# Posts by Damon

Total # Posts: 30,129

**Chemistry**

Balance your mols 2C2H2 + 5O2 --> 4CO2 + 2H2O PV=nRT n is mols V is volume P R T all the same for each

**sc chn kh needs genetics expertise **

I can not help with this.

**PreCal**

circumference = 2 pi R

**Physics**

as it crosses pole average speed = -20 m/s change in speed in one second = a(1) = -10 m/s so speed at top of pole = -15 m/s so how far did it fall to get to -19.5 -15 = -10 t t = 1.5 d= (1/2) g t^2 = 5(1.5)^2 19.361 h = (1/2)g t^2 = 5 * 2.25 = 11.25

**Phyisic**

amount dissolved (solubility) DEPENDS on temperature

**Math**

I do not know what a giant ine is but 2/8 + 5/8 = 7/8

**Maths**

http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html XN = 79+d(N-1) = -5 814 = N/2[ 2(79) + (N-1)d ] 814 = N/2 [ 79 -5 ] N = 1628/74 N= 22 then -5 = 79 +d(21) d = -84/21 = - 4

**Physics**

same m g h different m g h / t

**Physics**

M1 V1 + M2 V2 = ( M1 + M2) Vfinal You do not tell me if V2 is positive or negative, stern ramming or head on collision. initial ke = (1/2)M1 V1^2 + (1/2)M2 V2^2 final ke = (1/2)(M1+M2) Vfinal^2

**Physics**

v = k T^a M^b m/s = (kg m/s^2)^a (kg/m)^b m^1/s^1 = kg^a kg^b * m^a /m^b * 1/s^2a looks like a = .5 for s^-1 = s^-2a b must be -a for kg^(a+b) =kg^0 = 1 so b = -.5 now m m^1 = ?? m^.5/m^-.5 = m^(.5+.5)= m^1 yes so v = k T^.5 M^-.5 v = k sqrt (T/M)

**Math**

208.40 - x = 4 (150.20 - x)

**Physics**

at the top the velocity is indeed zero. so if you know the fall time, t" then v at the ground = - g t" the - sign is because v is down. YES, t = t' by symmetry it takes just as long to deaccelerate to the top as to accelerate down to the starting point. NO, t&...

**Physics**

assume hand at 17 meters height. t = rise time indeed t' = time to fall from top to 17 meters t" = time to fall from top to ground

**Physics**

suggest you check by doing it my way.

**Physics**

a = - g = -9.81 m/s^2 v = 15 - 9.81 t v = 0 at top so solve for t at top H = max height = 17 + 15 t - 4.9 t^2 now it falls from height H 17 = H - 4.9 t'^2 solve for t' , the fall time from top to 17, v here = -9.81 t' add t to t' to get flight time so far 0 = H...

**Physics help!**

f = k Q1 Q2 / d^2 change in Ke = 1/2 m v^2 = change in U = k Q1 Q2 (1/r2 -1/r1) r1 = 2*10^-14 + 8.22*10^-13 r2 = 2*10^-14

**Physics help!**

volume of drop = V = (4/3) pi r^3 mass of drop = m = 1000 * V gravity down = F = m g force up = Q E = F number of excess electrons = Q /e

**Science**

What would happen to your car engine if you connected the output of the oil pump directly to the inlet without routing it to the bearings etc.?

**Math**

-2 h^2 x / -4 h^2 = x/2

**7th Grade Math**

....and you should know that gravitational acceleration on the moon is about 1/6 that on the earth surface. Therefore earth weight = 6 times moon weight for the same mass.

**Math**

Rachel J/5 Rena (1/6)(4J/5) = 2 J/15 now remaining J - J/5 - 2 J/15 =15 J/15 - 3 J/15 - 2 J/15 = 10 J/15 = 2 J/5 now Rachel eats 2/5 *2 J/5= 4J/25 now have 2 J/5 - 4J/25 = 6 J/25 now Rena eats 2/3 so 1/3 * 6 J/25 = 2 J/25 left half of that = J/25 so 2 = J/25 J = 50

**Precalculus 11**

no actually you are multiplying first by 4b/4b and the second by 7a/7a so (4b/4b)[(8a-3b)/7a]-(7a/7a)[(2a+5b)/4b] ={4b[(8a-3b)]-7a[(2a+5b)]} /28ab

**Physics**

t = actual time racing along the road 22/60 = .367 hour tend - tbegin = (t +.367)hr and d/(t+.367) = 77.8 but d/t 89.5 so d = 89.5 t and 89.5 t/(t+.367) = 77.8 89.5 t = 77.8 t + 28.5 t = 2.438 hr driving t+ .367 = 2.805 hr = 2 hours and 49 minutes

**PHYSICS**

v = Vi + a t so at top 0 = 20 - 9.81 t solve for t at top T = time in air = 2 t h = h of hand + 20 t - 4.9t^2 solve for h v at hand = -20 by symmetry I do not know how high his hand was so can say nothing about ground

**physics**

radius = R = .5 m at top, m g = m v^2/R for contact force = 0 sso v^2/R = g so v = sqrt (g R) at top of circle which is height 2 R potential energy at top of plane = m g h potential energy at 2 R = 2 m g R loss of potential = m g h - 2 m g R = mg (h-2R) = Ke at top of circle...

