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1 year ago

Name two units for measuring the diameter of nucleus atom.

Physics

1 answer:

Evgesh-ka [11]1 year ago

6 0

The two units for measuring the diameter of nucleus atom are femtometre and metre.

How do you measure the size of the nucleus ?

Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.

femtometre (fm), which equals $10^{-15}$ metre.

A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.

Learn more about nucleus of an atom here :-

brainly.com/question/10658589

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Read 2 more answers

A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is

hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

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