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3 years ago

Does the nuclear mass or the charge of the nucleus determine what element an atom is?

Physics

2 answers:

riadik2000 [5.3K]3 years ago

6 0

Answer:

All atoms have a dense central core called the atomic nucleus. Forming the nucleus are two kinds of particles: protons. which have a positive electrical charge, and neutrons, which have no charge. All atoms have at least one proton in their core, and the number of proton determines which kind of element an atom is

Explanation:

GarryVolchara [31]3 years ago

6 0

Neither the mass alone nor the charge alone tells you what element the atom's nucleus is.

The only charged particles in the nucleus are the protons, which are positively charged. So every atom has a positively charged nucleus. The <em>NUMBER of protons in the nucleus</em> is the unique thing about each element.

It's not the mass, because there are also neutrons in the nucleus. A neutron has the same mass as a proton has, but no charge. AND, just to make it a little more complicated, every element can have atoms with a few <u><em>different</em></u> numbers of neutrons in the nucleus, so atoms of that element can have a few different masses. (These are called "isotopes" of that element.)

Bottom line:

-- The element is identified by the number of protons in the nucleus.

-- The nuclear charge is positive (that number), and

-- The nuclear mass is (that number) + (the number of neutrons in it).

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3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

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Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

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Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

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3 years ago

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Ira Lisetskai [31]

<h2>Answer</h2>

It will be single step endothermic reaction.

<h2>Expalantion</h2>

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Read 2 more answers

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o-na [289]

Answer:

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The gravitational power energy is

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U = 350 0.359

U = 125.7 J

c) in this case this point coincides with the reference system

y = 0

U = 0 J

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