**Algebra**

one flip in f(x) from 5 x^2 to - 6 x second from -6 x to +9 no flips in f(-x) = 2x^4 + 5x^2 + 6x + 9 so two or zero positive real roots check with http://www.emathhelp.net/calculators/algebra-1/descartes-rule-of-signs-calculator/

**math**

n , (n+2) 4/n - 1/(n+2) = 5/6 24/n - 6/(n+2) = 5 24(n+2) - 6 n = 5n(n+2) 24 n + 48 - 6 n = 5 n^2 + 10 n 5 n^2 - 8 n - 48 = 0 (5 n + 12)(n-4) so n = 4 n+2 = 6

**Algebra**

statement: reasons: Ray AC bisects <BAD (Given) m<BAC = 13x (Given) m<CAD = 6x + 21 ****GIVEN **** m<BAC = m<CAD (Definition of angle bisector) 13x = 6x + 21 SUBSTITUION OF VALUES 13x - 6x = 6x - 6x + 21 (Addition/Subtraction Property) 7x = 21 (Substitution ...

**Earth Science (check answers)**

A is true, and I am sure that is what her aunt said. However the question is what is NOT true.

**Earth Science (check answers)**

Yes, what I said agrees with her aunt luckily. Although my undergraduate degree is in physics my doctorate is in engineering and I was an engineering professor at MIT :)

**Earth Science (check answers)**

1. NO! it is not B The atomic number is number of protons. The atomic mass is protons plus neutrons. The mass of electrons is tiny in comparison 2. If they were less stable in combined form why would they combine?

**World History (Check answer)**

Yes, roads !

**math**

were the elements distributed uniformly in the set? Anything from 0 to ten

**Math**

So where is your attempt to do the problems?

**typo**

a = F/100

**Physics**

F = .4 * 100 * 9.81 a = 100 v = 15 - a t so t = 15/a average speed = 15/2 = 7.5 d = 7.5 t

**Science**

sqrt(9+16+144) 13

**Physics**

m v^2/r + m g = 7 m g v^2/r = 6 g r = (210)^2/(6*9.81) r = 749 meters

**Science**

1/f = 1/do + 1/di 1/20 = 1/30 + 1/di 60/20 = 60/30 + 60/di 3 = 2 + 60/di di = 60 cm

**Science**

C at 40 F at 20 object at 30, halfway between C and F so image is BEYOND C http://www.physicsclassroom.com/class/refln/Lesson-3/Image-Characteristics-for-Concave-Mirrors

**statistics**

95.45 % see: http://davidmlane.com/hyperstat/z_table.html I have no idea what the rule is.

**Physics**

height h at angle T = R - R cos T height h at 60 deg = R/2 height h at 120 deg = 3 R/2 kinetic energy (1/2)mv^2 + m g h = constant at 60 deg (1/2)v1^2 + g(R/2)=(1/2)v2^2 + g(3R/2) (v1^2 -v2^2) = 2 g R if it just makes it to the top then (1/2) m v1^2 = m g * 2R v1^2 = 4 g R (4 ...

**Physics**

If it is to make it around the circle then the velocity at the top must be positive. (1/2)m v^2 >/= m g (2R) if it is running around a track and can fall, then the centripetal acceleration at the top must be 1 g at least. mv^2/R >/= m g You do not say which.

**physics**

Ce is temp read error = e = Ce - C I assume error = e = m C + b then .7 = m(0) + b so b = .7 99.5 - 100 = -.5 = m(100) +.7 100 m = -1.2 m = -.012 at Ce = 30 e = 30 - C = -.012 C + .7 C-.012 C = 29.3 .988 C= 29.3 C = 29.7

**Physics**

Well the speed of light changes when passing through material, like air or water or glass, but not in a vacuum. What does temperature mean in a vacuum? Something really, really massive nearby can have an effect (Einstein, relativity, gravitational lens)

**Physics**

Huh ? Please Google speed of light in a vacuum.

