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2 years ago

Help with these please someone

Physics

1 answer:

Mice21 [21]2 years ago

3 0

Answer:

Five: B

Ten: A

Explanation:

Five

An alpha decay looks like this.

$A^x_y====> B^{x-4}_{y - 2}+alpha^4_2$

So whatever is produced must have a mass of 4 less than 234 and a number on the periodic table of 2 less than 92. In other words, B has a mass of 230 and a number on the periodic table of 90.

The answer should be Thorium which is B.

Ten

There is actually not enough information to do 10. You get it by making an assumption and seeing if it works.

In general a beta decay can (the most common one ) look like this.

$A^x_y====> B^{x}_{y-1}+Beta^0_{-1}$

Is there anything that looks like that?

The weight stays the same (234) and the atomic number of the mother element (on the left) is 1 lower than the given chemical on the right.

$Th^{234}_{90}====> Pa^{234}_{91}+Beta^0_{-1}$

The answer is Thorium A

Beta decays are very tricky. Be very careful how you handle them. One of three items can be what is decayed. I have assumed it was an electron, but there are two other possibilities. I won't confuse you by adding them. Just be aware that they exist.

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Answer:

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Explanation:

Given that

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$P_1V_1=P_2V_2=PV$

Now by putting the values

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$V=\dfrac{5.3}{P}$

Where P is in KPa and V is in $m^3$

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Take ln both sides

So $\ln(PV)=\ln C$

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By differentiating

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$\dfrac{dP}{dV}=-\dfrac{P}{V}$

When P= 50 KPa

$\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}$

It indicates the slope of PV=C curve.

It unit is $\dfrac{Pa}{m^3}$ .

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Answer:

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Fisrt statement:

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Answer:

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Complete statement of the question is :

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