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andrezito [222]
3 years ago
13

Which description of energy changes in nuclear reactions is correct?

Physics
1 answer:
Kryger [21]3 years ago
4 0
The answer for this question is letter "B.Fission releases energy, and its products have greater stability."

Fission and Fusion are both nuclear reactions that when they release energy, they make the nuclei more stable. So among the choices, option B is the most fitting for the definition. 
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PLS ANSWERR THIS QUESTION FOR ME !!
MA_775_DIABLO [31]

Explanation:

20 joule is your answer

Answer:

here

mass m =100kg

distance d=50m

acceleration due to gravity a =10m/s²

work =force×displacement

= ma/d=100×10/50=20joule

3 0
2 years ago
Read 2 more answers
What distance should be used to pattern a shotgun hunter ed?
Murljashka [212]
I think the distance that should be used is the distance that one expects to be from the game you are hunting. Before taking a shotgun for a gobbler or even for ducks or other animals, you need to see how your gun performs by patterning it at various ranges with the load you want to use.
7 0
3 years ago
A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below
Harlamova29_29 [7]
H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v =  the velocity with which the flower pot hits the ground.
Then
v² = 2gh
    = 2*(32 ft/s²)*(280 ft)
    = 17920 (ft/s)²
v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer:  133.9 ft/s or 91.3 mi/h

5 0
3 years ago
A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
2 years ago
Which term describes the quantity of energy transferred from a warmer object to a cooler object?
Mars2501 [29]

Answer: B

Explanation:

6 0
3 years ago
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