**Physics**

40 = u T so T = 40/u v = Vi - 9.8 t at top v = 0 t = Vi/9.8 at top. This is at time t = T/2 = 20/u so 20/u = Vi/9.8 u Vi = 196 now max range at 45 degrees so u = Vi Vi = sqrt 196 = 14 = sqrt(2 g h) 196 = 2 * 9.8 * h h = 10 meters why is range max at 45 degrees? http://www....

**physic**

particle 1 x = -21 + 8 t particle 2 x = 7 + 4 t so -21 + 8 t = 7 + 4 t 4 t = 28 t = 4 seconds x = 7 + 16 = 23

**huh**

Sorry, I do not think that A could make it back to the starting point. (even if no friction you would have to put external forces on)

**Finance - Family Planning**

we would get .75 * 7 % which is 5.25% if we chose the taxable account, so take the 6%

**Economics - Finance**

By the way, die right away to get the maximum if it stops at age 20. Best make it for 20 years after you die.

**Economics - Finance**

well, say we have the present value of an annuity. Usually the value goes up at interest rate r but in this case we have only inflation, negative interest (unrealistic of course, presumably your successor would deposit the insurance proceeds in an interest bearing account, ...

**physics similar questions**

coincidence? http://www.jiskha.com/display.cgi?id=1501514173

**physics**

in 2 seconds velocity decreases by 64.4 ft/s a = change in v/change in time = -64.4/2 = -32.2 ft/s^2 (which by the way is about g on earth) v = 67.8 - 32.2 (t-1) v = 100 -32.2 t

**physics**

(vi+v3)/2 = 12/3 = 4 and (v3+v6)/2 = 30/3=10 so v3 =8-vi and v3 = 20-v6 and v6=vi+6a so 8-vi = 20 -vi -6a or 6a = 12 a = 2 then vi + 3(2) = v3 and vi+v3 = 8 so vi+6 = 8-vi 2 vi = 2 vi = 1

**physics**

x = A cos (2 pi f t + pi} but cos (a+b) =cos a cos b -sin a sin b so x = A[ cos (2pift) cos pi - sin (2 pi f t) sin pi ] but cos pi = -1 and sin pi = 0 so x = -A oos 2 pi f t x = -.55 cos 35 t 2 pi f = 35 = omega 2 pi/T = 35 f = 35/2pi v = dx/dt = .55(35) sin 35 t v = 0 at t...

**physics**

T = sqrt(L/g) 5T = sqrt (4 L/gm) 5 sqrt (L/g) = sqrt (4 L/gm) 25 L/g = 4 L/gm gm = g [ 4/25 ]

**physics**

A = |A| cos thita i + |A| sin thita j B = x i + y j A dot B = 0 for perpendicular |A| x cos thita + |A| y sin thita = 0 if x = -sin thita and y = cos thita that works (also the opposite of course)

**physics**

average speed during stop = 15 m/s time to stop = 30/15 = 2 seconds a =change in speed/change in time= -30/2=-15 m/s^2

**Physics**

mu m g = 20 N mu (4)g = 20 mu = 5/g so about 0.5

**Maths,physics,biology,english,arabic,chemistry**

500 mL is HALF a liter. Therefore .5 + 40 = 20 grams

**Science**

http://socratic.org/questions/what-is-the-balanced-equation-of-c6h6-o2-co2-h2o

**diploma applied physics1**

3 in x direction 2 /sqrt 2 in x and 2/sqrt2 in y x force = 3 + 2/sqrt2 y force = 2/sqrt 2 }F} = sqrt [ (3 + 2/sqrt2)^2 + 9 ] tan angle above x axis = [ (2/sqt2)/(3 +2/sqrt2)]

**diploma applied physics1**

angle between = 2A F cos A + F cos A = F cos A = 1/2 angle we want = 2 A

**math**

total cents = 5 n + 25 q divide by ten ( 5 n + 25 q )/10 if the answer is an integer plus a fraction, truncate and just use the integer unless you have a saw to cut the next dime up.

**Math**

1 in^2 = 5 mi * 5 mi = 25 mi^2 2.5 in^2 (25 mi^2/in^2) = 62.5 mi^2

**Physics**

constant velocity no acceleration so F = m g Tension in cable = 400 * 9.80 = 3920 Newtons power = force * speed = 3020 * 0.6 m/s = 2352 Newton meters/sec or Watts

**stop it please**

I do not see your drawing and do not know what your geometry is. Is the beam up at 45 from horizontal from a hinge at the wall. Is cable AB attached to the wall above and at 30 from the wall? is the 1200 at point B? is the 300 at beam center? I have no idea.

