Sunday

November 23, 2014

November 23, 2014

Total # Posts: 19,411

**question**

I believe so.
*November 10, 2014*

**Math**

d = 4 t d + t = 20 5 t = 20 t = 4 d = 16 16-4 = 12
*November 10, 2014*

**Math help please!!!!!**

24/10 = 30/x 24 x = 300 x = 12.5 I am assuming that x and 10 are along the wall and not the ground.
*November 10, 2014*

**PHYSICS**

dW = p A dr dW = p 4 pi r^2 dr integrate W = p (4/3) pi (r final^3 - r initial^3) W = p (final volume - initial volume)
*November 10, 2014*

**Trigonometry**

sin^-1 (1.2 /4.4 )
*November 9, 2014*

**Physics**

F = m a 150 lbs = 68 kg Fscale up - m g down = m a Fscale up = m g + m a = 150 pounds + 68* 10 newtons = 150 pounds + 680 Newtons 680 Newtons = 153 pounds so 150 + 153 is about 300 pounds
*November 9, 2014*

**math**

65/60 is to the right of 1 on a number line. The result is positive. +
*November 9, 2014*

**Physics**

Thanks, I just went and looked, might do that but I am pretty busy here when I am not coaching our sailing team or whatever.
*November 9, 2014*

**Physics**

I will go look though.
*November 9, 2014*

**Physics**

I am a retired professor, do this for fun.
*November 9, 2014*

**Physics**

You are welcome.
*November 9, 2014*

**Physics**

Ah, here: http://earthsky.org/earth/tides-and-the-pull-of-the-moon-and-sun ( former clammer and lobsterman here :)
*November 9, 2014*

**Physics**

maximum high and low tides occur when the earth, moon and sun are co-linear. That is at full moon and new moon. Such tides are called "spring tides" and occur about twice a month.
*November 9, 2014*

**Physics**

Yes 0 free falling weightless :)
*November 9, 2014*

**Physics**

Backup link :) http://www.jiskha.com/display.cgi?id=1282164495
*November 9, 2014*

**Physics**

Like - "weightless" :)
*November 9, 2014*

**Physics**

No !!! ZERO no force on scale - in free fall, accelerating downward at 9.81 m/s^2
*November 9, 2014*

**Physics**

Total force on person = m g = m a in free fall. No external force on person from scale
*November 9, 2014*

**physics**

ke = .5 m v^2 + .5 I w^2 I = m r^2 w = v/r so ke = .5 m v^2 + .5 m r^2 v^2/r^2 ke = m v^2 v = 2.1 * 3.8 = 8.0 m/s so ke = 5.6 *64 Joules = m g h = 5.6 g h so 64 = g h h = 64/9.8 sin 17.7 = h/distance up ramp so distance up ramp = h.sin 17.7
*November 9, 2014*

**Physics**

see this similar reply http://www.jiskha.com/display.cgi?id=1415568126
*November 9, 2014*

**Chemistry**

How do you find the amount of water that reacted 2 mol of phenylmagesium bromine?
*November 9, 2014*

**Chemistry **

If there are 25 mg of water in a typical drop of water, how many drops of water would react completely with 2 mmol of phenylmagnesium bromide? How do I set up the problem?
*November 9, 2014*

**Physics**

sum of vectors, not product Pa =3.2[ 2.3 i + 4.2 j] = 7.36 i +13.44 j Pb =2.9[ -1.8 i +2.7 j] =-5.22 i +7.83 j P = (7.36-5.22) i +(13.44+7.83) j
*November 9, 2014*

**Physics**

initially p1 = 4 i + 2 j p2 = -3i + 3 j total p = 1 i + 5 j final p must be the same final p1 = px i + py j p2 = 6 i - 3 j px + 6 = 1 px = -5 vx = -5/2 = -2.5 py -3 = 5 vy = 5/2 = 2.5 so v = -2.5 i + 2.5 j
*November 9, 2014*

