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Elden [556K]
2 years ago
9

4. Changing the number of electrons changes the _____ of the atom, which is then called an _____.

Physics
1 answer:
statuscvo [17]2 years ago
5 0

Changing the number of electrons changes the <em>net electric charge</em> of the atom.

If the atom was originally neutral, then it becomes an <em>ion</em> .  (If it wasn't originally neutral, then it was already an ion.)

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How many changes can this undergo?Does it mean from this current state or how it can change.Please explain it to me.If you were
Fofino [41]

Let's assume this is a drawing of particles of a gas substance. This assumption is made upon the fact that these particles are not close and are represented in motion characteristic for gases. Gases can become solid by skipping the liquid phase. This process is called deposition. Also, a gas can become a liquid through the process of condensation as a result of energy loss at molecular level. Likewise, this is enabled thanks to heat loss or applied pressure.

5 0
3 years ago
Sierra did 500 J of work to move her couch. If she exerts 250 N of force on the couch, how far did she move it?
olya-2409 [2.1K]

Answer:

2 m

Explanation:

Work = force × distance

500 J = 250 N × d

d = 2 m

She moved the couch 2 meters.

5 0
3 years ago
What is the particles instantaneous speed at t=16 sec
Ira Lisetskai [31]

It's  3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.

Does that tell you anything ?
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8 0
3 years ago
a person is sitting on the last bench can see clearly see things written on book but cannot see them distinctly on board. what t
marysya [2.9K]

Answer:

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7 0
2 years ago
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
2 years ago
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