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podryga [215]
3 years ago
11

Why are isotopes not shown on the periodic table

Physics
1 answer:
Fofino [41]3 years ago
3 0

Hey there!

The periodic table is arranged by number of protons/electrons which is the same for all isotopes of an element. <u>A different isotope will have a different number of neutrons only, which does not qualify it for a separate space on the periodic table.</u>

I hope this helps!

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Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
Define friction State the law of frictionGive two factors that may affect friction
umka21 [38]

Friction:

Friction is the resistive force that acts on the body in a direction opposite to the direction of motion of the same object.

Laws of friction:

1. The frictional force acting on the object is proportional to the normal force acting on the object.

For example, if the object of mass m is moving along the horizontal surface, the normal force acting on the object is mg.

Thus, the frictional force acting on the object is,

\begin{gathered} F_r\propto F_N \\ F_r=\mu F_N \end{gathered}

where F_N is the normal force acting on the object,

\mu\text{ is the coefficient of friction}

The diagrammatic representation of the frictional force acting on the object is,

2. The frictional force acting on the object depends upon the nature of the surface in contact with the object.

3. The frictional force does not depend upon the area of contact of an object with the surface.

4. The kinetic friction acting on the object is independent of the velocity of the object.

5. The coefficient of static friction is more than the coefficient of kinetic friction.

Two factors on which the frictional force depends are:

1. The frictional force depends on the adhesion between the two bodies in contact.

2. The frictional force acting on the depends upon the roughness of the surface on which the object is moving.

3 0
10 months ago
PLEASE HELP ME I AM TIMED!
cupoosta [38]
I believe the answer is D
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Sdhdmzir d sjdurkshrjeidnrjddneuxneixfnsidnrjxcbfnxudnx
4 0
3 years ago
How do I know if i’m doing number 2 right?
Ksju [112]

We are given an object that is speeding up on a level ground.

Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.

The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.

Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.

We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.

8 0
1 year ago
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