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3 years ago

Calculate the wavelength of a radio wave with a frequency of 93.1 x 10^6 ^-1

2 answers:

7 0

Wave speed = frequency x wavelength

therefore wavelength = wave speed / frequency

the speed of radio waves ( and all electromagnetic waves in vacuum) is 3 x 10^8 meters per second

just put the numbers in now ☺

melomori [17]3 years ago

6 0

Answer: 3.22 m

Explanation:

The relationship between wavelength and frequency of the wave follows the equation:

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of the wave = $93.1\times 10^6s^{-1}$

c = speed of light = $3\times 10^8ms^{-1}$

$\lambda$ = wavelength of the wave

Now put all the given values in this formula, we get

$\lambda=\frac{3\times 10^8ms^{-1}}{93.1\times 10^6s^{-1}}$

$\lambda=3.22m$

Thus wavelength of a radio wave with a frequency of $93.1\times 10^6s^{-1}$ is 3.22 m.

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What is the energy of an electromagnetic wave with a frequency of 8•10^12 Hz?

aniked [119]

Hello!

Find the Energy of the Photon by Planck's Equation, given:

E (photon energy) =? (in Joule)

h (Planck's constant) = $6.626*10^{-34}\:J * s$

f (radiation frequency) = $8*10^{12}\:Hz$

Therefore, we have:

$E = h*f$

$E = 6.626*10^{-34}*8*10^{12}$

$E = 53.008*10^{-34+12}$

$E = 53.008*10^{-22}$

$\boxed{\boxed{E = 5.3008*10^{-21}\:Joule}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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3 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

(1) Preparatory phase

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

(2) Pay-off phase

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that BPG has a mixed anhydride and the bond at C1 is a very high energy bond. In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

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