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babunello [35]
3 years ago
6

How does an atom change if all of its electros are removed?

Physics
1 answer:
kogti [31]3 years ago
4 0

The atom becomes positively charged.

EXPLANATION :

When all the electrons are removed from the atom, the atom contains only protons and neutrons. The neutron is a neutral nucleon. ...

Hence, the correct answer is that atom will be positively charged.

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Physical values in the real world have two components: magnitude and
Nadusha1986 [10]

Answer:

Dimension.

Explanation:

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3 years ago
An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs
OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

E=\dfrac{n^2\pi^2h^2}{8mL^2}

For ground state, n = 1

E_1=\dfrac{\pi^2h^2}{8mL^2}.............(1)

For first excited state, n = 2

40=\dfrac{2^2\pi^2h^2}{8mL^2}.............(2)

Dividing equation (1) and (2), we get :

\dfrac{E_1}{40}=\dfrac{1}{4}

E_1=10\ meV

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0
3 years ago
1. boiling point of water
melomori [17]

Answer:

what is the question. . .

5 0
3 years ago
Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor
Lemur [1.5K]

Answer: A 120 metros por segundo

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7 0
2 years ago
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is
lbvjy [14]

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\

b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

8 0
3 years ago
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