# Posts by Damon

Total # Posts: 29,637

**Interferometry carried to an unimaginable (for me) extreme**

You better Google LIGO, but that is the whole thing. There is no simple reply really.

*October 8, 2016*

**physics**

6 g - T = 17 a T - mu(11 g) = 11 a what is a? d = (1/2) a t^2 1 = (1/2)a(1.48)^2 a = .913 m/s^2 6 g - T = 15.5 so T = 6 g - 15.5 so 6 g - 15.5 - 11 g mu = 10.0 let g = 9.8 and solve for mu

*October 8, 2016*

**Math**

LOL, cost / number of sprays

*October 8, 2016*

**physics**

vertical: Ftop = mg + Tt = m v^2/R Fbot = Tb-mg = m v^2/R so max is Tb = m g + mv^2/R so mg +Tt = Tb-mg 2 mg = Tb-Tt LOL, could have guessed Horizontal T = m v^2/R given Tb = 1.066 T mg + mv^2/R = 1.066 mv^2/R g = .066 v^2/R so v^2 = g R /.066

*October 8, 2016*

**Maths**

one at x = 0 then 10 more at 5, 10, 15, 20, 25 , 30, 35, 40, 45, 50 You always have one more fence post than panels of fencing

*October 8, 2016*

**Maths**

50/5 + 1 = 11 (one is at x = 0)

*October 8, 2016*

**Social Sudies**

fishing if the ^%$&* feds would let you fish.

*October 8, 2016*

**physics**

I do not see the figure but am at a loss to understand how the ledge has anything to do with it unless he hits it. u = 17 m/s until splash Vi = 0 h = 7.9 m h = 4.9 t^2 so t = 1.27 seconds v = 9.81 t = 9.81*1.27 = 12.5 m/s speed = sqrt (17^2 + 12.5^2)

*October 8, 2016*

**precalculus**

T = 3 so f = 1/T = .333 Hz omega = 2 pi f = 2.09 radians/s at t = t0, d = -10 Amplitude = 10 d = -10 cos [ omega (t - t0) ] that is -10 when t = t0

*October 8, 2016*

**Marbles**

26 in bag both times 10/26 * 7/26

*October 8, 2016*

**Stats**

http://davidmlane.com/hyperstat/z_table.html

*October 8, 2016*

**physics**

5*4*4 = 80 6*3*3 = 54

*October 7, 2016*

**Calculus - Approximation**

y = x^(1/3) dy/dx = (1/3) x^-(2/3) at x = 1, y = 1 and dy/dx = (1/3)=m y = 1 + (1/3) (x-1) y = (1/3) x + 2/3 at x = 1.1 y = (1/3)(1.1) + 2/3 y = (1/3) 3.1

*October 7, 2016*

**physics**

north speed = Vn = 150 cos 35 east speed = Ve = 150 sin 35 + 26 speed = sqrt (Vn^2+Ve^2) heading = tan-1(Ve/Vn)

*October 7, 2016*

**huh?**

If you are a multiple of ten you are not a prime number. 30 = 3 * 2 * 5

*October 7, 2016*

**Units?**

130 POUNDS and 49 kg ? I think not

*October 7, 2016*

**Physics**

PE = m g h same m so same gh h = (9.8/3.7)5 in feet

*October 6, 2016*

**Physics**

Well, sort of, but in fact momentum is not conserved in this collision. The first law, conservation of momentum, applies ONLY if there is no external force. That is not the case here. The ground exerts a FORCE on the ball, changing its momentum from +mv to -mv Force = rate of ...

*October 6, 2016*

**physics**

call mass m and not 13.9 to save work, it cancels (1/2) m v^2 = retarding force * d (1/2)m v^2 = .122 m g d (1/2)(1.81)^2 = .122(9.81) d solve for d

*October 6, 2016*

**Calculus 1**

I suspect a square A = L w L + w = constant = k L = k - w so A = (k-w)w = k w - w^2 dA/dw = k - 2w = 0 for max or min so w = k/2 or w = k = k/2 sure enough a square so 1200/4 = 300 = L = w A = 90,000

*October 6, 2016*

**Physics**

pressure = rho g h = 1000 kg/m^3 * 9.81 * 1.2 Force = pressure * area = 1000 * 9.81 *1.2 * .8 /(10,000)

*October 6, 2016*

**math**

360/6 = 60 If there are 6 outside there are 6 inside. You probably want an even number so each has a mate opposite and you want 360/n to be a whole number.

