Gareth56

So if I understand this correctly the "squashing" of the Earth, which is due to the rotation of the Earth, is primarily responsible for the tidal bulges we observe?

So the effect of the Moon is negligible when it comes to the tides?

It's to do with the gravitation differential. The earth does not block the moons gravity, therefore on the opposite side of the earth to where the moon is, the moon "under your feet" is adding to the earth's gravity at that point. This makes an area of higher gravity, and water runs "down" the gravitational gradient to seek equilibrium.

 

Then on the moon side, it's not so much that the moon is pulling the water, it's that the gravitational center of the earth moon system is displaced toward that side, making the area below the moon, the "lowest" point wrt to the center of gravity of the system, and again the water runs down the gravity gradient towards it.

 

At least that's how I understand it.

I'm not too sure how the moon can be under my feet (emboldened above) also the water is being "pulled" away from the Moon on the opposite side of the earth to where the Moon so again I'm not sure how the water can run "down"

Sorry but all that "gubbins" is way above my head.

I can understand why there is a tidal bulge on the side of the Earth that is facing the Moon because basically the Moon is pulling on the sea that is facing it, however what I cannot understand is why there is a there a similar bulge on the opposite side of the Earth as there is nothing on that side pulling on the sea.

Could someone put me out of my misery as this has confused me for years?

Thanks

I was just wondering what the individual coils have to push against when an initial push is delivered to the first coil?

If an astronaut went outside the space shuttle (all protective clothing worn of course) and attached one end of a slinky to the shuttle could the astronaut create a longitudinal(or transverse) wave in the slinky?

Can I ask why for example:-

8a^5 b^5 + 10a^5 b^5 = 18a^5 b^5 and not 18a^10 b^10

Thanks

When I read " We work out this example as if it were 3(x - y) - 5(2y - x). We begin by multiplying out using the distributive property"

Does "multiplying out using the distributive property" just mean multiplying everything inside the bracket by what's outside it, in this case 3 & 5?

Many thanks for the clarification. Both of you

Continuing through the algebra book I'm studying I came across this example but cannot decide if it's correct or not.

Sorry but I just cannot get to grips with the proper way to present this type of equation on this forum. Hopefully it'll make sense.

x^2 y^3

3z^-4

= x^2y^3z^4[/u]

3

Sorry i can't get the 3 directly underneath = x^2y^3z^4

Doesn't it work out that the 3 also comes up to the numerator? Or is it only the z^-4 that comes up if so why does the 3 stay below?

Is the fact that a Minus x Minus = Positive an axiom or is there a [simple] mathematical proof?

A typo on my part.

So we can [now hopefully ] agree that the mass of one oxygen atom measured in grams is NOT 5.36 × 10^−23g.

So when the website concerned said that mass of one oxygen atom measured in grams is 5.36 × 10^−23g, they are incorrect?

Also insane_alien I'm not too sure what you mean by "no, that would be the wrong value for 1 amu. 1 oxygen atom is a thousandth of that value."

Had I incorrectly quoted the value for the amu somewhere?

Ok still confused as to what the mass of an oxygen atom is.

I note from the literature that 1 atomic mass unit (amu) = 1.66053886 × 10^-27 kilograms and from the literature and above that the atomic mass of oxygen is 16amu. So am I correct in saying that, expressed in kilograms, the mass of an oxygen atom is 16 x 1.66053886 × 10^-27kg = 2.6569 x 10^-26kg? (to 5 s.f.)

If it isn't 2.6569 x 10^-26kg? (to 5 s.f.) would someone please indicate what the mass, expressed in kilograms, of an oxygen atom is.

Thanks.

Sorry slightly confused now; is the website therefore correct when it says that the mass of one oxygen atom measured in grams is 5.36 × 10^−23g ?

Doesn't 16 amu equate to 16 x (1.6753 × 10^−27kg) = 2.68 x 10^-26kg?

I'm obliged for the confirmation. The website I read this on was clearly wrong then! I've also just checked the website also states that the mass of an Oxygen atom equates 16.0044 amu and 1amu ~ 1.6753 × 10^−24g (which I somewhat agree with) and this means that the mass of one oxygen atom measured in grams is 5.36 × 10^−23g!!

I get 2.6561 X 10^-23g so conclude the website means Oxygen molecule and not atom!

If it's not too technical could someone explain why the closer an electron is to the nucleus of an atom, the less energy it has and the farther away from the nucleus, the more energy it has.

I've always understood that the closer an electron is to the nucleus the more energy is required to remove it.

Many thanks.

Is it possible to include a picture to assist in the question that one is asking about in this space? I have question that can really only be asked with reference to an illustration that why I ask.

Thanks.

Since posting my question my SOH has found out that they are Powerline & Aerial Tower Markers. So yes you're correct.

Nice video about them here

Balls

Thanks.

During a recent trip to Austria I noticed attached to the wires on some of the electricity pylons in the fields what looked like steel or plastic balls/spheres (perhaps 12" diameter, hard to tell from the ground) spaced at intervals along the wire. These balls/spheres were not on every wire just one length. I was wondering if anyone knows what these are used for. They don't seem to have any of these objects attached to the wires on pylons here in the UK.

Oh no how embarrassing. I'll go and stand in the corner

The answer part (a) according to the book is 485km. They have used 90deg as the angle in the range equation which to me is pointing straight up into space (if your standing on the Moon).

What I don't understand is why haven't they taken the horizontal range which is the normal way (I think) when the angle between the flight of the bullet and the ground would be zero.

Part (b) What would the muzzle velocity have to be to send the bullet into circular orbit just above the surface of the Moon? (You will need to look up the radius of the Moon)

For interest it's Q95 of Chapter 3 of the book.