Country Boy

It's not a matter of gravity suddenly "kicking in". A tiny mass produces a tiny gravitational force. Two very tiny masses will move toward each very slowly but with slowly increasing speed. Two larger masses will move toward each other with greater acceleration. It's simply a matter of when you are able to notice the acceleration. That depends on you not the mass.

Yes, an "endothermic" reaction ((1) endothermic reaction definition - Bing) is one that takes heat from the environment, so feels cold. An "exothermic" reaction ((1) exothermic reaction definition - Bing) is one that produces heat.

I have no idea whaty you are trying to say or what you are trying to ask. But certainly "200 ul" is NOT equal to "160 ul"! You say the ratio is "1/5000", "1/500", "1/250" OF WHAT? Apparently you want the "volume of antibody" to be 1/5000 or SOMETHING but 1/5000 or WHAT? My best guess at what you are trying to say: 1/5000 of something is 200 ul so the "something" is 200(5000)= 1000000 ul or 1 l. 1/500 or that 0. ul or l. 1/250 of that is 4000 ul or 0.004 l. (I assume your "ul" is a good replacemet for $\mu l$ or "microliters", one millionth of a liter.)

What do you mean by "new models"?

I thought that but didn't post it!😄

You should state that the three values of X are assumed to have equal probability so P(-1)= P(0)= P(1)= 1/3. Then E(g(x))= g(-1)/3+ g(0)/3+ g(1)/3= 0/3+ 1/3+ 2/3= 1. Zak100, it is not the "expected function" but the "expected value" of the function.

The verb is "affect", "effect" is a noun. Gravity "affects" time and thus has an "effect" on time.

"c/t" has unit of "distance over time squared" so is an acceleeration, not a distance.

Think about the angle your legs make with smaller steps rather than longer steps. Both legs with be more nearly vertical at all times so will have a larger normal force.

Sorry. I missed the word "open" in the original question. Yes, a non-empy OPEN set in a measure space is necessarily of positive measure.

The first statement is not true. For example, a singleton set has 0 measure. What is true that any set containing an interval has non-zero measure.

If "the surfaces of A and B are rough" then there is enough friction between them that they do not move relative to each other. Since A and B are in equilibrium the force between A and B must be F.

Alpha Centauri, the brightest "star" in the constellation Centaur, turns out to be a lot more complicated that was thought! What was called "Alpha Centauri" turned out to be a number of stars orbiting together. What was called "Proxima Centauri" because it is (currently) closest to earth turned out to be, itself, a double star. The two stars making up "Proxima Centauri" are called "Proxima A" and "Proxima B".

What about the zinc giving up electrons to the iron?

Eiot, geez, calm down! Even if what people say on wikipedia is wrong (and it often is), it doesn't follow that they were lying! Maybe they just don't know any better. And if you are going to tell us it was criminal, you had better tell us what law they broke!

I have no problem with a person not knowing something. I do have a problem with a person asserting something that is not true!

Where in theworld did you get the idea that that the distance would decrease? The tree is perpendicular to the ground so you have a right triangle. Your view-line from yourself, up the tree, and the object is the hypotenuse of a right triangle. The hypotenuse is always longer than the other two sides, not shorter!

Since it has been a month, a right triangle with one leg of length 200 m and the other of length 100 m has hypotenuse of length $\sqrt{100^2+ 200^2}= \sqrt{10000+ 40000}= \sqrt{50000}$ or about 223.6 m. That is the straight line distance from the hiker's end point back to his starting point. the direction is $\arctan(\frac{200}{100})= \arctan(2)$ which is about 63.3 degrees east of south (so 90- 63.3= 26.7 degrees southeast).

The standard transliteration for "[tex]\pi[/tex]" is "pi", not "pie".

One way to arrive at [tex]e^{ix}= cos(x)+ i sin(x)[/tex] is to use the MacLaurin series for [tex]e^x[/tex], cos(x), and sin(x). [tex]e^x= \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \frax{x^4}{4!}+ \frac{x^5}{5!}+ \cdot\cdot\cdot[/tex]. [tex]e^{ix}= 1+ ix+ \frac{(ix)^2}{2!}+ \frac{(ix)^3}{3!}+ \frac{(ix)^4}{4!}+ \frac{(ix)^5}{5!}+ \cdot\cdot\cdot[/tex]. Of course, [tex]i^2= -1[/tex], [tex]i^3= i^2(i)= -i[/tex], [tex]i^4= i^3(i)= -i(i)= 1[/tex] and then it starts again. [tex]= (1- \frac{x^2}{2!}+ \frac{x^3}{3!}- \frac{x^4}{4!}+ \cdot\cdot\cdot)+ i(x- \frac{x^3}{3!}+ \frac{x^5}{5!}+ \cdot\cdot\cdot[/tex]. And [tex]cos(x)= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+ \cdot\cdot\cdot[/tex\] and [tex]sin(x)= x- \frac{x^3}{3!}+ \frac{x^5}{5!}+ \cdot\cdot\cdot[/tex].

The notation doesn't make sense to me! [tex]\int (f(x)+ dy/2)dx= \int f(x)dx+ \frac{1}{2}\int dydx[/tex]. The first, [tex]\int f(x)dx[/tex], makes sense but the second, [tex]\frac{1}{2}\int dydx[/tex], looks like it should be a double integral, not a single integral.

And that's why I kick dead donkeys!

Have you ever kicked a live donkey?

I am wondering if you are not confusing "equilibrium" with "stable equlibrium". First imagine a ball rolling up a hill. If it does not have enough kinetic energy to get to the top of the hill, it will slow to a stop, then roll back down the hill. At the instant it stops, it is momentarily "at rest" but it is not "at equilibrium" because there is a net downward force. Second imagine a ball sitting in a valley between two hills. There is no net force because the downward force of gravity is offset by the upward force of the ground on the ball. The ball is "at rest" and "at equilibrium". If it were disturbed slightly, so that it moved a bit, it would roll back to the bottom. That is a "stable equilibrium". Finally imagine a ball sitting at the top of a smooth hill. Again the only forces on the ball are gravity downward and the ground upward so there is no net force so the ball is "at rest" and "at equilibrium". If it were disturbed slightly, so that it moved a bit, it would roll to the bottom of the hill. That is an "unstable equilibrium".

Suppose an object is moving along a straight line with position given by x= t^2- 8t where x is the directed distance in meters from some fixed point and t is the time in seconds. At t= 4 seconds its speed is 0. Is it "at rest"? Is it "in equilibrium"?