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slavikrds [6]
3 years ago
13

There are three naturally occurring isotopes

Physics
1 answer:
mote1985 [20]3 years ago
7 0

Explanation:

^{41}\text{Hi}: 41.000\:\text{g/mol}×0.283 = 11.603\:\text{g/mol}

^{42}\text{Hi}: 42.000\:\text{g/mol}×0.429 = 18.018\:\text{g/mol}

^{44}\text{Hi}: 44.000\:\text{g/mol}×0.288 = 12.672\:\text{g/mol}

Adding all the weighted values on the right hand side, we get

\text{Hi}: 42.293\:\text{g/mol}

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Well it occurs in igneous, metamorphic, and sedimentary rocks.
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3 years ago
Which particles contribute to the net charge and how does each change the net charge?
Licemer1 [7]

The two subatomic particles that contribute to the net charge of an ion are electrons and protons.

<h3>What is an atom?</h3>

Atom is the smallest possible amount of matter which still retains its identity as a chemical element, now known to consist of a nucleus surrounded by electrons.

The atom is made up of three components called subatomic particles as follows;

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The proton is the positively charged subatomic particle forming part of the nucleus of an atomwhile the electron is the subatomic particle having a negative charge and orbiting the nucleus.

This suggests that the two subatomic particles that contribute to the net charge of an ion are electrons and protons. That is;

Net charge = protons - electrons

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2 years ago
An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

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