Friday

July 3, 2015

July 3, 2015

Total # Posts: 10,681

**physics**

V^2 = Vo^2 + 2a*d a = (V^2-Vo^2)/2d V = 0.63 m/s Vo = 0 d = 0.45 m. Solve for a. F = M*a M = 69 kg Solve for F.
*February 25, 2015*

**Math SO HARD :C**

T + 8 = -20 T = -20-8 = -28o C @ Midnight.
*February 25, 2015*

**physical science**

Correct.
*February 25, 2015*

**Physics**

I = V/R = 9/230 = 0.0391 Amps = 39.1 mA.
*February 25, 2015*

**Algebra 1**

Multiply Eq1 by -2 and add: -10x - 2y = -18 +10x - 7y = 18 Sum: 0 - 9y = 0 Y = 0 In Eq1, replace Y with 0 and solve for X -10x - 2*0 = -18 X = 1.8
*February 24, 2015*

**maths**

X pizzas 4X Drinks 12x + 0.5*4x = 112 Solve for X.
*February 24, 2015*

**physics**

Thank you Damon! I think I understand what you meant when I read your post My triangle was correct though, yes? I just referred to them wrong?
*February 24, 2015*

**physics**

Wait that's not how it's supposed to look. Ok
*February 24, 2015*

**physics**

How do I know which trig function to use? I made my second triangle like this _____ \ | \ | \ | \|
*February 24, 2015*

**physics**

Why did you swap Betas trig functions?
*February 24, 2015*

**physics**

What do you mean, I did indicate on the question that it was west of north?
*February 24, 2015*

**physics**

A golfer takes two putts to sink the ball, one is (81.6 ft, 31.7 degrees N of E) and the other is (3.20 ft, 53.4 degrees W of N). What is the displacement of the single putt that would sink the ball on the first try? Well, what I did was make a diagram of the two vectors, I ...
*February 24, 2015*

**Science**

M*g = 90 N. M*9.8 = 90 M = 9.18 kg Density = 9.18kg/45mL = 0.204 kg/mL Note: A 90-N. object has a mass of 9.18 kg or 9,180 grams.
*February 24, 2015*

**Finance**

V = Vo(1+r)^n r = 5%/365/100% = 0.00013699 = Daily % rate expressed as a decimal. n = 365Comp./yr * 11Yrs = 4015 Compounding periods. V = 10,000(1.00013699)^4015 = $17,332.13
*February 24, 2015*

**Physical science**

Fp = Mg*sin35 = 315 Mg = 315/sin35 = 549 N. = Force of gravity. Fn = Mg*Cos35 = 549*Cos35 = 450 N. = Normal force.
*February 24, 2015*

**Physics**

d = 42m1[135o] - 34mi[270o] X = 42*Cos135 = -29.7 mi Y = 42*sin135 - 34*sin270 = 63.7 mi. Tan A = Y/X = 63.7/-29.7 = -2.14473 A = -65o = 65o N. of W. = 115o, CCW. d = Y/sin A = 63.7/sin 115 = 70.3 mi. d = V*t = 70.3 V = 70.3/t = 70.3/0.5h = 140.6 mi/h
*February 24, 2015*

**Physics**

Vp = -250m/s + i25m/s, Q2. Tan A = Y/X = 25/-250 = -0.10000 A = -5.71o = 5.71o N. of W.= 174.3o,CCW Vp = X/Cos A = -250/Cos(174.3)=251 m/s
*February 23, 2015*

**Physics**

V = Vo + a*t = 31m/s[30o] Vo - 4*4 = 31*Cos 30 + i31*sin 30 Vo - 16 = 26.8 + 15.5i Vo = 42.8 + 15.5i Tan A = Y/X = 15.5/42.8 = 0.36215 A = 19.91o Vo = 42.8/Cos 19.91 = 45.5m/s[19.9o] = Initial velocity.
*February 23, 2015*

