All categories

3 years ago

Explain how fusion and fission are similar to and different from each other. Thanks.

2 answers:

5 0

Fusion is the act of putting together ( fusing ) two items so that they become one. Fission is the act of dividing or splitting an item.

Margaret [11]3 years ago

5 0

<span>Nuclear fission and nuclear fusion are similar in that they both involve changes in the nucleus of the atom. Another similarity is that relatively high energies (for the size of the matter involved) are associated with those nuclear changes.</span>

You might be interested in

What is the equivalent resistance of a circuit that contains four 75.02

viktelen [127]

Answer:

option A is the correct answer

3 0

2 years ago

If a book has a mass of 3 kg, what is the book's weight in N?

sesenic [268]

Answer:

29.4 N

Explanation:

F = ma

F = (3 kg) (9.8 m/s²)

F = 29.4 N

5 0

3 years ago

Where does the engery of an earthquake originate

allochka39001 [22]

From convection of magma under the earths crust makes the plates slowly move and as they move over time they build up potential energy from the different plates grinding against each other and after so long the plates will lose there grip on each other and release the potential energy they've been building up for so long as kinetic energy causing what you know as an earthquake hope this helps please give brainliest

6 0

3 years ago

HELP ME DO THIS QUESTION PLEASE

raketka [301]

Answer:

i dont know

Explanation:

but what you can do is ask you mom or dad to get you a tutor to help you

7 0

3 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago