Monday

December 22, 2014

December 22, 2014

Total # Posts: 20,008

**science**

http://curious.astro.cornell.edu/question.php?number=300 Note that if you are in the Northern hemisphere You can see stars that are high in the northern sky all year. It is the ones close to the plane of the earth's orbit that you can only see when you are facing them at ...
*October 22, 2014*

**Sorry**

mistook / for times (bottom*derivative of top - top * derivative of bottom)/bottom^2 (x)/(2+x)= = [(x+2)(1)-x(1) /(x+2)^2 = [ x+2 -x ] /(x+2)2 = 2/(x+2)^2
*October 22, 2014*

**Derivatives**

(2+x) d/dx (x) + x d/dx (2+x) (2 + x) + x(1) 2 + 2 x 2(x+1)
*October 22, 2014*

**physics**

6.3 * sin 34 = 3.52 meters up m g h = 13.2 * 9.81 * 3.52 = 456 so I get -456 J I think you reversed digits
*October 22, 2014*

**physics...damon please help**

delta L/L = k delta T delta L/L = 6.7*10^-6 inches/inch degree F (117 deg F) L = 10 meters =394 inches so delta L = 394 * 6.7 *10^-6 * 117 = .309 inches = .784 cm = .0784 meters
*October 22, 2014*

**COLLEGE PHYSICS**

|v1| = 2|v2| v1^2 = 4 v2^2 so Ke v1 = 4 Ke v2 so 5 Ke2 = 7500 J so Ke2 = 1500 J and Ke1 = 6000 J
*October 22, 2014*

**Trig ?**

I do not understand your symbols, do not have whatever font you are using. I assume this is some kind of a phase shift in the argument of the cotangent.
*October 22, 2014*

**physics**

17 = (1/2) 1.9 t^2 solve for t v = 1.9 t
*October 22, 2014*

**Math**

II. When calculating interest , you need too ______? (1pt) a. Change the percentage to a decimal then multiply the decimal against the amount. --------------------------------- VIII .0450 - .0375 = .0075 .0075 * 1000 = 7.50 or answer c BUT this assumes interest PER MONTH (very...
*October 22, 2014*

**Algebra 1 ?????**

do you mean -2/5x - 9 < 9/10 ??? or (-2/5)(x - 9) < 9/10 or -2/(5x - 9 )< 9/10 or -2/[5(x - 9)] < 9/10 or what ?
*October 22, 2014*

**math**

You mean he checked 25 wheels ????? If so then what times 4 + what times 3 = 25? 1*4 + 21 sure 1*4 + 7*3 2*4 + 17 nope 3*4 + 13 nope 4*4 + 9 sure 4*4 + 3*3 4*5 + 5 nope 4*6 + 1 nope
*October 21, 2014*

**math**

1.65 * 100 = 165
*October 21, 2014*

**Precalculus**

PLEASE PROOF READ - I assume you mean 2.2ft. / HOUR a) A = pi r^2 b) 2.2 t c) 2.2 * 2 2.2 * 2.5 d)A(2) = pi * (4.4)*2 A(2.5) = pi (5.5)^2 e) A(t) = pi (2.2 t)^2 f) A(2) = pi (2.2*2)^2 A(2.5) = pi (2.2*2.5)^2 g) [ A(2.5) -A(2) ]/.5 h) [A(7.)-A(5)]/ 2.5 i) you will see that it ...
*October 21, 2014*

**precalculus**

A = Ai e^kt 2 = e^3k 3 k = ln 2 k = .231 30 = e^(.231 t) ln 30 = .231 t t = 14.7 hours
*October 21, 2014*

**precalculus**

already did this above
*October 21, 2014*

**algebra**

note a may not be 10 or -4 or the problem is undefined (zero denominator)
*October 21, 2014*

**algebra**

(a+10)/(a+4) by dividing top and bottom by (a-10)
*October 21, 2014*

**Physics**

4.63 * 4.10 work "in" in x direction -3.79 * 3.86 work "out" in y direction (braking) 2.16 * -2.45 work "out" in z direction now just add the results (not a vector) 4.63 * 4.10 - 3.79 * 3.86 - 2.16 * 2.45 Joules (scalar but does have sign)
*October 21, 2014*