**physics**

d = V*t = 2*0.3 km = 0.6 km 216 * t = 0.6 t = (0.6/216) h. t = (0.6/216)h * 3600s./h = 100 s. Note: The sound travels 600m(0.6km).
*February 23, 2015*

**Physics**

1. M*g = Wt. of roller. Fp = Mg*sin 30 = 2500 Mg = 2500/sin 30 = 5000 N. M = 5000/g = 5000/9.8 = 510.2 kg 2. Fn = Mg*Cos 30 = 5000*Cos 30 = 4330 N = Normal force = Force exerted by the surface.
*February 23, 2015*

**Physics**

sin23.3 = h/0.7 h = 0.7*sin 23.3 = 0.277 m. h = 0.5g*t^2 h = 0.277 m. g = 9.8 m/s^2 Solve for t.
*February 23, 2015*

**physics**

Incomplete, and too many errors.
*February 23, 2015*

**Math Vectors**

Vr = 400km/h[200o] + 100km/h[225o] X = 400*Cos200 + 100*Cos225 = -447 km/h Y = 400*sin200 + 100*sin225 = -208 km/h Tan Ar = Y/X = -208/-447 = 0.46532 Ar = 25o = Reference angle. A = 25 + 180 = 205o (Q3). V = X/Cos A = -447/Cos205 = 493 km/h. = Resultant velocity.
*February 22, 2015*

**Math**

a. Range = Vo^2*sin(2A)/g Range = 1200 Ft. Vo = 500 Ft./s. g = 32 Ft/s^2 Solve for A. b. V^2 = Vo^2 + 2g*h h = -Vo^2/2g Vo = 500 Ft./s. g = -32 Ft./s^2 Solve for h
*February 22, 2015*

**math**

Isaiah: X cards. x-x/9 = 560/2 8x/9 = 280 8x = 2520 X = 315 560 - 315 = 245 More than Andy.
*February 22, 2015*

**math**

X = 5 km/h Y = -3 km/h i. Speed = Sqrt(X^2+Y^2) = sqrt(5^2+3^2)= 5.83 km/h = 1.62 m/s. b. Tan A = Y/X = -3/5 = -0.60 A = -31o = 31o S. of E. d = 30m/Cos(-31) = 35 m. = Distance to cross. d = V*t = 35 m. t = 35/V = 35/1.62 = 21.6 s.
*February 21, 2015*

**Physics**

E = 58.0 Volts 1/Ct = 1/C1 + 1/C2 + 1/C3 1/Ct = 1/3.83 + 1/5.89 + 1/12.9 = 0.5084 Ct = 1.97 uF. Qt = Q1 = Q2 = Q3. Qt = Ct*E = 1.97 * 58 = 114.1 uC Q1 = C1*V1 = 114.1 uC V1 = 114.1/3.83 = 29.8 Volts.
*February 20, 2015*

**physics**

Mc = 14 kg = Mass of cart Md = Mass of dog food. d = 0.5a*t^2 = 2.30 m. 0.5a*2.55^2 = 2.3 3.25a = 2.3 a = 0.707 m/s^2. F = M*a M = F/a (Mc+Md) = F/a F = 12 N. a = 0.707 m/s^2 Solve for Md.
*February 20, 2015*

**math**

a. A = 4.1m * 2.2m = 9.02 m^2. A max = (4.1+0.01) * (2.2+0.01) = A min. = (4.1-0.01) * (2.2-0.01) = 8.957 m^2.
*February 20, 2015*

**Math**

X+X = 2x = Total bought. Red Tulips: x-24 Yellow: 4(x-24) 2x-24 Tulips remaining. x-24 + 4(x-24) = 2x-24 x-24 + 4x-96 = 2x-24 3x = 120-24 = 96 X = 32 2x = 64
*February 20, 2015*

**Algebra 2**

W^2 + 5W = 36 W^2 + 5W - 36 = 0
*February 20, 2015*

**physics**

Bill, please clarify your problem by adding the missing words.
*February 19, 2015*

