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2 years ago

What is the source of the radioactive nuclei present in spent fuel rods?

1 answer:

6 0

The answer is <span>nuclear fission. T</span>he source of the radioactive nuclei present in spent fuel rods is nuclear fission. Nuclear fission<span> is the </span>process<span> in which a large nucleus splits into two smaller nuclei with the release of energy. </span>

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2 someone was described in a novel as riding off into the sunset," then they'd be travelling west. o True o False

Vanyuwa [196]

Answer: west

Explanation: the sunset always sets in the west.

5 0

1 year ago

At this rate, what fraction of its mass would io lose in 4. 5 billion years?.

Pie

In 4.5 billion years, the amount of mass lost is 1.41912 * 10¹⁷ kg.

<h3>Equation</h3>

An equation is an expression used to show the relationship between two or more variables and numbers.

Given that:

Io loses about a ton (1000 kilograms) of sulfur dioxide per second to Jupiter's magnetosphere.

In 4.5 billion years, amount of mass lost = 1000 kg/s * (4.5 * 10⁹ * 365 * 60 60 * 24) = 1.41912 * 10¹⁷ kg.

In 4.5 billion years, the amount of mass lost is 1.41912 * 10¹⁷ kg.

Find out more on Equation at: brainly.com/question/13763238

4 0

1 year ago

PLZ hurry! Thank you

Alenkinab [10]

Answer:do it again

Explanation:

It will work

3 0

2 years ago

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

2 years ago

What is the theory behind a ticker tape lab?

Lana71 [14]

Purpose: experiments will use it to measure the straight-line accelerated motion of a human hand. The displacement data will be measured and velocity and acceleration will be calculated, run the ticker tape under the guides on the timer and under the carbon circle.Hope this helps! ; )

5 0

2 years ago