**math**

Oh, I misunderstood question I think.
*October 21, 2014*

**math**

72 min and 75 min (common multiple?) 72 = 12 * 6 = 3*3 *2*2*2 75 = 5*5*3 so we need 25 * 3 * 8 = 600 minutes = 10 hours
*October 21, 2014*

**Physics**

just did it below kinetic energy at bottom = potential energy at top
*October 21, 2014*

**Physics**

m g h = (1/2) m v^2 v^2 = 2 g h v = sqrt (2 g h) h = 3.27 (1 - cos 35.6) meters g = 9.8 m/s^2
*October 21, 2014*

**Physics**

work done by gravity = change in potential energy in falling from initial height h = L (1-cos theta) h = .8 (1 - cos 40.4) so work=m g h = 2.63*9.8*.8 (1 - cos 40.4)
*October 21, 2014*

**Physics**

2275 = 207.8 t +(1/2)15.6 t^2 solve quadratic for t then v = 207.8 + 15.6 t
*October 21, 2014*

**Physics**

d = Vo t + (1/2) a t^2 219 = 0 + 2.3 t^2 solve for t v = Vo + a t v = 0 + 4.6 t
*October 21, 2014*

**Physics**

here for example Vo = 14.8 a = 11.3 ft/s^2 t = 6.6 sd v = Vo + a t so v = 14.8 + 11.3 * 6.6 = 89.4 feet/second
*October 21, 2014*

**Physics**

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html You want the box labeled: "Description of motion in one dimension"
*October 21, 2014*

**inequalities**

10x -20 + 11 < 10 -12 + 2x 8 x < 7 x < 7/8
*October 21, 2014*

**Note**

0 = Vo - a t ignore that line, final speed is not zero and is not needed
*October 21, 2014*

**please help! physics**

95 km/hr *1000/3600 = 26.4 m/s in .5 seconds moves 13.2 meters then a = F/m = -2400/1200 = - 2 m/s^2 0 = Vo - a t d = 13.2 + 26.4 t - (1/2) (2) t^2 t = 2 so d = 13.2 + 48.8 = 62 meters d = 13.2 +
*October 21, 2014*

**Physics**

at bottom v = sqrt (2 g h)
*October 21, 2014*

**science**

They are all gasses but the most dense of the bunch is ethyne (Acetylene)
*October 21, 2014*

**physic**

Vi = 28 sin 60 =24.25 m/s v = Vi - 9.8 t h = Vi t - 4.9 t^2 = 24.25(3.3) - 4.9(3.3)^2 = 26.7 meters high not 29.2 when does v = 0 (top of arc)? 0 = 24.25 - 9.8 t t = 2.47 seconds so h at top = 24.25(2.47) - 4.9 (2.47)^2
*October 21, 2014*

**Physics**

m v^2/r = 1630 * (13.9)^2/385 b) I do not know what is holding your vehicle in the circle !!!! If on a string - tension force If held in a big cylinderical rotating can (carnival ride) - normal force If on a flat curving road - friction on the tires If in orbit in space - ...
*October 21, 2014*

**math**

I (we) did not use any resources such as the web or books other than our textbook. ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; and you want me to work on it ?
*October 19, 2014*

**Physics**

Total momentum = 0 Mb*2.2 - Me*2.9 = 0 Mb/Me = +2.9/2.2 the ratio of masses is unlikely to be negative and I think you have it upside down.
*October 19, 2014*

**Physics!**

force = rate of change of momentum (only m a if m is constant :) change of momentum = 79.1 (0 - -6.27) change of time = 2.06*10^-3 s so force = + 6.27*79.1 / .00206 Newtons
*October 19, 2014*

**Calculus**

You are welcome.
*October 19, 2014*

**Calculus**

In other words you got the slope right but did not get b right in y = m x + b
*October 19, 2014*

**Calculus**

you say y = 5 x + 2.5 then if x = pi/4, y = 5 pi/4 + 2.5 You say but what is f(x) at x = pi/4 ??? sin^2 = 1/2 2 tan = 2 so f(x) = 2.5 You better take 5 pi/4 off your answer
*October 19, 2014*

