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katrin2010 [14]
3 years ago
14

What type of energy is released when a proton is nocked out of an atoms nucleus?

Physics
2 answers:
Ne4ueva [31]3 years ago
4 0
I think the answer would be: atomic energy. Hope i helped.

yarga [219]3 years ago
3 0
I think it would be atomic energy, which would most likely be in very short waves to affect the protons.
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A moving particle is subject to conservative forces only. When its kinetic energy decreases by 10 J, what happens to its mechani
satela [25.4K]

Answer:

Its mechanical energy is the same.

Explanation:

If forces are only conservative, the mechanical energy will be the same.

It can be different if energy get transformed in another kind of energy like elastic energy for example, although the amount of energy is always the same.

If we just have mechanical energy not geting transformed we have:

Em=K+U

Em: Mechanical energy

K: Kinetic energý

U: Potential energy

Then if Kinetic energy decreases 10J, Potential energy will grow up 10J to keep the same amount of mechanical energy.

8 0
3 years ago
Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1
Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0
3 years ago
A 23.8 kg space rock is pushed with a force and accelerates at 8.97 m/s/s. How much force was applied?
kotegsom [21]

m=23.8kg a=8.97m/s^2 Fnet=? Fnet=ma=(23.8kg)(8.97m/s^2)=213.486N

5 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
The equation that relates two poles to the force between them and their separation is the ____ square law.
Ivenika [448]
A-exponential
You welcome
4 0
3 years ago
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