2 years ago

Decide which examples are Nuclear Fission reaction.

Physics

2 answers:

MrMuchimi2 years ago

4 0

a,c,d, is your answer I think I hope it helps

Grace [21]2 years ago

4 0

Answer is the letter A

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The heart has structures that are called_____. This allows the blood to move in one direction.

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Answer:

valves

The heart has four valves - one for each chamber of the heart. The valves keep blood moving through the heart in the right direction.

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2 years ago

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The more ______ an object moves, the more __________ it has (Kinetic Energy)

belka [17]

Answer:

__________

Explanation:

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3 years ago

Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha

Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

$M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}$

So, we have:

$v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}$

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

$10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}$

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3 years ago

As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a

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Im pretty sure it’s a because it makes more sense you know?.

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2 years ago

Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha

leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q / ε₀

E = Q / 4πε₀r²

E ∝ 1 / r² .

6 0

3 years ago