**Physics**

Keb = (1/2)55 (7.5^2) = 1545 J = 1.55kJ I agree with your book :)
*October 19, 2014*

**Physics**

55*22 = 1210 = 55 Vb cos 76+ 45 Vs cos 12 45*26 = 1260 = 55 Vb sin 76 - 45 Vs sin 12 1210 = 13.3 Vb + 44.0 Vs 1260 = 53.4 Vb - 9.36 Vs times 44/9.36 1210 = 13.3 Vb + 44.0 Vs 5922 = 251 Vb - 44.0 Vs ------------------------- 7132 = 264.3 Vb Vb = 27 km/hr = 7.5 m/s humm, we ...
*October 19, 2014*

**physics**

force in = mg sin 12 = m v^2/r v^2/r = g sin 12 turns 180 deg = pi radians r = v^2/(g sin 12) pi r = distance flown = pi v^2/(g sin 12) time = distance / speed = pi r/v = pi v/(g sin 12 here v = 450/3.6 and g = 9.81
*October 19, 2014*

**physics**

m = 5 t^.8 - 3 t + 20 dm/dt = (5*.8) t^-.2 - 3 so at max or min 0 = 4/T^.2 - 3 T^.2 = 4/3 = 1.333 .2 log T = log 1.333 Log T = .6246 T = 10^.6246 = 4.21 seconds find m at t = 4.21 and find dm/dt at 3 and 5 (signs better be different)
*October 19, 2014*

**Math**

You are welcome.
*October 18, 2014*

**Math**

A + B = 500,000 .14 A + .10 B = 59,000 multiply second equation by 10 1.0 A + B = 500,000 1.4 A + B = 590,000 --------------------- subtract -.4 A = -90,000 A = 225,000 etc
*October 18, 2014*

**Math**

A + B = 260,000 A = B + 77,000 so B + 77,000 + B = 260,000 or 2 B = 183,000 etc
*October 18, 2014*

**Math**

Yes, it is a quadratic equation
*October 18, 2014*

**Math**

been there, done that see below
*October 18, 2014*

**math**

LOL, well if 6 is the answer then -16(36) + 96 (6) better be zero is it ? :)
*October 18, 2014*

**Math**

yes, quadratic equation
*October 18, 2014*

**Math**

speed before Forde = (s-5) speed after Forde = s time to Forde = (t+96) note 4 days = 96 hr time after Forde = t then distance = speed * time 70 = (s-5)(t+96) 60 = s t or t = 60/s 70 = (s-5)(60/s + 96) 70 = 60 + 96 s - 300/s - 480 490 = 96 s - 300/s 245 = 48 s - 150/s or 48 s^...
*October 18, 2014*

**Physics**

v = Vi + a t d = Vi t + (1/2) a t^2 2240 = 199 t + 4.25 t^2 solve quadratic for t then v = 199 + 8.5 t
*October 18, 2014*

**Physics**

d = (1/2) a t^2 176 = 4.35 t^2 solve for t v = a t = 8.7 t
*October 18, 2014*

**PHYSICS 1301 (COLLEGE)**

When people do not even go back and see if what they typed makes any sense, I just skip to the next question.
*October 18, 2014*

**Calculus**

v = -gt = -9.8*2 100 = 4.9 t^2 t = 4.52 seconds to ground so v = -9.8*4.52
*October 18, 2014*

**chemistry**

.55 a = .45(715)
*October 18, 2014*

**calculus**

4 cos anything is never bigger than 4 so there is no overlap here
*October 18, 2014*

**Calculus**

2 * L + 4*w = 800 so L = 400 - 2w A = L * w A = (400 -2w)w A = -2 w^2 +400 w dA/dw = 0 at max = -2 w + 200 w = 100 L = 400 -200 = 200
*October 18, 2014*

**Calculus**

decrease at start due to startup costs and paying off initial investment in staff and equipment increase over long term because we have to go deeper and deeper into the vein as we mine out the easy stuff near ground level.
*October 18, 2014*

