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IgorC [24]
4 years ago
5

This typically occurs on the sun? fusion fission very few chain reactions half life

Physics
1 answer:
melisa1 [442]4 years ago
5 0

Answer:

Fusion

Explanation:

The thermonuclear fusion(only occurs at extreme conditions of temperature and pressure like interiors of stars) of Hydrogen takes place in The Sun's core.

Every four hydrogen atoms combine to form a Helium atom with a release of 26.7Mev of energy.

On a large scale this amount of energy can be very huge.

It is calculated that tis proces has being going on for the past 5\times 10^{9} years and with the amount of Hydrogen left will be sufficient to keep it going for the same amount of time in the future.

After this process ends,the Sun's core which will be mostly Helium,will start to cool and collapse under its own gravity.

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Is this a scam/virus?
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Frictional force increases with the increase in the _______________ of the surface.
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Look at this model of an atom. Using a periodic table, which element does it represent?
Elodia [21]

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3 years ago
In a race, a runner traveled 12 meters in 4.0 seconds as she accelerated uniformly from rest. The magnitude of the acceleration
mestny [16]
<span>The runner is moving by uniformly accelerated motion, starting from rest (so, his initial velocity is zero). The law of motion of the runner is
</span>x(t) =  \frac{1}{2}  at^2<span>
where x(t) is the distance covered after time t, and a is the acceleration of the runner. By re-arranging the formula, we get
</span>a= \frac{2S}{t^2}<span>
We know the runner has covered a distance of S=12m in t=4.0 s, and if we plug these numbers into the equation, we find the acceleration of the runner: 
</span>a= \frac{2S}{t^2} = \frac{(2*12m)}{(4s)^2} =1.5 m/s^2<span>
</span>
5 0
3 years ago
Compute the resistance in ohms of a silver block 10 cm long and 0.10 cm2 in cross-sectional area. ( = 1.63 x 10-6 ohm-cm)
Vedmedyk [2.9K]
The resistance of the silver block is given by
R= \frac{\rho L}{A}
where
\rho=1.63 \cdot 10^{-6} \Omega \cdot cm is the silver resistivity
L=10 cm is the length of the block
A=0.10 cm^2 is the cross-sectional area of the block

If we plug the data into the equation, we find the resistance of the silver block:
R= \frac{(1.63 \cdot 10^{-6} \Omega \cdot cm)(10 cm)}{0.10 cm^2}=1.63 \cdot 10^{-4} \Omega
3 0
4 years ago
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