Posts by Damon
Total # Posts: 31,285
Where is your attempt to solve this problem? Where have you listed: specific heat of Cu specific heat of water heat of fusion of water heat of vaporization of water I assume the steam is at 100 deg C
Google specific gravity
d is distance apart call x distance from 4 M then (d-x) is distance from M 4M/x^2 = M/(d-x)^2 4(d-x)^2 = x^2 4(d^2 -2dx + x^2 ) = x^2 4 d^2 -8dx +3 x^2 = 0 (2d-3x)(2d-x)=0 x = 2 d (beyond M away from 4M) x = (2/3) d (between them but closer to M)
Maybe Nickie meant to type log base 8 x^3 = 4 then x^3 = 8^4 x^3 = (2^3)^4 = 2^12 x = 2^4 = 16
Universal gravitation, fields
no, 1/3 of distance from little one, 2/3 from 4M 1/(1/3)^2 = 4/(2/3)^2 9 = 4 /(4/9)
Did you change the length of the sides?
Chemistry - Liquid Methane?
https://courses.lumenlearning.com/wsu-sandbox2/chapter/intermolecular-forces/ Look particularly at Figure 5 and related text.
Yes, of course it has to be solid. Also: The weight of water displaced by the ice when you stand on it must be at least as much as the weight of the ice floe plus your weight. That defines how big the rigid chunk of ice has to be (assuming you are far from shore in deep water...
The Global Ocean
You are welcome.
The Global Ocean
Yes, (I live in a fishing port. We appreciate the effect. Cold water holds more oxygen. :) https://www.google.com/search?q=upwelling+definition+oceanography&ie=utf-8&oe=utf-8&client=firefox-b-1-ab
The Global Ocean
You are correct, although in fact the question is not very well phrased. Water particles in deep water waves (water depth greater than half a wavelength) describe circular orbital paths, decreasing exponentially in radius with depth. The particle velocity is up on the front (...
In ten words or less?
How on earth do you expect us to respond to this? Try to write something and perhaps someone here will comment on it.
If you can do the midpoint you can do the rest. Draw pictures.
Assume g is the same still or moving :) W = m g
Calculus - Damon today at 3:57pm P = 84.3 e^(6.5 t) + 415.7 if t = 1 P(1) = 84.3 e^6.5 + 415.7 = 56,487
been there, done that
Please check where I did this below
graph them first zeros (x-4)(x-1) = 0 x = 1 and x = 4 and x = 1 and x = 1 (vertex on x axis, in lower quadrants(sheds water) SO We are looking only at some region between x = 1(where they intersect, and somewhere between x = 1 and x = 4 where they hit again Where is that? x^2...
You are welcome.
F = m A so m = 20 N /12.5 m/s^2 in kilograms
P = 84.3 e^(6.5 t) + 415.7 if t = 1 P(1) = 84.3 e^6.5 + 415.7 = 56,487
dP/dt = 548 e^(6.5 t) ???? if I understand you that means P = (548/6.5) e^(6.5 t) + constant but we know when t = 0, P = 500 so 500 = 84.3 + c c = 415.7 so P = 84.3 e^(6.5 t) + 415.7 when is it 1500? 1500 - 415.7 = 84.3 e^(6.5 t) 12.86 = e^6.5 t ln 12.86 = 6.5 t
Just did this, look below
I already told you.
he makes 2400 + 0.10(s) so 3,000 </= 2400 + 0.10 s 0.10 s >/= 600 s >/= 6,000
You said: "...he cuts at most 10 yards per week..."
Given 4 possible answers to each of five questions, what is the probability that I will guess all five incorrectly?
Really Ancient Egyptian History
I do not know enough to help, but based on what little I know your answers seem to be pretty much random guesses. Nefertiti had a 12 meter America's Cup yacht named after her, but I am not sure about the rest.
V= i R 12 = 4 R sure enough 3
There is no force to change the horizontal component of velocity. u = 40, like period. If the helicopter continues at the same speed and heading, the package will hit exactly beneath it. When you drop a bomb, turn :) It falls for 3 seconds v = g t = 9.81 * 3 meters/second
North= 175 sin 30 + 150 cos 20 + 190 *0 East = 175 cos 30 - 150 sin 20 - 190 distance = sqrt(North^2+East^2) tan angle = North/East note, negative, quadrant 2, angle is angle above West direction
1 B 2 D assume you mean Wir 3 A 4 B 5 C
tan (angle) = y/x = -2.3/5.3 In fourth quadrant so angle = -23.46 degrees about 23 degrees below +x axis
Now wait a minute. Are you just posting random questions to keep a bunch of retired teachers occupied?