**Physics**

Circumference = pi*Dia = 3.14 * 32.6cm = 102.4 cm = 1.024 m. Vo = 247rev/60s. * 1.024m/rev=4.22 m/s. V = 378rev/60s * 1.024m/rev = 6.45 m/s. V = Vo + a*t V = 6.45 m/s Vo = 4.22 m/s t = 6.54 s. Solve for a. d = Vo*t + 0.5a*t^2
*February 19, 2015*

**physics**

See previous post: Thu, 2-19-15, 6:56 PM.
*February 19, 2015*

**Math**

See previous post: Thu, 10-23-14, 6:36 PM.
*February 19, 2015*

**physics**

A shorter method: R1 = 100 Ohms R2 = 449 Ohms Ri = Internal resistance. V1 = I1*R1 = 4 * 100 = 400 Volts. V2 = I2*R2 = 1.09 * 449 = 489.41 Volts. Ri = (V2-V1)/(I1-I2) = (489.41-400)/(4-1.09) = 30.73 Ohms. E = I1*R1 + I1*Ri = 400 + 122.9 = 522.9 Volts.
*February 19, 2015*

**physics**

E = Battery terminal voltage(e.m.f.). Ri = Internal resistance. R1 = 100 Ohms. R2 = 449 Ohms. I*R1 + I*Ri = E 4*100 + 4*Ri = E Eq1: -E + 4Ri = -400 I*R2 + I*Ri = E 1.09*449 + 1.09*Ri = E Eq2: -E + 1.09Ri = -489.41 Subtract Eq2 from Eq1: Eq1: -E + 4Ri = -400 Eq2: -E + 1.09Ri...
*February 19, 2015*

**physics**

V = 4.93revs/1.47s * 6.28rad/rev = 21.1 rad/s = Angular velocity. V = Vo + a*t V = 21.1 rad/s. Vo = 0 t = 1.47 s. a = rad/s^2 Solve for a.
*February 19, 2015*

**Math 222**

g^2 - 6g - 55 = 0 -55 = 5*(-11) g+5= 0 g+(-11) = 0 g-11 = 0 (g+5)(g-11)= 0
*February 19, 2015*

**physics**

Vo = 20m/s[40o] Xo = 20*Cos40 = 15.32 m/s. Yo = 20*sin40 = 12.86 m/s. a. Y^2 = Yo^2 + 2g*h Y = 0 Yo = 12.86 m/s. g = -9.8 m/s^2 Solve for h. b. Y = Yo + g*Tr Y = 0 Yo = 12.86 m/s. g = -9.8 m/s^2. Solve for Tr.(Rise time or time to reach max. ht.) c. V = Xo + Yi = Xo + 0 = Xo d...
*February 19, 2015*

**physics**

a. V = Vo + a*t V = 25.1 m/s Vo = 0 t = 0.48 s. Solve for a. b. F = M*a M = 910 kg a = Value calculated in part a.
*February 19, 2015*

**Chemistry**

Human bodies maintain their balance (homeostasis) through a series of equilibria. An important equilibrium involves calcium ion because Ca2+ is important for muscle contractions in the heart as well as for strong bones. Bone is a complex tissue consisting of living cells and ...
*February 19, 2015*

**Chemistry**

A 500.0 mL sample of saturated nickel(II) phosphate solution was analyzed spectrophotometrically and found to contain 1.88 × 10-5 g of nickel. (a) Determine the molarity of the nickel(II) phosphate solution. (b) Calculate the solubility product constant, Ksp for nickel(...
*February 19, 2015*

**Chemistry**

The solubility of insoluble substances changes depending on the nature of the solution. Below are two solutions in which Cu(OH)2 is dissolved; in each case, the solubility of Cu(OH)2 in those solutions is not the same as it is in pure water. (a) For each Cu(OH)2 solution below...
*February 19, 2015*

**Physics**

See previous post: 9-16-13, 8:40 PM.
*February 18, 2015*

**physics**

T^2 = 4pi^2*(L/g) L/g = T^2/4pi^2 = 4.17^2/39.5 = 0.440 0.75/g = 0.44 g = 0.75/0.44 = 1.70 m/s^2. It is on the moon!
*February 18, 2015*