*October 6, 2016*

**Statistics**

http://davidmlane.com/hyperstat/z_table.html

*October 6, 2016*

**Algebra**

-4.5 x - 2 y = -12.15 3.25 x - 1 y = -0.75 mult by 2 -4.5 x - 2 y = -12.15 6.50 x - 2 y = -1.50 ---------------------subtract -11 x + 0 = -10.65 so x = 10.65/11 etc

*October 6, 2016*

**Physics**

m g h same g same h (17.3 / 84.7) 2*10^3

*October 5, 2016*

**Physics**

.08 * 60 * 9.81 Newtons

*October 5, 2016*

**Physics**

pressure = rho g h = 1000 kg/m^3 * 9.81 * 1.2 Force = pressure * area = 1000 * 9.81 cm^2/m^2 *1.2 * .8 /(10,000cm^2/m^2)

*October 5, 2016*

**Physics**

zero

*October 5, 2016*

**Physics**

as it leaves the hand, ME = (1/2)mv^2 and mgh = 0

*October 4, 2016*

**algebra**

take 4.50 off half x and end up with 4.50 (1/2) x = 9 x = 18

*October 4, 2016*

**Physics**

F-mg = m a a) mg b) m(g+2) c) mg d) m(g-1.5) e) mg

*October 4, 2016*

**History**

I agree with you.

*October 4, 2016*

**Maths**

j-3 = 2(x-3) a +3 = 5(j+3) or a = 5 j +12 but j = 2x-3 so a = 5(2x-3) + 12 a = 10 x -3

*October 4, 2016*

**Physics**

a east b v = Vi + a t = 12.8 -(7.7/7.4)9.4

*October 4, 2016*

**Math**

230.12 ----> 200 if it were 250 I would say 300

*October 4, 2016*

**Math**

b works but so does d By the way two things have to happen. a(b+c) = 60 and if ab = 36 then ac must be 24

*October 4, 2016*

**Math**

yes

*October 4, 2016*

**Math**

both ways are fine.

*October 4, 2016*

**glen view**

You mean 20*20*8 - 3200 m^3 ?

*October 4, 2016*

**not at all sure about your answers**

I do not know where to start. I think you had best go back and read the relevant chapters in your text again.

*October 3, 2016*

**Physics**

change in velocity 20+30 = 50 m/s change of momentum = .15 * 50 kg m/s = impulse F = change in momentum / time = .15 * 50 / .01 Newtons

*October 3, 2016*

**PHYSICS**

total mass = 97 + 76 total force = 1176 - 839 so acceleration of system = a = (1176-839)/(97+76) Now look at 97 kg person for example 1176 N to left (for example) T to right net force = (1176-T) a = what we found above so 1176 - T = 97 *a solve for T

*October 3, 2016*

**PHYSICS**

Do this the same way as the tug of war problem.

*October 3, 2016*

**Algebra**

-2 y + 6 = 2 y - 6 4 y = 12 y = 3 So did you try to do it? What is the problem?

*October 3, 2016*

**math**

carbon is 3/25 = 12/100 = 12% No idea. You did not tell me the diameter of the stick.

*October 3, 2016*

**Physics**

well, let us assume he drives 280 km out. he takes 280/40 outward or 7 hr he takes 280/70 back or 4 hr so he takes a total of 11 hours to go 560 km so 560/11 km/hr (you are allowed to pick any other distance. I grabbed a common multiple because I do not have a calculator handy :)

*October 3, 2016*

**Physics**

PE = m g h d PE/dt = m g dh/dt h = 6 t sin 13 dh/dt = 6 sin 13 so d PE/dt = 6 m g sin 13

*October 2, 2016*

**physics-sir damon help**

wind components of speed: north -35 cos 40 = -26.8 east -35 sin 40 = -22.5 heading angle clockwise from N = h north air speed = 220 cos h east air speed = 220 sin h speed made good = v direction = 110 cos 110 = -.342 sin 110 = .940 north = v cos 110 = -.342 v east = v sin 110...

*October 2, 2016*

**Physics**

Note always do orthogonal (perpendicular) components of vectors like x and y or north and west. Do not fool around with northwest, split it up into north component and west component immediately. Otherwise you will be doing trig forever.