**physics**

no question was asked
*November 9, 2014*

**Physics**

m g h + (1/2)m v^2 = total energy = constant h is height above hanging straight down mg(2-2 cos 30)+.5m*9 = total energy = m g (2 -cos max angle when v = 0) notice m has nothing to do with this problem, cancels = mg(2-cos45) +.5 m v^2
*November 9, 2014*

**Physics**

Ke - work done against friction = (1/2)kx^2 (1/2)mv^2 - .3 m g (.2+x) = (1/2)(800)x^2 (1/2)*2*9 -.3*2*9.8(.2+x) = 400 x^2 solve quadratic for x why are you talking about m g h ??? it fell on horizontal plane?
*November 9, 2014*

**Math**

b did both draw Venn diagram fri---both- weekend |20-b | b |29-b| 20-b + b + 29-b = 40 49 - b = 40 b = 9 11 fri , 9 both , 20 weekend
*November 9, 2014*

**Calculus**

You are welcome.
*November 9, 2014*

**Calculus**

Do you mean? y= (x^3+8)/(x+2) ???? or what you typed? If my guess is correct then (bottom dtop/dx -top d bottom/dx)/bottom^2 =[(x+2)(3x^2)-(x^3+3)(1)]/(x+2)^2 =[3x^3+6x^2-x^3-3]/(x+2)^2 = (2x^3+6x^2-3)/(x+2)^2
*November 9, 2014*

**calc**

x y = 50 2 x + y = p 100/y + y = p dp/dy = 0 for min = -100/y^2 + 1 y = 10 x = 5 p = 20 feet of fence
*November 9, 2014*

**Algebra**

if x = 6 then f(x/2) = f(3) = 4 y = 4/2 = 2 (6,2)
*November 9, 2014*

**Algebra**

if x = 3 then f(x) = 4 so y = 4 + 4 = 8
*November 9, 2014*

**Physics, please!**

I used g = -10, you do it with your g
*November 8, 2014*

**Physics, please!**

I guess he kicked it horizontally? Hi = 22 0 = Hi - (1/2)(10) t^2 4.4 = t^2 t = 2.1 seconds goes 17 meters horizontal in 2.1 seconds u = 17/2.1 = 8.1 m/s
*November 8, 2014*

**Physics**

F = mu m g work = F d = mu m g d Ke lost = (1/2) m Vo^2 Vo^2 = 2 mu g d mu = Vo^2/(2gd)
*November 8, 2014*

**Algebra 2**

4x(x-6)-x(x-2)=-24 4 x^2 - 24 x - x^2 + 2 x + 24 = 0 3 x^2 -22 x + 24 = 0 18 and 4 I bet (3x - 4)(x - 6) ??? yes x = 4/3 or 6
*November 8, 2014*

**Algebra 2**

6 x^2 - 13 x + 6 = 0 we want 6 * 2 and a 1 to get 13 or 9 and 4 (3 x - 2)( 2x - 3 ) ah, 9 and 4 x = 2/3 or 3/2
*November 8, 2014*

**Algebra**

speed up = u -c speed down = u+c d = speed * time 3 = (u-c)1.5 3 = (u+c)1 1.5 u - 1.5 c = 3 1.5 u + 1.5 c = 4.5 --------------------- add 3 u = 7.5 u = 7.5/3 3 = 7.5/3 + c c = (9-7.5)/3 = 2.5/3 = .83333333333333333333333333
*November 8, 2014*

**S.S.**

Civil Rights Act of 1964? People are still discriminating against black people, particularly in the South
*November 8, 2014*

**Algebra 2**

we know that 7*7 = 49 and 49+1 = 50 so we know what to do 7 p^2 - 50 p + 7 = 0 (7 p - 1) ( p - 7) = 0
*November 8, 2014*

**Calculus (math)**

I tried method 2 but gave up about an hour ago. I do not think there is any trick, just plug and chug.
*November 8, 2014*