**physics**

a. T = 2min/60cycles = 120s/60cycles = 2s/cycles b. F = 1/T = 1cycle/2s. = 0.50 cycles/s. = 0.50 Hz.
*February 18, 2015*

**physical science**

T = 50km * 1h/5km * 60min/h = 600 min.
*February 18, 2015*

**physics**

Incomplete.
*February 18, 2015*

**Physical Science**

a. h = 0.5g*t^2 h = 20 m g = +9,8 m/s^2 Solve for t. b. V^2 = Vo^2 + 2g*h Vo = 0 g = +9.8 m/s^2 h = 20 m. Solve for V.
*February 17, 2015*

**Physics**

Please clarify your problem: Does V mean Voltage or velocity? Did you give all of the INFO? If: V = Supply voltage. V1 = Voltage across R1. V2 = Voltage across R2. R1 and R2 connected in series. Then, V = V1 + V2
*February 17, 2015*

**Physics**

I = 7.0 mA V = 14.7 Volts. v1 = 3.5 Volts. V2 = 1.4 Volts. a. V3 = V-V1-V2 b. R3 = V3/I I = 0.007 Amps
*February 17, 2015*

**Physics**

See previous post.
*February 17, 2015*

**physical**

Vo = 18.7m/s[51o] Xo = 18.7*Cos51 = 11.8 m/s. Yo = 18.7*sin51 = 14.5 m/s. Y^2 = Yo^2 + 2g*h = 0 h = -Yo^2/2g = -14.5^2/-19.6 = 10.8 m. Y^2 = Yo^2 + 2g*d = 0 + 19.6*(10.8-2.5)= 162.68 Y = 12.8 m/s = Ver. component. V^2 = Xo^2 + Y^2 = 11.8^2 + 12.8^2 = 303.08 V = 17.4 m/s. = ...
*February 17, 2015*

**science**

a = (V-Vo)/t = Acceleration in Ft./s^2 V = 20 Ft/s. Vo = 5 Ft/s.
*February 17, 2015*

**physics**

V = sgrt(X^2+Y^2)
*February 17, 2015*

**physics**

V = 5m/s[30o] Vy = 5*sin30 = 2.5 m/s
*February 17, 2015*

**Algebra**

Eq1: x + y = -8 Eq2: x - y = 4 Sum: 2x = -4 X = -2 In Eq1, replace x with -2: -2 + y = -8 Y = -6.
*February 16, 2015*

**physics**

M*g = 27 * 9.8 = 264.6 N. = Wt. of the lawnmower. Fn = 264.6 + Fap*sin36=264.6 + 0.588Fap Fk = u*Fn = 0.66*(264.6+0.588Fap) = 174.64 + 0.388Fap (Fx-Fk) = M*a (Fap*Cos36-(174.64+0.388Fap) = M*a (0.81Fap-174.64-0.388Fap = M*0 0.42Fap - 174.64 = 0 0.42Fap = 174.64 Fap = 416 N. = ...
*February 16, 2015*

**Physics**

W = Mg * d = 2.80*9.8 * (35-10) = 686 J.
*February 16, 2015*

**Physics**

a = (V-Vo)/t =(250-0)/12 =
*February 16, 2015*

**Physics**

a. T^2 = 4pi^2(L/g)=4*3.14^2(0.76/9.8) = 3.06 T = 1.75 s.
*February 15, 2015*

**physical science**

a. h = 0.5g*t^2 = 4.9*4^2 = 78.4 m b. h = 0.5g*t^2 h = 103 m. g = 9.8 m/s^2 Solve for t.
*February 15, 2015*

**Physics**

Incomplete.
*February 15, 2015*

**Business math**

Ordinary interest assumes a business year of 360 days. Exact interest uses 365 days per year. a. I = Po*r*T = 4400*(0.1025/360)*135 = $169.125 b. 4400 + 169.125 = $4569.13
*February 15, 2015*

