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Neporo4naja [7]
3 years ago
8

Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller element

s, A also gives off a beta particle, B gives off gamma rays, C gives off neutrons, and D gives off alpha particles. Which element would make a possible fuel for a nuclear reactor
Physics
1 answer:
Zolol [24]3 years ago
5 0

Answer:

<em>Element C will be best for a nuclear fission reaction</em>

Explanation:

<em>Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process.</em> For the reaction to be continuous in a chain reaction,<em> the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission</em>. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

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The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan
ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

I = \frac{P}{4\pi r^2}

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

3 0
3 years ago
Arrange the following substances in order of decreasing magnitude of lattice energy.
kolbaska11 [484]
The formula is F = ( q1 * q2 ) / r ^ 2 
<span>where: q is the individual charges of each ion </span>
<span>r is the distance between the nuclei </span>
<span>The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. </span>

<span>From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. </span>
<span>For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. </span>

<span>So Far: ( BaO and SrO ) > ( NaF and KCl ) </span>

<span>The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. </span>
<span>For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. </span>

<span>Therefore: </span>
<span>Answer: SrO > BaO > NaF > KCl </span>
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Answer:

Explanation:

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Answer:

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Explanation:

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