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2 years ago

What is the modern periodic table of elements?

Physics

2 answers:

barxatty [35]2 years ago

8 0

I agreee ^ with that person

Kamila [148]2 years ago

3 0

C is the correct answer

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The half-life of iodine-131 is 8.04 days. It’s decay constant equals

Svetllana [295]

Answer:

Explanation:

we know that half life of an element is

T=0.693/λ

where λ is decay constant in order to find decay constant

λ=0.693/T

λ=0.693/8.04

λ=0.086

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2 years ago

How many protons are in this Atom if it has a balanced charge?

Dmitry [639]

Answer:

the answer is A

Explanation:

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3 years ago

Light waves can be easily blocked but ______ waves pass through all substances? ( fill in the blank)

IRINA_888 [86]

Compressional waves can travel through all states of matter.

8 0

3 years ago

2CO + 2NO → 2CO2 + N2 is it balannced

natita [175]

If the same atoms appear on both sides, then it's balanced.

In this reaction, there are 4 Oxygens, 2 Carbons, and 2 Nitrogens on each side. So numerically, <em>it's balanced</em>. But I don't know enough chemistry to say whether the reaction is possible.

4 0

3 years ago

Read 2 more answers

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

3 years ago