**Chemistry**

heat it 25 deg C 550 * 2.03 * 25 Joules
*October 18, 2014*

**physics please help**

In the end torque = F*.4 - friction torque = I d omega/dt
*October 18, 2014*

**physics please help**

Not knowing that frictional torque, I am at a loss. Please proof read question.
*October 18, 2014*

**Physics**

69 v1 = 47 v2 so v2 = (69/47)v1 19 = (v1+v2)t 19 (v1 + 1.47 v1) t so v1 t = distance of first = 19/2.47 = 7.69 meters
*October 18, 2014*

**Physics**

same old conservation of momentum at first p = 2 *1 + m * 4.4 at the end p = (2+m) * 1.5 so (2+m)(1.5) = 2 + 4.4 m 3 + 1.5 m = 2 + 4.4 m etc
*October 18, 2014*

**DIFF CALCULUS**

now the volume is another story because now we will have to make some assumption, for example constant slope up from 3 m to 1 m. In that case we have a trapezoid and V = 6 * 12 * average depth = 6 * 12 * 2 - 144 m^3
*October 18, 2014*

**DIFF CALCULUS**

Part b is easy and requires no assumptions about the shape of the pool bottom. The rate of change of height = volume added per second / surface area of water. (draw a picture) or dh/dt = .25 m^3/min / (12m*6m) = .00347 m/min or 21 cm/hour
*October 18, 2014*

**Algebra typo**

22x maybe? what is the question anyway? factor it? put in an equal sign and solve it find zeros of parabola ? by the way x^2 + 22 x + 121 = (x+11)^2 (x+11)^2 -y^2 = (x+11-y)(x+11+y) because a^2-b^2 = (a-b)(a+b)
*October 18, 2014*

**physics**

if v = 0, p = 0 v ground = sqrt (2 g h) momentum at ground = m v = 4 sqrt (2 g h)
*October 18, 2014*

**physics**

falls from top for 6.65/2 = 3.325 seconds h = (1/2) (9.8) (3.325)^2
*October 17, 2014*

**MATH HELP PLEASE**

no sqrt 36 = 6 but what is sqrt 21 ?
*October 17, 2014*

**Last ratio/equation**

m/w = 10/12 so w = 12 m/10 then w/c = 10/15 12 m /10 / c = 10/15 12 m/c = 100/15 = 20/3 m/c = 20/36 = 5/9
*October 17, 2014*

**equations**

a/b = 5 so b = a/5 use that in the second a/5 / c = 2/3 multiply by 5 a/c = 10/3
*October 17, 2014*

**equation**

divide both sides by 3b 3a/3b = 5b/3b simplify a/b = 5/3
*October 17, 2014*

**Science-Physics**

y velocity = 8.9 x velocity = 19.3 tan theta = (8.9/19.3)
*October 16, 2014*

**Science (Physics)**

acceleration = change in velocity / change in time same as speed on the way up, but down not up or in other words it falls from 3 meters up m g h = (1/2) m v^2 so v = -sqrt(2 g h) = -sqrt ( 2 * 9.81 * 3)
*October 16, 2014*

**Algebra**

LOL - It happens
*October 16, 2014*

**Algebra**

for heaven's sake subtract: 1.10 - 1.05 = ?????? 1.15 - 1.10 = ??????
*October 16, 2014*

**physics**

I hope and pray we are talking about buoys and not boys. Anyway, here is a solution for totally immersed spheres. If the sphere waterlines are halfway down the sphere, then double the resistance. (you have half as much conductance) http://www.df.unipi.it/~macchi/TEACHING/...
*October 16, 2014*

**normal distribution**

http://davidmlane.com/hyperstat/z_table.html
*October 16, 2014*

**Algebra 2**

Now that point (5,6) 5 is beyond x = 3.5 so on the way down y = (7/2)(5)- (1/2)(25) = 10/2 = 5 will not make it over wall
*October 15, 2014*

**Algebra 2**

y = (7/2) x - (1/2)x^2 reasonable domain is above ground, y>0 so solve for zeros of x x^2 -7x = 0 x = 0 to x = 7 range (vertex is when x = half 7 or3.5 ymax = (7/2)(7/2) -(1/2)(7/2)^2 = (1/2)(49/4) = 49/8 = 6.125 so range is 0 to 6.125
*October 15, 2014*