How much heat you have to put into a unit mass of something to raise its temperature one degree. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html
momentum vector P = m V where m is the scalar mass and V is the vector velocity.
cos A = 50/56 find A then sin A = h/56
.25 x + .65 y = .57 * 50 x+y = 50
water mass = 1000 kg/m^3 * volume volume of 10.3 high by 1 meter square base = 10.3 m^3 so mass of water sitting on that square = 10.3*10^3 kg weight = m g = 9.81*10.3*10^3 = 101 *10^3 = 1.01*10^5 N/m^2 or Pascals do the HG the same way.
15 Nm if at 90 degree angle. Otherwise all bets are off.
You mean graph y = 2x^3 + 9x^2 + 4x - 15 I suspect http://www.wolframalpha.com/widgets/view.jsp?id=eddad051a15adadc20c85a7b1b455fa1
who cares what the mass is? period of long rod pendulum = 2 pi sqrt (2(0.17)/3g) period of regular pendulum with mass at end = 2 pi sqrt (L/g) so L/g = 0.17 * .667 / g L = .113 = 11.3 cm
v at ground = sqrt(2gh) (1/2) m v^2 at ground = (1/2)m(2gh) keep (3/4) of that so Ke = (3/8)m(2gh) that will be spent going back up to Hend m g Hend = (3/8) (m g) (2h) so Hend = 6/8 h = (3/4) h = 30 m of course :) If it keeps 3/4 of the energy it will go 3/4 of original height.
Forces Within and Deciphering Earth's History
I am inclined to agree, 10^9 years.
well, first the roll d = (1/2)a t^2 50 (1/2)(4) t^2 t^2 = 25 t = 5 seconds conveniently speed = a t = 4*5 = 20 m/s at start of dive so u = 20 cos 24 for the whole trip down =18.3 m/s Vi = -20 sin 24 = initial vertical speed = - 8.13 m/s H = Hi + Vi t - 4.9 t^2 0 = 30 - 8.13 t...
x = 1 sin (2 pi f t) v = 2 pi f cos(2 pi f t) so 2 pi f = 10
3x = 21 x = 7 7x = 49
cows/ducks = 3/8 cows/24 = 3/8 cows = 3 * 24/8 = 9
remember startedd out backwards
a = 2 t + 2 so v = t^2 + 2 t + constant -3 = 0 + 0 + constant so v = t^2 + 2 t -3 x(t) = (1/3)t^3 + t^2 - 3 t + constant let x(0) = 0 so x(t) = (1/3)t^3 + t^2 - 3 t at t = 5 x = (1/3)(125) + 25 - 15 = indeed 155/3 = 51 2/3 HOWEVER v was - at start. It asked for the distance, ...
NO - I did this last night The thing moves backwards from zero then forward again You found the displacement, not the distance.
Oh, ice floats? :)
LOL, I bet it is one kilogram :)
I get .619c 6.51 = 5.11/sqrt(1 -beta^2) sqrt(1-beta^2) = .785 1-beta^2 = .616 beta^2 = .384 beta = v/c = .6195 or v = .62 c
Bombay(Mumbai) faces the Indian ocean, west I see a big bridge across the Zambezi river: https://www.google.com/search?q=google+map+Zambia&ie=utf-8&oe=utf-8&client=firefox-b-1-ab Yes, Victoria Falls https://www.google.com/search?q=victoria+falls+bridge&ie=utf-8&oe=utf-8&client...
So far :)
SD is what???
standard deviation of 12. 6. 7 <-----???? anyway go here: http://davidmlane.com/hyperstat/z_table.html
starts at 55m but call it zero because relative to top of tower is asked v = 4.85 - 9.81 t h = 4.85 t - 4.9 t^2 = 4.85*2.31 - 4.9(2.31^2) = 11.2 - 26.1 = -14.9 about 15 meters below top of tower
You have got to be kidding
Even if I had the math model of the vehicle and motors, this would be take hours. You are to make body angle proportional to linear or angular accelerations, velocites and magnitudes it seems. You seem to have a math model, put in your force or torque proportional to those ...
conserving momentum (Initial is zero): 2.55*10^6 v + 254 (100) = 0 v is negative because if it fires out the bow, the sub moves backwards.
Math (Please help!)
You are welcome. Practice factoring !
Math (Please help!)
25/45 and 15/27 ? (25/5) / (45/5) = 5/9 (15/3) / (27/3) = 5/9 so The two fractions are the same.
horizontal force component = 775 cos 20 so a = F/m = (775/67.5)cos 220 meters/ second^2 force*time = impulse = momentum change = m v so 775 cos 20 * 0.340 = 67.5 v
because it is log base10 (H+ ion concentration) If 2 = log H+a - logH+b then what is the ratio of H+a to H+b? 2 = log (H+a/H+b) because subtracting logs is division https://www.thoughtco.com/definition-of-ph-in-chemistry-604605https://www.thoughtco.com/definition-of-ph-in-...
