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babymother [125]
3 years ago
14

Compare the practical uses and limitations of nuclear fission and fusion. Include in your answer a detailed description of the t

wo processes, comparing and contrasting them in terms of the types of atoms involved, the conditions needed to get the reaction started, and the types of products produced.
Physics
1 answer:
shtirl [24]3 years ago
8 0
In fission, an atom is split into two or more smaller, lighter atoms. Fusion, in contrast, occurs when two or more smaller atoms fuse together, creating a larger, heavier atom. For the former, the energy released by fission is a million times greater than that released in chemical reactions, but lower than the energy released by nuclear fusion. On the other hand, fusion process<span> is three to four times greater than the energy released by fission. Hope this would help. Have a nice day.</span>
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Waves of energy in the electromagnetic spectrum travel through space and into the earths atmosphere. Which waves pass more crest
andrew-mc [135]
The answer is D.......
3 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
A snowboarder goes down the hill with a slope of 28° if friction acts on him as he slides down which of the following is the cor
xxMikexx [17]

Answer:

A

Explanation:

All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A

6 0
3 years ago
The force of attraction between a ball is F=.........×10^-¹¹
DIA [1.3K]

Answer:

4.45×10¯¹¹ N

Explanation:

From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

Mass of bowling pin (M₂) = 1.5 Kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) = 3 m

Force of attraction (F) =?

The force of attraction between the ball and the bowling pin can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 4 × 1.5 / 3²

F = 4.002×10¯¹⁰ / 9

F = 4.45×10¯¹¹ N

Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N

8 0
3 years ago
The formula K2S indicates that
jekas [21]
There are two atoms of potassium bonded to one atom of sulfur.
4 0
3 years ago
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