**math**

d = 120km/h * 2.6667h=Distance traveled. L = d/4 = Liters of fuel used. Cost = $59 * L =
*February 15, 2015*

**science**

V^2 = Vo^2 + 2g*h Vo = 0 g = +9.8 m/s^2 h = 50 m. Solve for V.
*February 15, 2015*

**Physics**

M*g = 150 N. Fp = 150*sin20 = Force parallel. Fn=150*Cos20 = Normal or perpendicular.
*February 15, 2015*

**Physics**

Fx = 280*Cos33o Fy = 280*sin33o
*February 15, 2015*

**Physics**

Xo = 8.57 m/s Yo = -2.61 m/s a. X = Xo + a*t = 8.57 + (-0.105)*6.67 = 7.87 m/s Y = Yo + a*t = -2.61 + 0.101*6.67 = -1.94 m/s b. V^2 = X^2 + Y^2 = 7.87^2 + 1.94^2 = 65.7 V = 8.11 m/s.
*February 13, 2015*

**Physics Honors**

Y^2 = Yo^2 + 2g*h = 0 Yo^2 - 19.6*23.2 = 0 Yo^2 = 454.72 Yo = 21.3 m/s = Ver. component of initial velocity. Yo = Vo*sin A = 21.3 m/s 48*sin A = 21.3 sin A = 0.44,425 A = 26.4o = Angle at which golf was hit. Vo = 48m/s[26.4o] Xo = 48*Cos26.4 = 43.0 m/s. V = Xo + Yi = 43 + 0i...
*February 13, 2015*

**science**

B. 0.29/5 = W/115
*February 12, 2015*

**science**

Wt. = (170/10) * 25 =
*February 12, 2015*

**Physics**

a. Energy = 0.5*C*V^2 V doubled: Energy = 0.5*C*(2V)^2 = 0.5*C*4V^2 So the energy is increased by a factor of 4. b. Yes.
*February 12, 2015*

**Physics Dynamics**

M*g = 70 * 9.8 = 686 N. = Wt. of sled. Fn = Mg-F*sin20 =
*February 12, 2015*

**Physics Dynamics**

Range = Vo^2*sin(2A)/g = 38 32^2*sin(2A)/9.8 = 38 104.5*sin(2A) = 38 sin(2A) = 0.36364 2A = 21.3o A = 10.7o
*February 12, 2015*

**Physics Dynamics**

Tan A = 35/80 = 0.4375 A = 23.6o a. Tan23.6 = Vc/2 Vc = 2*Tan23.6 = 0.874 m/s = Velocity of the current. b. Direction = 23.6o to the left.
*February 12, 2015*

**Physics**

Vo = 21m/s[40o] Xo = 31*Cos40 = 16.1 m/s. Range=Vo^2*sin(2A)/g = 21^2*sin(80)/9.8 = 44.3 m. Range = Xo*T = 44.3 m. 16.1T = 44.3 T = 2.75 s. = Time in air.
*February 12, 2015*

**Physics**

V^2 = Vo^2 + 2g*h Vo = 2.18 m/s. g = +9.8 m/s^2 h = 70.8 m. Solve for V.
*February 9, 2015*

**College physics**

d = 239mi * 1600m/mi = 382,400 m. = 382.4 km. Work = F*d = 369 * 382.4 = 141,106 kJ.
*February 9, 2015*

**Physics**

Wt. = M*g Work = Mg * h Therefore, the person with the most wt. will do the most work if both covered the same distance up. If they did not cover the same distance, you will have to calculate the work done by each to make a comparison. Power is the rate of doing wt. P = Mg * h...
*February 9, 2015*

**math**

h = -16x^2 + 64x + 80 = 0 Use Quadratic Formula. X = -1, and 6. Select the positive value: X = 6 s. = Time to reach gnd.
*February 9, 2015*

