Test Me on Electromagnetic Spectrum Physics GCSE!!!?

problem-free is interior the variety of a wave, and the spectrum in many cases refers to diverse wavelengths of light which have diverse appearances. you are perfect that electrical energy is the stream of electrons, and magnetism is analogous. An electromagnet has a magnetic field created by technique of the bypass of electrical energy by ability of a coil. although problem-free and electrical energy are in truth diverse, there are different parallels. for instance, electric powered voltage might want to correctly be measured with an oscilloscope and could appear as if a wave

1. is prime spectrum $Spec(R)$ countable?

For the first question, no: take $R=mathbbC[x]$ to be a polynomial ring in one variable. The prime ideals are $0$ and those generated by $x-a$ for $a in mathbbC$, of which there are uncountably many.For the second question, the closed points of $mathrmSpec(R)$ are precisely the maximal ideals. I do not know what topology you are using on the set of primary ideals. Can you clarify?

2. 1H-NMR spectrum for furfural

They are not coupling with the aldehyde proton. Because the three protons around the ring are locked in such close proximity to one another, they will all couple each other. Let us consider these protons:$hspace6cm$As by the $(n1)$ rule of multiplicity, a proton will couple with each nearby, chemically unique proton and be split into $(n1)$ peaks. These split signals can then also be split into $(n1)$ more peaks by other nearby, chemically unique proton. Since all three protons are in the same region of space around the ring, they all couple each other, splitting each proton's signal into $(11)(11)=4$ peaks, making each signal a doublet of doublets (a quartet).If you would like a detailed analysis of the NMR of furfanal, you can read about it here.

3. which of the following is NOT apart of the electromagnetic spectrum?

Sound waves. Is not there a survey that said that something like 60% of the US think that sound waves are EMR?

4. Where am I on the political spectrum?

Too contradictory to fit in any nice box Macro wise the following shows your contradictions no death penalty is left no abortion is right yes for gay rights is left drugs illegal is right

5. What is the origin of the term "spectrum” in mathematics?

I do not know the full story, but I found the following interesting tidbits in History of Functional Analysis by Dieudonn.On page 171 he writes the following about physicists in the 1920s: "It finally dawned upon them that their "observables" had properties which made them look like hermitian operators in Hilbert space, and that, by an extraordinary coincidence, the "spectrum" of Hilbert (a name which he had apparently chosen from a superficial analogy) was to be the central conception in the explanation of the "spectra" of atoms."Dieudonn earlier writes (page 150): "Although Hilbert does not mention Wirtinger's paper, it is probable that he had read it (it is quoted by several of his pupils), and it may be that the name "Spectrum" which he used came from it; but it is a far cry from the vague ideas of Wirtinger to the extremely general and precise results of Hilbert."He's referring to the same paper by Wirtinger referred to in Gjergji Zaimi's answer.

6. Thom spectrum in the definition of power operations

If a group $G$ acts on a vector space $V$, and $X$ is a space where $G$ acts properly discontinuously, then the map $$ (V times X) / G to X/G $$ can be given the structure of a vector bundle: it inherits this structure from the vector space $V$. The pullback of this bundle to $X$ is the trivial bundle $X times V$, but it is probably not trivial on $X/G$.(If you prefer to think about vector bundles in terms of classifying spaces, the action of $G$ on $V$ is a group homomorphism $G to GL(V)$, and there's an induced map $BG to BGL(V)$. The space $X/G$ has a map to $BG$ classifying the principal $G$-bundle $X to X/G$, and the composite $X/G to BG to BGL(V)$ classifies this associated bundle.)This recipe for the associated bundle also gives you a recipe for the Thom space: the Thom space is $$(S^V wedge X_)/G$$ where $S^V$ is the one-point compactification. We sometimes write this as $S^V wedge_G X_$.In the case of this vector bundle $bar

ho$ that you've written down, there a $Sigma_k$-equivariant isomorphism between $Bbb R^k$ and $Bbb R oplus V$, where $Sigma_k$ acts on the factor of $Bbb R$ trivially. The associated sphere is $S^k$, and this decomposition determines a $Sigma_k$-equivariant isomorphism of one-point compactifications $S^k simeq S^1 wedge S^V$. The Thom space of the bundle $

ho$ associated to $Bbb R^k$ then satisfies an identity: $$Th(

ho) cong S^k wedge_Sigma_k X_ cong S^1 wedge (S^V wedge_Sigma_k X_) cong S^1 wedge Th(bar

ho)$$ On the level of Thom spectra, we can desuspend both sides and turn this into an identity $$ X^bar

ho = Sigma^infty Th(bar

ho) simeq S^-1 wedge Sigma^infty Th(

ho) simeq Sigma^-1 X^

ho. $$The version where we multiply the vector bundle by an integer $m$ is roughly the same, except that we get more trivial factors and we have to take care that we get everything with virtual bundles correct when when $m