**Calculus**

if you were supposed to use feet instead of meters as your other problem that I just saw, use 32 ft/s^2 for g not 9.81 m/s^2
*November 8, 2014*

**Calculus**

a = g = -9.81 m/s^2 v = -gt = -9.81 t h = Hi - (1/2) g t^2 = Hi - 4.9 t^2 b) h at t = 0 = Hi h at t = 3 = Hi-4.9 t^2 = Hi - 44.1 m/s average = (-44.1)/3 = - 14.7 m/s c) at 0 v = 0 at 3 v = -9.81*3 = -29.4 m/s at 6 v = -9.81*6 = -58.9 m/s d) well I am not about to Google Hi Hi...
*November 8, 2014*

**calculus**

I mean e^x as x --> big negative is 1/e^big positive
*November 8, 2014*

**calculus**

e^-a = 1/e^a
*November 8, 2014*

**calculus**

e^x as x --> big negative is 1/e^big negative
*November 8, 2014*

**calculus**

Thought I answered this -x/e^x e^x >> x for large x
*November 8, 2014*

**Math**

C --> y S --> 3 y J --> 3 y + 50 y + 3 y + 3 y + 50 = 7 y + 50
*November 8, 2014*

**Calculus**

Oh good grief, feet :( old text had initial speed 16 ft/s up ? I guess ? h = 32 + 16 t - 16 t^2 0 = 32 + 16 t - 16 t^2 solve quadratic t = [ -16 +/- sqrt ( 256+2048) ] /-32 t = 2 seconds v = dh/dt = 16 - 32 t = 16 - 64 = -48 ft/s
*November 8, 2014*

**Algebra 2**

4 x^2 + 2 x - 2 = 0 2 ( 2 x^2 + x - 1) = 0 (2x-1)(x+1) = 0 x = 1/2 or -1 5 x^2 + 34 x - 7 = 0 (5x-1)(x+7) = 0 etc
*November 8, 2014*

**Algebra 2**

http://www.gregthatcher.com/Mathematics/GaussJordan.aspx x= 1, y=2, z = 3
*November 8, 2014*

**Calculus**

x --> + oo of -x/e^x
*November 8, 2014*

**Physics**

sqrt (251^2 + 80^2)
*November 7, 2014*

**Physics**

How long to fall 10 meters? 10 = (1/2)(10) t^2 solve for t, time in air then 4 meters = u t solve foe u
*November 7, 2014*

**Science Help!!!!**

Yes, first it is burned in the engine to make CO2 The CO2 is used in photosynthesis to make wood and leaves and stuff, releasing the O2 for us to breathe.
*November 7, 2014*

**physics**

Problem entirely vertical assuming your straight line is horizontal Vi = 10 sin 46 v = Vi - g t delta h = Vi t - (1/2) g t^2
*November 7, 2014*

**CAL**

f'(x) or d f(x)/dx y' or dy/dx maybe?
*November 6, 2014*

**Calculus**

-4 cos^3 x^4 sin x^4 (4 x^3) = - 16 x^3 cos^3 x^4 sin x^4 how that is tan x2 whatever that is I can not imagine
*November 6, 2014*

**Math**

log 9 - log 300 2 log 4 + 4 log 3
*November 6, 2014*

**Science**

F2 + F1 = .450 (.750) F2 - F1 = .240 (.750) -------------------------- 2 F2 = .690(.750) etc
*November 6, 2014*

**Physics**

force down = 9.81*12 max friction force = 9.81 * 12 * mu forward force - friction force = m a 15 - 9.81*12 * mu = 12 (.2)
*November 6, 2014*

**Math**

I do not think so I think h = 300 - 16 t^2 she is accelerating DOWN 100 = 300 - 16 t^2 t^2 = 200/16 t = 10 sqrt 2 / 4 = 2.5 sqrt 2
*November 6, 2014*

