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AURORKA [14]
1 year ago
13

in a certain nuclear reactor, neutrons suddenly collide with carbon nuclei, which have 12 times the mass of neutrons. in a head-

on elastic collision with a stationary carbon nucleus, what fraction of its initial speed does the neutron have after the collision?
Physics
1 answer:
castortr0y [4]1 year ago
5 0

carbon atomic mass Equals m2

neutron mass = M,

as mg 12m1 = the speed of the neutron following a collision

m1 - m2 / m1 + m2 * u1 = v1

= [(-12m1+m1) / 13m1] * u1

= [(-12+1) 13] * u1

v1 = -11/13*u1

Neutron speed equals 11/13 starting speed.

Response: 11/13

A nuclear reactor is a device used to start and regulate nuclear chain reactions involving fission or nuclear fusion. Nuclear reactors are used in nuclear power plants to produce electricity and in nuclear propulsion for ships.

The neutron, symbol n or n0, is a subatomic particle with a neutral charge and slightly more mass than a proton. Atomic nuclei consist of protons and neutrons.

The element with atomic number 6 carbon has small, straightforward atoms. Six positively charged protons make up the nucleus, and there are six electrons outside the nucleus that are split between two shells.

To know more about nuclear reactor, click on the link below:

brainly.com/question/23160065

#SPJ4

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A 0.537-kg basketball is dropped out of a window that is 5.88 m above the ground. The ball is caught by a person whose hands are
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Answer:

a) 20.54 J

b) 30.97 J

c) 10.43 J

d) -20.54 J

Explanation:

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

h = Height

W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88-1.98)\\\Rightarrow W=0.537\times 9.81\times 3.9\\\Rightarrow W=20.54\ J

Work done by the ball's weight is 20.54 J

W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88)\\\Rightarrow W=30.97\ J

Gravitational potential energy of the basketball, relative to the ground, when it is released is 30.97 J

W=mgh\\\Rightarrow W=0.537\times 9.81\times (1.98)\\\Rightarrow W=10.43\ J

Gravitational potential energy of the basketball, relative to the ground, when it is released is 10.43 J

Change in gravitational potential energy

\Delta U=10.43-30.97=-20.54\ J

Change in gravitational potential energy is given by -20.54 J

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