**maths**

a. KE = 0.5*M*V^2 = 130 J. 0.5*11*V^2 = 130 V^2 = 23.64 V = 4.86 m/s. b. V^2 = Vo^2 + 2g*h = 4.86^2 + 19.6*3 = 82.4 V = 9.1 m/s c. V^2 = Vo^2 + 2g*h = 0 + 19.6*15 = 294 V = 17.15 m/s.
*February 9, 2015*

**engineering science n2**

V = Vo + g*t V = 0 Vo = 20 m/s. g = -9.8 m/s^2 Solve for t.
*February 9, 2015*

**physics**

See previous post: Sun, 2-8-15, 8:14 PM.
*February 8, 2015*

**Physics Dynamics**

M*g = 5 * 9.8 = 49 N. = Wt. of box. Fn = 49 - 45*sin40 = 20.1 N. = Normal force. Fk = u*Fn = 0.2 * 20.1 = 4.01 N. = Force of kinetic friction. a = (Fx-Fk)/m = (45*Cos40-4.01)/5 = 6.1 m/s^2
*February 8, 2015*

**physics**

d = V*t d = 18.4 m V = 44.8 m/s Solve for t.
*February 8, 2015*

**Physics**

Series Connection: When one lamp burns out, all of the lamps go out. So you are in the dark until the lamp is replaced.
*February 8, 2015*

**physics**

7grams * 5.66*10^21atoms/gram * 46e/atom = 1.82252*10^24 Electrons. b. 1.82252*10^24e * 1.6*10^-19C/e = 2.916*10^5 C. = Your answer.
*February 8, 2015*

**physics**

1/f = 1/di + 1/do LCD = di*do 1/f = do/di*do + di/di*do 1/f = (do+di)/di*do Invert both sides: f = di*do/(do+di)
*February 8, 2015*

**physics**

KE = KE1 + KE2 = 0.5*M*V^2 + KE2 = 0.5*0.6*3^2 + 2.5 = 5.2 J. KE = 0.5*0.6*V^2 = 5.2 V^2 = 17.33 V = 4.16 m/s.
*February 8, 2015*

**physics**

a = (V-Vo)/t = (9.3-0)/1.39 = 6.69 rad/s^2
*February 8, 2015*

**science**

P = F * d/s = Mg * d/t = 5000 * 25/50 = 2500 J./s = 2500 W.
*February 8, 2015*

**physics**

a. KE = 0.5*M*V^2 = 0.5*66*4.3^2 = 610.2 J. = Energy lost due to friction. b. M*g = 66 * 9.8 = 646.8 N. = Wt. of runner = Normal(Fn). Fk = u*Fn = 0.6 * 646.8 = 388.1 N. = Force of kinetic friction. KE = Fk*d = 610.2 388.1*d = 610.2 d = 1.57 m.
*February 8, 2015*

**physics**

Vo = 111km/h = 111000m/3600s. = 30.8m/s W = KE = 0.5M*Vo^2 =
*February 7, 2015*

**Physics**

2.76km[0o] + 0.85km[270o] X = 2.76*Coso + 0.85*cos270 = 2.76 N. Y = 2.76*sin0 + 0.85*sin270 = Fr = sqrt(X^2+Y^2)
*February 7, 2015*

**Physics**

24N[60o] + 9N[0o] X = 24*Cos60 + 9*Cos0 = 21 N. Y = 24*sin60 + 9*sin0 = 20.8 N. Fr^2 = 21^2 + 20.9^2 = 873.64 Fr = 29.6 N.
*February 7, 2015*

**Physics**

Pellet A: V = Vo + g*Tr Tr = -Vo/g = -27.3/-9.8 = 2.79 s. = Rise time. Tf1 = Tr = 2.79 s. = Time to Fall back to the edge of the cliff. Tf2 = Fall time from edge of cliff to gnd. Ta = Tr + Tf1 + Tf2 = 2.79 + 2.79 + Tf2 = 5.57 + Tf2. = Time in air, pellet A Pellet B: Tb = Tf = ...
*February 7, 2015*