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3 years ago

How high will a 0.345 kg rock go if thrown straight up by someone who does 111 J of work on it? Neglect air resistance.

1 answer:

7 0

To solve this problem we will apply the concepts related to energy conservation. Here we will understand that the potential energy accumulated on the object is equal to the work it has. Therefore the relationship that will allow us to calculate the height will be

$W = PE$

$W = mgh$

Here,

m = mass

g = Acceleration due to gravity

h = Height

our values are,

$m = 0.345 kg$

$g = 9.8m/s^2$

$W = 111J$

Replacing,

$W =mgh$

$111 = (0.345)(9.8)h$

$h = 32.83m$

Then the height is 32.83m.

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What type of fat may help lower the risk of heart disease

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Polyunsaturated fatty acids which is Omega-3 fatty acids.

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3 years ago

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point

seropon [69]

since centripetal acceleration is always towards the center of the circle

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also we know that

$a_c = \frac{v^2}{r}$

$11.6 = \frac{2.5^2}{R}$

$R = 0.54 m$

so the coordinates of the center will be

$(x, y) = (3.20 , (3.5 + 0.54))$

$(x, y) = (3.20, 4.04)$

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3 years ago

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

3 years ago

Prompt Write about what you have learned about parts of the atom and the

Kay [80]

Answer:

The table can be used to predict the properties of elements, even those that have not yet been discovered. Columns (groups) and rows (periods) indicate elements that share similar characteristics. The table makes trends in element properties apparent and easy to understand. The table provides important information used to balance chemical equations. Atoms are important because they form the basic building blocks of all visible matter in the universe. There are 92 types of atoms that exist in nature, and other types of atoms can be made in the lab. The different types of atoms are called elements. Hydrogen, gold and iron are examples of elements comprised of unique types of a single kind of atom.

Explanation:

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2 years ago

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Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$