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3 years ago

8

If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years

1 answer:

OLEGan [10]3 years ago

5 0

20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.

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I need help!!!!!!!

wlad13 [49]

One simple use of the elements of the electromagnetic spectrum that we use during our everyday lives is our daily use of microwave radiation. microwave radiation is absorbed by water molecules which heats up and cooks the food whilst killing bacteria. Another would be ultraviolet radiation which we use daily in things such as light bulbs. The sun also uses this. Lastly, we use radio waves constantly. May it be tv programs, radio, or our cell phones.

4 0

3 years ago

What is 1/12+7/9 hj

Umnica [9.8K]

Explanation:

$\frac{1}{12} + \frac{7}{9} \\ \\ = \frac{1 \times 3}{12 \times 3} + \frac{7 \times 4}{9 \times 4} \\ \\ = \frac{3}{36} + \frac{28}{36} \\ \\ = \frac{3 + 28}{36} \\ \\ = \frac{31}{36} \\ \\ \huge \purple{ \boxed{\therefore \: \frac{1}{12} + \frac{7}{9} = \frac{31}{36} }} \\$

6 0

3 years ago

Read 2 more answers

To make a given sound seem twice as loud, how should a musician change the intensity of the sound?

Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

$\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}$

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

$I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1$

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0

3 years ago

What is E of a hydrogen atom in the 3p state?

notsponge [240]

Answer:

E=-1.51 eV.

$L=\hbar\sqrt{2}$

Explanation:

The nth level energy of a hydrogen atom is defined by the formula,

$E_{n}=-\frac{13.6}{n^{2} }$

Given in the question, the hydrogen atom is in the 3p state.

Then energy of n=3 state is,

$E_{n}=-\frac{13.6}{(3)^{2} }\\E_{n}=-1.51eV$

Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.

Now, the value of L can be calculated as,

$L=\hbar\sqrt{l(l+1)}$

For 3p state, l=1

$L=\hbar\sqrt{1(1+1)}\\L=\hbar\sqrt{2}$

Therefore, the value of L of a hydrogen atom in 3p state is $L=\hbar\sqrt{2}$ .

4 0

3 years ago

An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of

Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

$3\times \frac{\lambda}{2} = 0.39$

$\lambda = 0.26 m$

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

$d = \frac{\lambda}{4}$

$d = \frac{0.26}{4} = 0.065 m = 6.5 cm$

So the distance will be 6.5 cm

3 0

3 years ago