**Physics easier**

http://outreach.phas.ubc.ca/phys420/p420_04/sean/
*November 6, 2014*

**Hurricane Project for Physics Class**

Physics (hard) http://www.psfc.mit.edu/library1/catalog/online_pubs/iap/iap2013/emanuel.pdf
*November 6, 2014*

**Hurricane Project for Physics Class**

Physics (hard) http://www.psfc.mit.edu/library1/catalog/online_pubs/iap/iap2013/emanuel.pdf
*November 6, 2014*

**Hurricane Project for Physics Class**

Click on hurricane in the upper right corner
*November 6, 2014*

**Hurricane Project for Physics Class**

http://coast.noaa.gov/hurricanes/#app=c64c&88cd-selectedIndex=1
*November 6, 2014*

**Calculus**

no v = w^2 h = (2300/3) h but h = (2300 -w^2)/4w
*November 5, 2014*

**Calculus**

v = w^2 h A = 2300 = w^2 + 4 w h 4 w h = 2300 - w^2 h = (2300 -w^2)/4w so v = w^2 [ (2300 -w^2)/4w ] v = 2300 w/4 - w^3/4 4 dv/dw = 2300 -3 w^2 = 0 for max w^2 = 2300/3 w = 27.7 cm etc
*November 5, 2014*

**Math**

2 w + 2 L = 18 so L = 9-w A = w L = 9 w - w^2 w^2 - 9 w = -A find vertex by completing the square for max A w^2 - 9 w + (9/2)^2 = - (A-81/4) (w -9/2)^2 = - (A-81/4) w = 9/2 for max area by the way that means 2 w = 9 and therefore 2 L = 9 so it is a square with sides = 9/2
*November 5, 2014*

**Physics**

No. To start it moving you must overcome the static friction. After that it gets easier. 648 * .86
*November 5, 2014*

**algebra**

sum of arithmetic series a1 = 1 d = 2 an = a1 + (n-1) d so a700 = a1 +(699)(2) sum = (a1 + a700)/n
*November 5, 2014*

**calculus careful with parentheses**

do you mean what you wrote 7 + 3 x^.5 or do you mean 3 [x+7]^.5 ??????????? anyway (x+1)(derivative of second) + (second) (1)
*November 5, 2014*

**Calculus I**

using T for theta dR/dT = (Vo^2 /9.81) cos2T (note max range when T = 45 degrees :) ) 100 dT/T = 2 dT = .02 T = .02 * pi/6 dR = (Vo^2 /9.81) cos2T dT 100 dR/R = cos 2T dT / sin 2T
*November 5, 2014*

**physics**

m g h = (1/2) m v^2 so h = (1/2) (v^2/g) mass has nothing to do with it
*November 4, 2014*

**RIGHT TRIANGLE TRIG**

You are welcome.
*November 4, 2014*

**RIGHT TRIANGLE TRIG**

sin 65 = h/150 assuming that the string hypotenuse is straight
*November 4, 2014*

**Physics- linear acceleration**

Find out how far the wagon went in 19 seconds d = (1/2) a t^2 = (1/2)(.3)(19)^2 each revolution or 2 pi radians the wheel rolls 2 pi r meters so divide d by (2*pi*.374) to find how many turns then multiply that by 2 pi to get radians (1/2)(.3)(19)^2 /.374
*November 4, 2014*

**Math, Please Help! I don't understand!**

I guess x is right and y is up final x = initial x + change in x final y = initial y + change in y beginning x = -2 moves one right -2 + 1 = -1 beginning y = -5 moves 3 up -5 + 3 = -2 b) (-1 , -2)
*November 4, 2014*

**Algebra**

f(f(x)] = p(px+q) + q f[f(f(x)) = p[p(px+q) + q] + q = p^2(px+q) + pq + q p^3 x + p^2 q + pq + q = 8 x + 21 = (p^3-8)x + p^2 q + pq + q = 21 this can not be a function of x or it would not be constant 21 so p^3-8 = 0 p = 2 4 q + 2q + q = 21 q = 3
*November 4, 2014*

