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melisa1 [442]
2 years ago
5

How many protons, electrons, and neutrons are present in an atom of cr-24?

Physics
1 answer:
mote1985 [20]2 years ago
8 0

The number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

<h3>What is atom?</h3>

An atom is the smallest unit of matter, consisting of the positively charged nucleus and the electrons which move around it. The atom can not be divided further.

The atom of a matter is made by three elements-

  • 1) Neutron-Neutron is the element of an atom, which has zero charge.
  • 2) Proton-Proton is the element of an atom, which has a positive charge.
  • 3) Electron-Electron is the element of an atom, which has a negative charge.

The atom of cr-24 has the atomic number 24. The atomic number of atom is equal to the number of proton and electron in an atom. Thus,

\rm Proton=24\\\rm Electron=24

Now, if this atom has the mass number of 52 (most common). Then the number of neutron is,

\rm Neutrons=Mass\; Number-Atomic\; Number\\\rm Neutrons=52-24\\\rm Neutrons=28

Thus, the number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

Learn more about the atom here;

brainly.com/question/17545314

#SPJ4

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A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
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Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

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\displaystyle 0=v_osin\theta-\frac{gt}{2}

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We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

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