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3 years ago

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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Which pair below describes isotopes of the same element? A) an atom with 6 protons and 8 neutrons - an atom with 8 protons and 6

iogann1982 [59]

An isotope is an atom with a different number of neutrons than another atom of the same element. Since atoms of the same element all have the same number of protons, choice B(6pro &6neu vs. 6pro&7neu) is an example of isotopes

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aliya0001 [1]

All matter is made up of atoms

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What is the moment of inertia of a disc of mass 5kg and radius 10cm?

vodka [1.7K]

Answer:

500

Explanation:

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HELP

blagie [28]

Answer:

Explanation:

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3 years ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

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2 years ago