Where is your attempt to do these problems? They are straightforward if tedious applications of the binomial distribution.
yes, ratio between corresponding sides is 4
3 x = n x - (1+2+3 ...... n) 1+2+3....+n is arithmetic sequence https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html a = 1 and d = 1 sum = (n/2)(2 + n-1)= (n/2)(1+n) = .5(n^2+n) so 3x = nx -.5(n^2+n)
t 00 02 04 06 08 10 12 14 T 73 73 70 68 73 80 86 89 There are a million ways :) One way: T(10)= 80 T(6)= 68 change in T = 12 change in t = 4 delta T/delta t = 3 degrees/hour average rate of change between 6 am and 10 am is used as slope at 8 am
p under = .4 p not under = .6 p all five not under = .6^5 add p of 2 under + 3 under + 4 under + 5 under or more easily 1 - (p 0 under + p 1 under) p 0 under = .6^5 = .0778 p 1 under = C(5,1) * .4^1*.6^4 =.2592 so 1 - .2592 -.0778 = .663
WHAT ?? You are not thinking about the questions. How much industry and resulting pollution do you have in the desert or jungle ?
No. I told you how to think about this. If the tribes are not getting along it is hard for the nurses to help.
I personally would pick D
But you said eliminate, not pick.
If you are supposed to eliminate one, eliminate B. It makes no sense at all.
Wait a minute. I missed that word "eliminate" Indeed eliminate B. It does the opposite off encourage trade.
It asked what HELPS trade, not what stops it !
Well it is hard to grow food in the Sahara and the nomadic people there tend to be wicked tough. The jungle tribes are kind of scary as well.
I would not.
Well, oil/methane is mainly Middle East although a factor in some North African countries but not much lumber there. Coal? I associate coal mining more with the Northern Hemisphere. If you live near the Sahara or in the Sub Saharan jungles I suppose land and water would be ...
I have no idea where you are at school or what your test is about. Moreover our job is to help students learn, not take tests for them.
I think the worst is when the country splits up into warring factions and people are not able to communicate and medical help and information can not get to them. How many countries in Africa with an HIV/AIDS crisis export oil?
I am inclined to agree with you.
SS- I think this got deleted
B seems to fit the sentence descriptions best.
SS- I think this got deleted
Where did it say parliament in the sentence about Nigeria?
multiply both sides by 4
I suppose so. Certainly if you look at the size of Algeria and South Africa and Egypt and ......
That is right. You will get a different number of half waves inside there and it depends what you call m = 1 and m = 0. I would use the convention your teacher used. If you use the + sign then m = 0 has half a wavelength in the film but if you start at m = 1 you want the - ...
The point is that it differs by half a wavelength :)
The frequency is the same on both sides of the transition. There is no box to store waves in so as many have to pass a point in an hour on the left as on the right. Same f and same w and same 1/f = period T The tension is the same both sides w = 2 pi f = the same both sides ...
1960 to 1970 ... 100(.06-.04)/.04 = 50% 1970 to 1980 ... 100(.15-.06)/.06 = 150% 1980 to 1990 ... 100(.25-.15)/.15 = 67% 1990 to 2000 ... 100(.33-.25)/.25 = 32% average = sum/4 = 75% Always up, but 1970 to 1980 is so big an increase it dominates the statistics. (That is when ...
The period has nothing to do with the mass. It is proportional to the square root of the length 1.97 * sqrt(2.5)
Force = m g = 53.2*9.81 area = pi r^2 so 2.37*10^8 = 53.2*9.81/(pi R^2)
It is water. The density is 1 gram/cc or 1000 kg/m^3 Q = flow rate , volume/second. Q is constant, the same amount of water goes by each point each seconds (this is continuity) therefore V * area = constant 2.77 * pi (1.11^2) = v * pi (0.91^2) or v = 2.77(1.11/0.91)^2...
1.71 mm = 1.71*10^-3 m y = 1.71*10^-3 * sin (44 pi t) v = 1.71*10^-3 *44 pi *cos(44 pi t) max v = 44 pi * .00171 meters/second
5/7 times x = 140 5 x = 140 * 7 x = 140 * 7 / 5 x = 28 * 7 x = 196
5.00 micro C (5.00*10-6) is Q, the charge. It is GIVEN. No, you do it.
Bob Pursley already did this for you.
What is the kinetic energy at the bottom? Ke = (1/2) m v^2 = .5 * 2 * 20.1^2 What is the average speed going up? Vav = 20.1 / 2 = 10.05 m/s So how long did it take to go up? t = 4.1/2 = 2.05 seconds upward so how high did it go? h = 10.05 * 2.05 so how much work did gravity do...