*October 2, 2016*

**Physics**

west distance = 63*2.1 + 53 (sqrt2/2)*3.3 = 255.97 right north distance = 53*(sqrt2/2)*3.3 = 123.67 right so 284.28 is correct for the DISTANCE NOT DISPLACEMENT' Distance is scalar , km Displacement is vector , distance AND direction tan (angle north of west) = 124/256 so ...

*October 2, 2016*

**Physics**

a. 40 m/s down potential energy = m g h same h so same potential energy so same kinetic energy (1/2)m v^2 b. how long up 0 = 40 - g t t = 40/9.81 2 t = time up + time down = 80/9.81 a parabola, sheds water y = 40 t - 4.9 t^2

*October 2, 2016*

**physics**

m v^2/r v less --> force less

*October 2, 2016*

**Physics 12**

710 cos 32 horizontal 62*9.81 - 710 sin 32 vertical I doubt it though, tension seems too low

*October 2, 2016*

**Math (Limits)**

Yes, if g(x) = x then if x is pi so is g(x)

*October 2, 2016*

**Math (Limits)**

I do not know what g(x) is perhaps it is g(x) = 15 x + 27 x^15 for all I know in that case g(pi) = 15*pi + 27 * pi^15

*October 2, 2016*

**physics**

NOTE - acceleration is in m/s^2 NOT m/s Its horizontal velocity component does not change since there is no horizontal force u = 4 m/s Vertical Vi = 0 v = Vi + a t v = 0 + 4 * 2 = 8 m/s v = 8 - 12 (1) = -4 m/s so in the end u = 4 and v = -4 speed = 4 sqrt 2 direction = 45 ...

*October 2, 2016*

**Physics**

d = Vi t + (1/2) a t^2 = 20.5 mi/hr * 7.67 s + .5(7.96 mi/hr-s)(7.67^2 s^2) = [20.5(7.96) +.5(7.96)(7.67^2)]mi s/hr * (1 hr/3600s) = [20.5(7.96)+.5(7.96)7.67^2)]/3600 miles

*October 2, 2016*

**Physics**

HUH? What do you mean? if y = x^2 y does not equal m x + b I suppose you could plot log y = 2 log x that is easiest to do on log/log graph paper.

*October 2, 2016*

**Physics**

Now Google Newton's third law !!!

*October 2, 2016*

**Physics**

There is one equation, Newton's second law. Force = rate of change of momentum or if the mass is constant F = m A which is mass times acceleration F = .5kg * 4 m/s^2 = 2 Newtons 2 Newtons = 20 A A = 2/20 = 0.1 m/s^2 If you are taking physics, you better read the chapter!

*October 2, 2016*

**chemistry**

look up heat of vaporization of water.

*October 2, 2016*

**Math (Domain and range)**

also -oo to +oo

*October 2, 2016*

**physical science**

if you go 50 km in half an hour, how far do you go in an hour?

*October 2, 2016*

**Math**

10 boys 20 girls 30 total students 10/30 = 1/3 are boys

*October 2, 2016*

**Science**

If this is a wire or rod or something R = resistivity * length/area so four times the length is four times the resistance

*October 2, 2016*

**maths(Quantitative Method)**

sorry, put 1 on the diagonal, zeros off, to get inverse.

*October 2, 2016*

**maths(Quantitative Method)**

Yes, Steve's method gives you the inverse matrix as well as the solution

*October 2, 2016*

**maths(Quantitative Method)**

+2 -3 +1 +1 -1 +1 -4 +3 +3 -1 +0 +2 I assume this is from |+2 -3 +1| |a| +1 |-1 +1 -4| |b|=+3 |+3 -1 +0| |c| +2 solve for a b and c google gauss jordan

*October 2, 2016*

**AP physics**

direct straight line through origin with slope b AP Physics?

*October 2, 2016*

**Physics**

just do thee same old hanging problems with g = 9.81 + 1.7 or g = 9.81 + A obviously tension higher in top string

*October 2, 2016*

**Physics**

tension = T newtons bottom block: 6.7 g - T = 6.7 a top block on table: T = 4.1 a so 6.7(9.81) - 4.1 a = 6.7 a etc

*October 2, 2016*

**Physics**

weight ccomponent down slope F = m g sin 13.2 so m g sin 13.2 = m a so a = 9.81 sin 13.2 ( what else is new?) d = (1/2) a t^2 1.7 = (1/2) a t^2 solve for t then v = a t

*October 2, 2016*

**Physics**

force = change in momentum / time average speed = s = 10^5(6+2.6)/2 so time = t = .044/s F = 9.11*10^-31 (6-2.6)10^5 / t compare to m g

