = 78 Solve for Vb, the velocity of the bat.
*April 13, 2015*

**science**

M1*V1 + M2*V2 = M1*V + M2*V 40*2 + 4*0 = 40V + 4V 44V = 80 V = 1.82 m/s.
*April 13, 2015*

**Physics**

Incomplete.
*April 12, 2015*

**Maths help**

See previous post: Sat, 4-11-15, 4:23 AM.
*April 12, 2015*

**Maths help**

See previous post: Sat, 4-11-15, 4:24 AM.
*April 12, 2015*

**CIRCUIT**

E = 3 + 7 +V3 = 12 Solve for V3
*April 12, 2015*

**Physic**

P = 75 + 25 = 100 W., Total. P = 120 * I = 100 I = 100/120 = 0.833A R = E/I = 120/0.83333 = 144 Ohms.
*April 12, 2015*

**Physic**

See previous post: Thu, 4-9-15, 4:55 AM.
*April 12, 2015*

**Physic**

E = 120 V. R1 = 4.8k R2 = ? R1+R2 = 4(R1*R2)/(R1+R2) (R1+R2)^2 = 4(R1*R2) R1^2+2R1*R2+R2^2 = 4R1*R2 23.04+9.6R2+R2^2 = 19.2R2 R2^2 - 9.6R2 + 23.04 = 0 Use Quadratic formula R2 = 4.8k Check: Parallel Connection P1 = E^2/2.4k = 120^2/2.4k = 6,000 mW = 6 W. Series connection P2...
*April 12, 2015*

**Math, Circuits**

R1 = 3,000 Ohms = 3k Ohms R2 = 0.5k R3 = 10k I = 15 mA V1 = I*R1 = 15mA * 3k = 45 V.
*April 12, 2015*

**CIRCUITS**

Rt = R1+R2+R3 = 20k + 20k + 20k = 60k Ohms. = Total resistance. I = E/Rt = 120/60k = 2 mA(milliamps). V1 = V2 = V3 = I*R1 = 2 * 20k = 40 Volts Therefore, the voltage across each resistor is 40 Volts.
*April 12, 2015*

**Science**

M*g = 7 N. M = 7/g = 7/9.8 = 0.71 kg a = (Fap-Ff)/M = 1.4-1)/0.71=0.56 m/s^2 V = Vo + a*t = 0 + 0.56*4 = 2.24 m/s
*April 11, 2015*

**MATHSS**

Eq1: R1 + R2 = 1600 Ohms. Eq2: R2 = 2R1 - 560 Ohms. In Eq1, replace R2 with 2R1-560 R1 + 2R1-560 = 1600 3R1 = 2160 R1 = 720 Ohms In Eq1, replace R1 with 720 Ohms. 720 + R2 = 1600 R2 = 880 Ohms
*April 11, 2015*

**MATHSS**

Theta = 4 Rad. Wo = 3.2? Rad/s a = 0.6rad/s^2 Theta = Wo*t + 0.5a*t^2 = 4 rads. 3.2t + 0.5*0.6t^2 = 4 0.3t^2 + 3.2t - 4 = 0 Use Quadratic Formula. t = 1.13 s (Wo = 3.2 rad/s). t = 0.125 s (Wo = 32 rad/s).
*April 11, 2015*

**Mathematics**

Leslie: 4.5Laps/0.75h = 6 Laps/h Kala: 3.5Laps/0.75h = 4.67 Laps/h 6Laps/h - 4.67Laps/h = 1.33 Laps/h. Therefore, Leslie can run 1.33 more laps in 1 hr. than Kayla.
*April 10, 2015*

**PHYSICS**

Impulse = M*V = 3 * 2 = 6
*April 10, 2015*

**Physics**

Vr = 14 * pi*r^2 * h = 3.14*(0.295/2)^2 * 6.5 = 6.22 m^3 = Volume of the raft. Vr*Dr = 6.22m^3 * 600kg/m^3 = 3732 kg = Mass of the raft. Vw*Dw = 6.22m^3 * 1000kg/m^3 = 6220 kg = Mass of the water displaced. 6220 - 3732 = 2488 kg can be added to the raft. 2488kg * 1person/75kg...
*April 9, 2015*

**Maths**

a. X stamps, Total. 124 stamps, Australia 0.3x stamps, France 0.3x-12 stamps, Japan X = 124 + 0.3x + 0.3x-12 0.4x = 112 X = 280 b. 280 + n = 1.25*280 = 350 n = 70 Stamps from Lisa's aunt. (0.3x+n)/350 = (84+70)/350 = 0.44 = 44% from France.
*April 9, 2015*

**Physics**

a. Fr = 10N[60o]CCW + 15N[0o] X = 10*Cos60 + 15 = 20 N. Y = 10*sin60 = 8.7 N. Q1. Tan A = Y/X = 8.7/20 = 0.433 A = 23.4o N. of E. Fr = X/Cos A = 20/Cos23.4=21.8 N.[23.4o] N. of E. = 23.4o CCW b. Equilibrant: X = -20 Y = -8.7 Q3. Tan A = Y/X = -8.7/-20 = 0.433 A = 23.4o S. of W...
*April 9, 2015*