**Algebra 2 well, maybe**

well, maybe y = a + b x + c x^2 3 = a + b(1) + c(1) 5 = a + b(2) + c(4) 6 = a + b(3) + c(9) so 2 = b + 3 c 1 = b + 5 c ---------- 1 = -2c c = -1/2 1 = b -5/2 b = 7/2 3 = a + 7/2 -1/2 = a + 3 so a = 0 y = 0 + 7 x/2 - x^2/2 y = (7/2) x - (1/2)x^2 or 2 y = 7 x - x^2
*October 15, 2014*

**Algebra 2**

y = a + b x^2 3 = a + b (1)^2 = a + b 5 = a + b(2)^2 = a + 4 b b = 3 - a 5 = a + 4(3-a) 5 = a + 12 - 4a 3 a = 7 a = 7/3 b = 9/3 - 7/3 = 2/3 so y =7/3 +2/3 x^2 check last point y = 7/3 + (2/3)(9) = 7 1/3 I do not think your table is a parabola
*October 15, 2014*

**Pre-Calc/Trig...**

v = Vi - g t at top v = 0 0 = 4 -32 t t = 1/8 second (4 ft /sec is really slow. I bet you mean 40) h = 10 + 4 (1/8) - 16 (1/64) = 10 + .5 - .25 = 10.25 ft
*October 15, 2014*

**science**

both the same ( Newton #3 action equal and opposite to reaction) F = G * 6 * 7.3 / .65^2 where G = 6.67 * 10^-11
*October 15, 2014*

**Math**

maybe ln y = ln 35.15 + 1.04 x
*October 15, 2014*

**Algebra 2**

332/2 = 166 = v+c 318/2 = 159 = v-c 2 v = 166+159 = 325 etc
*October 15, 2014*

**Math 112**

(86 + 88 + z)/3 = 90 86 + 88 + z = 270 etc
*October 15, 2014*

**Pre-Calc/Trig...**

(2x-1)(x+3) = 0 --> 1/2 and -3 x^2 - x - 1 = 0 x = [1 +/- sqrt (1+4) ]/2 = 1/2 +/- (1/2) sqrt 5
*October 15, 2014*

**Physics Help Please!!!!**

the mass of the shuttle is decreasing because fuel is being expelled F = m a same F less m more a Same happens with your car, but so slowly it does not matter :)
*October 15, 2014*

**physics... PLEASE HELP**

F = m a = 9 * 10^13 * 1.7 * 10^-27 = 15.3 * 10^-14 = 1.5 * 10^-13 Newtons
*October 15, 2014*

**physics**

R = 1.3 * 10^7 * 8 * 10^-9 ohms/m^2 i = V/R b. Same V, double R, half i
*October 15, 2014*

**physics... PLEASE HELP**

The relation between the energy and the momentum is E = p c where p is momentum so our momentum density here is our energy density/c
*October 15, 2014*

**physics... PLEASE HELP**

well, it is late but I might be able to point a direction average intensity = average power / (pi r^2) Peak = twice average power if a sine wave [ average of sin^2 = (1/2) ] energy density usually in Joules per square centimeter where it hits I assume per second. That would be...
*October 15, 2014*

**Physics - Second Law Question**

same as the other question (3-1.96)10^5 Newtons up
*October 15, 2014*

**Physics - Second Law Question**

4.9 - 4.5 = 0.4 0.4*10^-3 = 4 * 10^-4 down because weight is bigger than drag up
*October 15, 2014*

**Physics**

NO!! Usually in a given problem the SUM of kinetic and potential energy is more or less constant. I say more or less because energy is easily lost from your nice neat problem through friction or escape of heat or whatever messy thing happens to make your car not coast forever ...
*October 15, 2014*

**Math**

(7/8)hr/(3/4) puzzle = 7/6 hours/puzzle or 6/7 puzzles/hour
*October 15, 2014*

**Pre-Calc/Trig...**

b^2-4ac 5^2 -4(2)(-3) that will be positive and therefore its square root is a real number. Thus two real roots, nothing imaginary about it 1^2 - 4(-1)(1) everything positive again, two real roots
*October 15, 2014*

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