*October 2, 2016*

**Math**

x = e^(2y+4) - 4 ln(x-4) = 2y+4 y = [ ln(x-4) - 4 ] / 2 x better be bigger than 4

*October 2, 2016*

**Physics**

north: 3 -1 +2 +2

*October 2, 2016*

**Physics**

it is not accelerating therefore: Sum of forces in the x direction = 0 Sum of forces in the y direction = 0 Sum of moments around any point = 0 by the way 7 kg down is 7*9.81 Newtons down and tension in the line

*October 2, 2016*

**Physics**

1. Like the gravitational field of the earth if the object is the moon or the friction of tires on road if a car in a turn :)

*October 2, 2016*

**mathematics**

If I understand you first person gets x second gets .5 x third gets .25 x 1.75 x = 3500

*October 2, 2016*

**differential equation**

Oh, yes, the diff eq which everyone already knows the solution for da/dt = -ka da/a = - k dt ln a = -kt a = A e^-kt

*October 2, 2016*

**please type the whole problem**

How many after 2.5 ? x = Xi e^-kt x = 75 e^-2.5 k ln(x/75) = -2.5 k k = ln(x/75)/(-2.5) solve for k so ln .5 = -k t solve for t

*October 2, 2016*

**Physics**

m v^2/r

*October 1, 2016*

**Physics**

angle pushes car toward center, centripetal

*October 1, 2016*

**physics**

you do not say what the question is average rpm during stop = 1150 turns to stop = 1150 *2.9/60 original radians/s = 2300 revs/min*2 pi rad/rev *1min/60 s = 241 radians/s angular acceleration = -241/2.9 whatever

*October 1, 2016*

**Math**

10 x + 15 y = 2500 y = 195-x 10 x + 15(195-x) = 2500 10 x + 2925 - 15 x = 2500 5 x = 425 x = 85 y = 195-x = 110

*October 1, 2016*

**Math**

70*400 = 28,000

*October 1, 2016*

**Pre Cal help please!!**

given f(x) = { 2x^2 + 5 x < or equal to 2 3 - x^2 x > 2 What???? Perhaps you mean: given f(x) = 2 x^2 + 5 when x < or equal to 2 3 - x^2 when x > 2 if x is slightly bigger than 2 3 - 4 = -1 if x is slightly smaller than 2 OR if x is 2 then 2(4) + 5 = 8+5 = 13

*October 1, 2016*

**Physics**

24.5 x = 85.5(11.6-x)

*October 1, 2016*

**Physics**

I guess an intercontinental ballistic rocket is something "thrown over a large distance (on earth)" so 5. What a nutty question though :)

*October 1, 2016*

**maths**

(x^3+y^3)(x^3-y^3) (x^3+3x^2y+3xy^2+y^3)*(x^3-x^2y+3xy^2-y^3)

*October 1, 2016*

**Trigonometry**

in an hour he goes 1/24 of circumference (1/24)(2 pi *12,000/pi) = 24,000/24 = 1000

*October 1, 2016*

**Physics**

fast one takes t = 76/5.2 seconds slow one goes d = 2(76/5.2) so 76 - 2(76/5.2)

*October 1, 2016*

**simpson**

http://www.wolframalpha.com/input/?i=solve+dy+%3D+sin(x%5E2)+dx

*October 1, 2016*

**Physics**

Ac = v^2/r = 36/2 = 18 m/s^2 m F = M Ac = 51 *18 Newtons weight = 51 * 9.81 Newtons almost 2 g

*October 1, 2016*

**value**

change of momentum = .03*600 = 18 average speed = 300 time = .2/300 Force = change of momentum/change of time = 18 *300/.2 Newtons = 90*300 =27,000 agree

*October 1, 2016*

**maths sir-reiny help calculus**

Click on the step by step in Wolfram

*October 1, 2016*

**maths sir-reiny help calculus**

http://www.wolframalpha.com/input/?i=solve+x+y+dx+%3D+(x%5E2+-x+y+%2By%5E2)dy

*October 1, 2016*

**Algebra**

first scoop has 19 choices second has 18 choices cone has 3 choices 18*19*3

*October 1, 2016*

**Probability**

first ace 4/54 second ace 3/53 third ace 2/52 first king 4/51 second king 3/50 multiply them

*October 1, 2016*

**whoops**

30/54

*October 1, 2016*

**sorry**

30/54

*October 1, 2016*