**Physic**

R1 = 1.4k Ohms R2 = 2.5k Ohms R3 = 3.7k Ohms P1 max = I1^2*R1 = 500 mW I1^2*1.4 = 500 I1^2 = 357.14 I1 = 18.9mA, max = Total current. Req. = R1 + (R2*R3)/(R2+R3) = 1.4 + (2.5*3.7)/(2.5+3.7) = 1.4 + 1.49 = 2.89k Ohms E max = I1*Req = 18.9 * 2.89 = 54.6 V. =
*April 9, 2015*

**math**

b. h = Vo*t + 0.5g*t^2 Vo = 192 Ft/s g = -32Ft/s^2 h = 432 Ft. Solve for t(Use Quadratic formula). c. h max = -(Vo^2)/2g = -(192^2)/-64 = 576 Ft. d. V = Vo + g*Tr = 0 Tr = -Vo/g = -192/-32 = 6 s. = Rise time Tf = Tr = 6 s. = Fall time. Tr+Tf = 6 + 6 = 12 s. to hit gnd.
*April 9, 2015*

**math**

h = Vo*t + 0.5g*t^2 Vo = 192Ft/s t = 10 s. g = -32Ft/s^2 Solve for h.
*April 9, 2015*

**Physics**

M*g = 330 * 9.8 = 3234 N. = Wt. of sled. Fp = 3234*sin6 = 38 N. = Force parallel to the incline. Fn = 3234*Cos6 = 3216 N. = Normal = Force perpendicular to the incline. Fk = u*Fn = 0.135 * 3216 = 434.2 N. = Force of kinetic friction. a = (Fap-Fk)/M = (1780-434)/330 = 4.08 m/s^...
*April 9, 2015*

**Physics**

Vo = 101km/h = 101,000m/3600s = 28.1 m/s = Initial velocity. See Related Questions: Tue, 10-28-14, 2:33 AM.
*April 9, 2015*

**math**

P = Po - Po*r*t = 11,090.41 120,000 - 120,000*r*12 = 11,090.41 -1,440,000r = 11,090.41 - 120,000 r = 0.0756 = 7.56% per annum
*April 8, 2015*

**Physics**

M*g = 59.9 * 9.8 = 587 N. = Wt. of girl and sled. Fp = 587*sin30 + 131 = 424.5 N. = Force parallel to the incline. Fn = 587*Cos30 = 508.4 N. = Normal = Force perpendicular to the incline. Fk = u*Fn = 0.217 * 508.4 = 110.3 N. = Force of kinetic friction. a=(Fp-Fk)/M = (424.5-...
*April 8, 2015*

**DPS PHYCIS**

Fr = 10[30] + 10[60o] X = 10*Cos30 + 10*Cos60 = 13.66 Units Y = 10*sin30 + 10*sin60 = 13.66 Units Fr = sqrt(X^2+Y^2)
*April 8, 2015*

**Physics**

Wavelength=V*P = V*(1/F) = V*(1/3.8) = 0.48 m. V/3.8 = 0.48 V = 1.82 m/s
*April 8, 2015*

**Physics**

1. V = 3400rev/min * 6.28rad/rev * 1min/60s = 356 rad/s V = Vo + a*t V = 356 rad/s Vo = 0 t = 50 s. Solve for a(rad/s^2).
*April 7, 2015*

**physics**

M1*V1 + M2*V2 = M1*V + M2*V 15*5 + 3*V2 = 15*1.5 + 3*1.5 3V2 = 27 - 75 = -48 V2 = -16 m/s = 16 m/s to the left.
*April 7, 2015*

**Physics**

a. V = Vo + a*t V = 51.5 m/s Vo = 41.5 m/s a = 2.65 m/s^2 Solve for t. b. Same procedure and answer as part a.
*April 7, 2015*

**Grambling**

a. D = 4.58 + 6.25i - 8.65 - 1.84i = -4.07 + 4.41i, Q2 D = sqrt(4.07^2+4.41^2) km. b. Tan A = Y/X = 4.41/-4.07 = -1.08354 A = -47.3o = 47.3o N. of W.
*April 7, 2015*

**Grambling**

V^2 = Vo^2 + 2a*d V = 11.4 m/s Vo = 83.8 m/s d = 750 m. Solve for a. It should be negative.
*April 7, 2015*

**Physics**

h = 6.03*sin21.5 = 2.21 m. V^2 = Vo^2 + 2g*h = 4.87^2 + 19.6*2.21 V = Final velocity Vo = 4.87 m/s g = 9.8 m/s^2 Solve for V.
*April 7, 2015*

**Physics**

PE max = Mg*h-Fk*h = 0.5*M*V^2 M = 98.5 kg g = 9.8 m/s^2 h = 4 m V = 1.81 m/s. Solve for Fk. Divide both sides by M, and solve for Fk
*April 7, 2015*