**math**

or 0 = 25 - 32 t t = .781 h = 5 + 25(.781) - 16(.781)^2 = 14.8

**math**

how about complete the square to find the vertex? 16 t^2 -25 t = -h + 5 t^2 -1.5625 t = -(1/16)h + .3125 (t - .78125)^2 = -(1/16 )h + .3125 +.61035 (t - .78125)^2 = -(1/16 )h + .923 (t - .78125)^2 = -(1/16 )(h - 14.8) t = .781 h = 14.8

**Science**

v =vi - 9.81 t at top 0 = 15 - 0.981 t t = 15/9.81 = 1.53 seconds rising h = 17 + 15(1.53) -4.9(1.53)^2 = 17 + 22.9 - 11.5 = 28.5 meters maxh how long to fall from there and hoe fast does it hit? 0 = 28.5 - 4.9 t^2 t falling = 5.81 seconds so totsl time = 5.81+1.53 = 7.34 s ...

**math**

1 -- * * * * * 6 3 -- * * * 4 * * 5 -- * 2 * * * * 3/(3*6 ) = 1/6

**Physics**

average speed = 402/4.7 = 85.5 m/s top speed = 2*85.5 = 171 m/s a = (171-0)/4.7 = 36.4 m/s^2

**physics**

approach speed = 60 + 40 = 100 km/hr so crash in 4 hours 60*4 = 240 km from T1

**Physics**

Maybe you mean: g = 4 pi^2 (L/T^2) simple pendulum length L, all mass m at end, small angle m g L theta = m L^2 omega^2 theta omega^2 = g/L but omega = 2 pi f = 2 pi/T 4 pi^2/T^2 = g/L g = 4 pi^2 L / T^2

**Physics**

Either way :)

**Physics**

acceleration = a = [ (2234-0) res/60s ] /.8s = 46.54 revs/s^2 revs = (1/2)a t^2 = (1/2)(46.54)(.64) = 14.9 revs

**Physics**

g = v^2/R

**Calculus**

I guess maybe assume straight shoreline running NW to SE

**Algebra 2**

in general though see: http://www.mathsisfun.com/algebra/matrix-determinant.html

**Algebra 2**

you could do (1*-3) - 7(0) = -3

**Calculus**

slope = 5 at x = 2 so y = 5 x + b -4 = 5 (2) + b -4 = 10 + b b = -14 so y = 5 x -14

**Math**

47+29 = 76 first one 47/76 second on 46/75 third one 45/74 fourth one 43/73 multiply :)

**math**

I assumed you put the first card back. If not then it becomes 12/51

**math**

The first draw is some suit. The probability that the second draw is that suit is 1/4 = 0.25

**Physics**

55 degrees with what?

**Physics 30 meter rod?**

Please proofread. Where does the string attach to the rod? What on earth do you mean by normal force on a string? Strings are usually in tension.

**Physics**

top at 3/2 = 1.5 s = t v = Vi - g t 0 = Vi - g t so g t = Vi = 9.81 * 1.5 = 14.7 h = Vi t -4.9 t^2 = 14.7 *1.5 - 4.9(2.25) = 22 - 11 = 11 meters note, same as average speed = 14.7/2 times 1.5 seconds

**that depends**

on what the total current is current = i = electron charge * electrons/second so electrons/second total = i/Qe per unit area i/(Qe area)

**Quadratic equations**

4.9 t^2 -24.5 t -29.4 = 0 t^2 - 5 t - 6 = 0 (t-6)(t+1) = 0 t = 6 seconds (the -1 second is before release :)

**10 volts * C = q**

but C times ten volts = q per capacitor

**physics**

C = q/V so q = C V so each stores 2.5 * 10^-12 C so we need 1.2*10^-6/2.5*10^-12 = .48*10^6 =480,000

**Physics - Vectors/Forces**

(6/10) 10 g = 6 g = 6*9.81 = 58.86 Newtons

**Math**

http://calc101.com/webMathematica/long-divide.jsp#topdoit 3 x - 1 and remainderr -4/(x+2) draw a diagonal

**Math**

[ 6*3 sqrt 3 - 3 sqrt 3 /4 ]^2 9*3 [6-1/4]^2 27 * 33.06 893

**Physics**

75 * g * 0.60

**Math**

5 * 5 = 25 .6 * 25 = 15 x^2 = 15 x = 3.873 meters

**Physics**

By the way those of us who do navigation have trouble with x, y rather than east, north

**Physics**

initial east,x, momentum = 7150(5) -1650(20)cos 30 = 7,171 initial north,y, momentum = 0 - 1650(20)sin 30 = -16,500 Those will not change during the collision so tan of angle south of east = 16500/7171 angle south of east = 66.5 deg Their answer, is 293.5 COUNTERCLOCKWISE from...

**Math**

15 x -20 y = 40 15 x + 9 y = 69 -------------------subtract -29 y = -29 y = 1 3 x - 4 = 8 3 x = 12 x = 4 x+y = 4 + 1 = 5