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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.
You might be interested in
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, F = ke

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

F = ma

a = \dfrac{F}{m}

a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

7 0
3 years ago
Whats is the resistance of a bulb with a current of 60a in a circuit with a 12v battery
lara [203]

Answer:

60a in a circuit with a 12v battery

Explanation:

60a in a circuit with a 12v battery

6 0
2 years ago
How does the Sun's gravity influence tides?
nevsk [136]

Answer: <em>Around each new moon and full moon, the sun, Earth, and moon arrange themselves more or less along a line in space. Then the pull on the tides increases, because the gravity of the sun reinforces the moon's gravity. ... This is the spring tide: the highest (and lowest) tide. Spring tides are not named for the season.</em>

8 0
3 years ago
What forces act upon an object dropped in a vacuum
Vlada [557]
Gravity is the only one, since there's no air resistance.
8 0
3 years ago
The mass of the Earth is 6 × 1024 kg, the mass of the Moon is 7 × 1022 kg, and the center-to-center distance is 4 × 108 m. How f
Karolina [17]

Answer:

The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

Explanation:

Let the reference point be the center of the Earth

X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m}

Where;

Xcm is the distance from center of the Earth =?

Me is the mass of the Earth = 6 × 10²⁴ kg

Xe is the center mass of the Earth = 0

Mm is the mass of the moon = 7 × 10²² kg

Xm is the center mass of the moon =  4 × 10⁸ m

X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m} =  \frac{M_e(0) +M_mX_m}{M_e+M_m} = \frac{ M_mX_m}{M_e+M_m}}\\\\X_C_m = \frac{7 X 10^{22}*4X10^8}{6X10^{24}+7X10^{22}} =\frac{28 X10^{30}}{607X10^{22}}\\\\  X_C_m = 4.613 X10^6 m

Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

8 0
3 years ago
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