**math help**

Tank #1 = T min. Tank #2 = (T+60) min. T*(T+60)/(T+T+60) = 40 min (T^2+60T)/(2T+60) = 40 T^2+60T = 80T+2400 T^2 - 20T - 2400 = 0 Use Quadratic Formula. T = 60 min. T + 60 = 60 + 60 = 120 min.
*April 7, 2015*

**Please help asap science**

A. Ri = 0.4 + 0.4 = 0.8 Ohms B. I = E/(Ri+Rm+Ra+Rb) = 24/6.95 = 3.45A
*April 7, 2015*

**Physics**

D. P = 12 * 1.4 = 16.8W. = Greatest energy consumption.
*April 6, 2015*

**geometry**

2x+5 + 5x-79 + 122 = 180 7x = 180 + 79 = 259 X = 37 m<YWZ = 5x-79 = 5*37 - 79 = 106o
*April 6, 2015*

**physics**

a. 5000 Watts = 5,000 J/s = 5000 Joules every second.
*April 6, 2015*

**Physics**

a. Circumference = pi * 2r = 3.14 * 0.9 = 2.83 m. Arc = 2.83m/360o * 170o = 1.33 m. b. V = 0.25rad/s * 2.83m/6.28rads = 0.113 m/s. c. If the velocity is constant, the acceleration is zero.
*April 6, 2015*

**physics**

Max ht. occurs at 45o. Vo = 4m/s[45o] Yo = 4*sin45 = 2.83 m/s = Ver. component of initial velocity. Y^2 = Yo^2 + 2g*h Y = 0 Yo = 2.83 m/s. g = 9.8 m/s^2 Solve for h.
*April 6, 2015*

**physics**

See previous post: Mon. 4-6-15, 8:58 AM.
*April 6, 2015*

**physics**

V = Vo + a*t = 30 m/s 0 + a*20 = 30 a = 1.5 m/s^2 d = 0.5a*t^2 = 0.5*1.5*10^2 = 75 m.
*April 6, 2015*

**Physics**

V = g*t V = 20*1 = 20 m/s. V = 20*2 = 40 m/s. V = 20*3 = 60 m/s. Increase = 20 m/s with each second.
*April 5, 2015*

**math**

(1/2)/3 = 1/2 * 1/3 = 1/6 of the Half- pie. 1/2 * 1/6 = 1/12 of the whole pie.
*April 4, 2015*

**physics**

Momentum before the collision=Momentum after the collision: M1*V1 + M2*V2 = M1*V + M2*V 0.075*80 + 13*0 = 0.075*V + 13*V 6 + 0 = 13.075V V = 0.459 m/s
*April 4, 2015*

**physics**

M!*V1 + M2*V2 = M1*V + M2*V 0.075*80 + 13*0 = 0.075*V + 13*V 6 + 0 = 13.075V
*April 4, 2015*

**Physics**

M*g = 13 N. = Wt. of puck. M = 13/g = 13/9.8 = 1.33 kg. Fn = 13*cos37 = 10.4 N. Fk = u*Fn = 0.08 * 10.4 = 0.831 N. sin37 = h/3 h = 3*sin37 = 1.81 m. PE = Mg*h - Fk*d = 13*1.81 - 0.831*3 = 21.04 J. At bottom of ramp: KE = PE = 0.5*M*V^2 = 21.04 J. 0.5*1.33*V^2 = 21.04 V^2 = 21....
*April 3, 2015*

**Physics**

V^2 = Vo^2 + 2a*d = 0 + 2a*2pi = 4a*pi V = 2*sqrt(pi*a) = 2*1.772*sqrt a = 3.54*sqrt a
*April 3, 2015*

**math**

Sam has X stamps. Jill has (576-X) Stamps. X-x/7 = 6x/7 = Sam's bal. 576-x + x/7 = 576 - 6x/7 = Jill's bal. 1/4(576-6x/7) = 144 - 6x/28 3/4(576-6x/7) = 432 - 9x/14 = Jill's new bal. 6x/7 + 144-6x/28 = 144 + 24x/28-6x/28 = 144 + 9x/14 = Sam's new bal. Jill's...
*April 2, 2015*

**Physics**

M*g = 475 * 9.8 = 4655 N. P = F*d/t = Mg * d/t = 4655 * 10/24 = 1940 J/s = = 1940 W. = 1.94 kW
*April 2, 2015*

**math help`**

Any pair of parallel lines will have equal slopes. I don't know what you mean by "some is always -1".
*April 2, 2015*

**math**

Multiply Eq1 by -2 and add the Eqs: -4x + 6y = 4 +4x + y = 24 Sum: 7y = 28 Y = 4 In Eq1, replace Y with 4 and solve for X
*April 2, 2015*

**algebra**

d1 = d2 60 + 60t = 80t 20t = 60 t = 3 h.
*April 2, 2015*

**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*