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SCORPION-xisa [38]
4 years ago
11

Which of the following is true of both nuclear fusion and nuclear fission?

Physics
2 answers:
Furkat [3]4 years ago
7 0

a.  They both require a high energy input.  <em>No.  Fission doesn't require ANY energy input to get going. </em>  

b.  They both occur without byproducts.   <em>No.  The by-products of fission are the biggest real problem in commercial nuclear power . . . where to put the bad stuff that's left over. </em>

c.  They both have a high energy output.  <em>Yes !</em>

d.  They both involve large atoms breaking into smaller atoms.  <em>No.  'Fission' involves large nucleii breaking into smaller ones.  'Fusion' is exactly the opposite ... combining small nucleii into more complex ones. </em>

maw [93]4 years ago
5 0

Answer:

c. They both have a high energy output.

Explanation:

- Nuclear fusion is the process in which two lighter nuclei combine together into a heavier nucleus

- Nuclear fission is the process in which a heavier nucleus breaks apart into smaller nuclei

In both cases, a huge amount of energy is released. In fact:

- In the nuclear fusion, the mass of the heavier nucleus is slightly lower than the sum of the masses of the two initial nuclei: this means that part of the initial mass has been converted into energy, according to the equation

E=mc^2

and we can see that even for small amount of mass, m, the energy released is huge

- In the nuclear fission, the mass of the final products is less than the mass of the initial heavier nucleus, so again, part of the initial mass has been converted into energy, according to the equation

E=mc^2

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The 25 kg wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it
N76 [4]

Answer:

Explanation:

radius of gyration of wheel k then

k² = r²/2

r² = 2 k²

r = √2 k

= 1.414 x .3 m

r = .4242 m

Moment of inertia of wheel

= mass x radius of gyration ²

= 25 x .3 x .3

= 2.25 kg m²

Friction force acting on it ( sliding )

= μmg , μ being coefficient of kinetic friction

This friction force will create linear acceleration in forward direction

Acceleration produced

= μg

= .6 x 9.8

= 5.88 m / s ²

This will also rotate the wheel , angular acceleration being

linear acceleration / radius

= 5.88 /.4242

= 13.86 radian / s²

3 0
3 years ago
A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from
kogti [31]

Answer:

32

Explanation:

7 0
4 years ago
PLEASE HELP &lt;3
aliya0001 [1]

1) The landing spot of the projectile is given by d=v_x \sqrt{\frac{2h}{g}}

2) The common variable is the time

Explanation:

1)

The motion of a projectile consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

The landing spot can be determined in the following way:

- First of all, we analyze the vertical motion to find the time of flight of the projectile. This can be done by using the suvat equation

h=ut+\frac{1}{2}at^2

where

h is the vertical displacement of the projectile, which corresponds to the height from which the projectile has been fired, above the ground

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

t=\sqrt{\frac{2h}{g}}

- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity v_x. Therefore, the horizontal distance covered is given by

d=v_x t

And substituting the time of flight,

d=v_x \sqrt{\frac{2h}{g}}

2)

Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by

x=v_x t

where v_x is the horizontal velocity and t is the time.

The vertical motion is accelerated, so the vertical component of the displacement is given by

y=\frac{1}{2}gt^2

where g is the acceleration of gravity and t is the time.

Therefore, from the two equations we see that the common variable is t, the time.

Learn more about projectile:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste
grin007 [14]

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

I = m L^2        

where L is the length of road, we will take whole length of rod because mass is at  the end of it.      

I = 1.5 \times 0.85^2  

I = 1.084 kg.m²                        

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

8 0
3 years ago
Which example has only potential energy and no kinetic energy?
Damm [24]
D.a ball sitting on a shelf
5 0
4 years ago
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