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AfilCa [17]
2 years ago
10

Name two units for measuring the diameter of nucleus atom.

Physics
1 answer:
Evgesh-ka [11]2 years 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

#SPJ1

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An elephant's legs have a reasonably uniform cross section from top to bottom, and they are quite long, pivoting high on the ani
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Answer:

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Explanation:

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3 years ago
A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the
iren [92.7K]

Answer:

work = 1728

Power = 134

Explaination:

by using the formula,

Work(W)= Force(F)×Distance(D)

<h2> and</h2>

Power(P)= Work(W)/Time taken(T)

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4 years ago
Jake is in chemistry class. He makes a list of the chemicals his instructor described and the properties of each.
koban [17]
Silver: bonds with other atoms because of the weak forces of the valence electrons 
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6 0
3 years ago
Read 2 more answers
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration
umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

F=qvB     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

F=qvB=ma       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0
3 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
3 years ago
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