**Algebra**

1. slope = 3 y = 3 x + b so to find b 2 = 0 + b b = 2 y = 3 x + 2 2. same way but slope = -1
*November 4, 2014*

**Physics**

find change in momentum .145(27+32) force = change in momentum/time = change in momentum /(1.5*10^-3 )
*November 4, 2014*

**Physics**

131 * 9.9 .00932 * 618
*November 4, 2014*

**Calculus**

x^2 - 6 = - y vertex on y axis, opens down (sheds water) so just do the right half half A = x y hA = -x (x^2-6) = -x^3 + 6 x dhA/dx = 0 at max = -3 x^2 + 6 x^2 = 2 x = sqrt 2 include left half length along x axis = 2 sqrt 2 if x = sqrt 2 y = 6 - x^2 = 6-2 = 4
*November 4, 2014*

**Calculus**

Oh, well I guess I found your point :) (15/26,5/13)
*November 4, 2014*

**Calculus**

rewrite as y = -(6/4)x + 1/4 line perpendicular has slope +4/6 = +2/3 y = (2/3) x + b we want it through 0, -1 -1 = (2/3) 0 + b b = -1 so y = (2/3) x - 1 where does that hit original line? (2/3) x - 1 = -(3/2) x + 1/4 4 x - 6 = -9 x + 3/2 13 x = 15/2 x = 15/26 then y = 5/13 ...
*November 4, 2014*

**physics**

change in momentum = .001 (300 - 700) time in block = .08/(average speed = .08 / [(700+300)/2] Force = change in momentum/time in block
*November 4, 2014*

**Physics**

F = m a solve for a then v = 0 + a t to get t: d = 3.3 = (1/2) a t^2
*November 4, 2014*

**Physics**

PLEASE CHECK FOR TYPOS slope of the air slope of air WHAT?
*November 4, 2014*

**least to greatest fractions**

6/9 is 2/3 if that is what you are asking. I have no idea what the second and third lines are about 4/40 is 1/10 2/5 is 4/10 maybe 7/10 is next ? 3/7 is 6/14 8/14 is 8/14 5/7 is 10/14 so next would be 12/14 = 6/7
*November 3, 2014*

**math**

y = 5 x + 100 your brother has the sign of the slope wrong
*November 3, 2014*

**Algebra**

q = [ (x+113)/2 ]^1/3 q = [ (137+113)/2 ]^1/3 q = 125^1/3 = 5
*November 3, 2014*

**Math**

-1.4 b = -28 b = 28/1.4 = 20
*November 3, 2014*

**Algebra**

y = x^5 - 1/3 inverse x = y^5 - 1/3 y^5 = x + 1/3 = -31/96 + 32/96 = 1/96 if y^5 = 1/96 then y =1/2.5 = .401
*November 3, 2014*

**Calculus**

cost of x tables = [160-.5(x-250)]x for x>/=250 c = 160 x + 125 x - .5 x^2 c = 285 x - .5 x^2 dc/dx = 285 - x max when x = 285 then c = .5*285^2 = 40,612.50 and c = 0 when x = 0 of course but also when x = 285*2 = 570
*November 3, 2014*

**Algebra**

f^-1(1) = 2 (when y = 1, x = 2) f^-1(2) = 5 f^-1(5) = 3
*November 3, 2014*

**Physics**

F = m a 12.1 = 5.9 a solve for a d = (1/2) a t^2 2.6 = (1/2) a t^2 we know a, solve for t v = a t
*November 3, 2014*

**Algebra**

x = 3 y^3 + 2 y^3 = (x-2)/3 so y = [ (x-2)/3 ]^(1/3) = f-1(x) so [ (x-2)/3 ]^(1/3) = 4 [ (x-2)/3 ] = 64 x - 2 = 192 x = 194
*November 3, 2014*

**algebra**

4 q = 5 * 18 2 q = 5 * 9 q = 45/2 = 22 1/2
*November 3, 2014*

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